Stoichiometry

Class 11 Chemistry, unit 4, book pages 80–95

Practise 68 MCQs Take the unit quiz Review 19 cards

Stoichiometry is the study of the quantitative relationships between the reactants and products of a chemical reaction. This unit uses the mole, the molar volume of a gas and the molarity of a solution to convert between masses, volumes and numbers of particles, and uses the balanced equation to find how much product forms. It then explains limiting and excess reactants, theoretical, actual and percent yield, and why accurate amounts matter in making medicines.

Learning outcomes (10)

Express balanced chemical equations in terms of moles, representative particles, masses, and volumes of gases at STP.
Explain the concept of limiting reagents.
Calculate the maximum amount of product and amount of any unreacted excess reagent.
Calculate theoretical yield, actual yield, and percentage yield when given appropriate information.
State the volume of one mole of a gas at STP.
Use the volume of one mole of gas at STP to solve mole-volume problems.
Calculate the gram molecular mass of a gas from density measurements at STP.
Derive measurements of mass, volume, and number of particles using moles.
Calculate the quantities of reactants and products involved in a chemical reaction using stoichiometric principles. (Some examples include calculations involving reacting masses, volumes of gases, volumes, and concentrations of solutions, limiting reagent and excess reagent, percentage yield calculations).
Explain with examples, the importance of stoichiometry in the production and dosage of medicine.

What is stoichiometry?

A chemical equation shows that elements and compounds react together in definite proportions. The study of the relationships between the amounts of the reactants and the products is called stoichiometry.

Stoichiometry: The study of the relationships between the amounts of the reactants and the products in a chemical reaction.
Origin of the word: It comes from the Greek words stoicheion, meaning element, and metron, meaning measurement.

Such a study is essential whenever quantitative information about a chemical reaction is needed, and it is important for predicting the yields of the products of chemical reactions.

4.1MoleOften examined

Chemists use the mole as the SI unit to weigh and count atoms, molecules and ions. A mole is the amount of substance that contains $6.023 \times 1 0^{23}$ representative particles. This experimentally determined number is called Avogadro's number.

Just as a dozen represents 12 identical things, a mole represents $6.023 \times 1 0^{23}$ particles of a substance. The 'representative particles' are atoms, molecules or ions, whichever the substance is made of.

A mole of any substance holds Avogadro's number of its representative particles
Substance	Representative particle	One mole contains
Carbon	atom, $C$	$6.023 \times 1 0^{23}$ carbon atoms
Sodium	atom, $Na$	$6.023 \times 1 0^{23}$ $Na$ atoms
Water	molecule, $H X_{2} O$	$6.023 \times 1 0^{23}$ $H X_{2} O$ molecules

This relationship allows us to convert moles into representative particles and vice versa.

N = n \times 6.023 \times 1 0^{23}

N = number of particles, n = number of moles

The atomic mass, formula mass and molecular mass of a substance, taken in grams, are equal to one mole of the substance. This relationship allows us to convert the mass of a substance to moles and vice versa. For example:

Mass of one mole of some particles
Representative particle	Mass of one mole
$O$ atoms	16 g
$O X_{2}$ molecules	32 g
$H X_{2} O$ molecules	18 g
$Na X^{+}$ ions	23 g
$NaCl$ formula units	58.5 g

n = \frac{mass in grams}{molar mass in g mo l ^{- 1}}

4.1.1Molar volumeOften examined

One mole of any gas at STP (standard temperature and pressure) occupies a volume of $22.414 d m^{3}$ . This volume is called the molar volume, $V_{m}$ . With the help of this relationship, we can convert the mass of a gas at STP into its volume and vice versa.

Molar volume ( $V_{m}$ ): The volume occupied by one mole of any gas at STP, equal to $22.414 d m^{3}$ .

V_{gas at STP} = n \times 22.414 d m^{3} mo l^{- 1}

$22.414 d m^{3}$ of any gas at STP is one mole
Gas at STP	Volume	Amount	Mass	Number of molecules
Any gas	$22.414 d m^{3}$	1 mole	molar mass	$6.02 \times 1 0^{23}$
$H X_{2}$	$22.414 d m^{3}$	1 mole	2 g	$6.02 \times 1 0^{23}$
$NH X_{3}$	$22.414 d m^{3}$	1 mole	17 g	$6.02 \times 1 0^{23}$

Example 4.1: volume of chlorine at STP

Determine the volume of 2.5 moles of chlorine molecules at STP.

Given

$n = 2.5 mol$ chlorine, $Cl X_{2}$
$V_{m} = 22.414 d m^{3} mo l^{- 1}$ at STP

To find

One mole of any gas occupies $22.414 d m^{3}$ at STP, so $22.414 d m^{3}$ of $Cl X_{2}$ at STP is 1 mole.
Multiply the volume of one mole by the number of moles.
$V = 2.5 mol \times 22.414 d m^{3} mo l^{- 1} = 56.035 d m^{3}$

V = 56.035 d m^{3}

4.1.2Molar mass and density of gases

Density is defined as the mass per unit volume of a substance. All gases have the same molar volume, so the density of a gas depends on its molar mass: a gas with a higher molar mass has a higher density and vice versa. The density of a gas can therefore be calculated from its molar mass and molar volume, and if the density of a gas at STP is determined, its molar mass can be calculated.

Derivation Molar mass of a gas from its density at STP

Aim. Find the molar mass $M$ of a gas from its density $ρ$ at STP.

Assumptions.

The gas is at STP.
The density $ρ$ is in $g d m^{- 3}$ .

Density is mass per unit volume, so $1 d m^{3}$ of the gas at STP has a mass of $ρ$ grams.
$1 d m^{3} of gas = ρ g$
One mole of any gas occupies $22.414 d m^{3}$ at STP, so multiply the mass of $1 d m^{3}$ by 22.414.
$22.414 d m^{3} of gas = ρ \times 22.414 g$
The mass of $22.414 d m^{3}$ is the mass of one mole, which is the molar mass.
$M = ρ \times 22.414 g mo l^{- 1}$

Result

M = ρ \times 22.414 or ρ = \frac{M}{22.414}

Gases with a larger molar mass have a larger density at STP, because one mole of each fills the same $22.414 d m^{3}$ .

Example 4.2: molar mass from density

Calculate the gram molecular mass of a gas which has a density of

1.43 g d m^{- 3}

at STP.

Given

$ρ = 1.43 g d m^{- 3}$ at STP

To find

gram molecular mass $M$

$1 d m^{3}$ of the gas at STP has a mass of $1.43 g$ .
$1 d m^{3} at STP = 1.43 g$
Scale up to the volume of one mole, $22.4 d m^{3}$ (the book rounds the molar volume here).
$22.4 d m^{3} at STP = 1.43 \times 22.4 = 32.032 g$
The mass of $22.4 d m^{3}$ at STP is the molar mass.

M = 32.032 g mo l^{- 1}

Concept assessment exercise 4.1 (1): moles of oxygen

How many moles of oxygen molecules are there in

20.0 d m^{3}

of oxygen gas at STP?

One mole of any gas occupies $22.414 d m^{3}$ at STP, so divide the volume by the molar volume.
$n = \frac{V}{V _{m}} = \frac{20.0 d m ^{3}}{22.414 d m ^{3} mo l ^{- 1}} = 0.892 mol$

n = 0.892 mol of O X_{2}

Concept assessment exercise 4.1 (2): volume of hydrogen

What volume does

0.6

mole of

H X_{2}

gas occupy at STP?

Multiply the number of moles by the molar volume.
$V = n \times V_{m} = 0.6 mol \times 22.414 d m^{3} mo l^{- 1} = 13.45 d m^{3}$

V = 13.45 d m^{3}

4.1.3Stoichiometric calculations and mole ratioOften examined

A balanced chemical equation tells us the amounts of the reactants and products in several equivalent ways. Consider the reaction:

2 H X_{2} (g) + O X_{2} (g) 2 H X_{2} O (g)

We get the following information from this chemical equation.

What the balanced equation for water tells us
Quantity	Reactants	Product
Moles	2 moles of $H X_{2}$ combine with 1 mole of $O X_{2}$	to produce 2 moles of $H X_{2} O$
Molecules	$2 \times 6.02 \times 1 0^{23}$ molecules of $H X_{2}$ react with $6.02 \times 1 0^{23}$ molecules of $O X_{2}$	to produce $2 \times 6.02 \times 1 0^{23}$ molecules of water vapour
Mass	4 g (2 moles) of $H X_{2}$ combine with 32 g (1 mole) of $O X_{2}$	to produce 36 g (2 moles) of $H X_{2} O$
Volume at STP	$2 \times 22.414 d m^{3}$ of $H X_{2}$ combine with $1 \times 22.414 d m^{3}$ of $O X_{2}$	to produce $2 \times 22.414 d m^{3}$ of $H X_{2} O$ vapour

Example 4.3: hydrogen from magnesium

When

100 g

of magnesium is treated with dilute hydrochloric acid, what volume of hydrogen can be collected at STP?

Given

$m_{Mg} = 100 g$
$M_{Mg} = 24 g mo l^{- 1}$
$V_{m} = 22.4 d m^{3} mo l^{- 1}$ at STP

To find

$V_{H X_{2}}$ at STP

Write the balanced equation.
$Mg (s) + 2 HCl (aq) MgCl X_{2} (aq) + H X_{2} (g)$
Convert the mass of magnesium to moles.
$n_{Mg} = \frac{100}{24} = 4.17 mol$
Use the mole ratio from the equation: 1 mol of Mg produces 1 mol of $H X_{2}$ .
$1 mol Mg \to 1 mol H X_{2} \to 1 \times 22.4 d m^{3} H X_{2} at STP$
Multiply by the moles of magnesium.
$V = 4.17 \times 22.4 d m^{3} = 93.408 d m^{3}$

V_{H X_{2}} = 93.408 d m^{3} at STP

Example 4.4: dissolving sodium chloride

Interpret the dissolving of

NaCl

in water,

NaCl H X_{2} O Na X^{+} (aq) + Cl X^{-} (aq)

, in terms of moles, formula units and mass.

In moles: 1 mole of $NaCl$ dissolved in water gives 1 mole of $Na X^{+}$ ions and 1 mole of $Cl X^{-}$ ions.
$1 mol NaCl \to 1 mol N a^{+} + 1 mol C l^{-}$
In particles: $6.02 \times 1 0^{23}$ formula units of $NaCl$ give $6.02 \times 1 0^{23}$ $Na X^{+}$ ions and $6.02 \times 1 0^{23}$ $Cl X^{-}$ ions.
$6.02 \times 1 0^{23} formula units \to 6.02 \times 1 0^{23} Na X^{+} + 6.02 \times 1 0^{23} Cl X^{-}$
In mass: the molar masses add up to that of the salt.
$58.5 g NaCl \to 23 g N a^{+} + 35.5 g C l^{-}$

Concept assessment exercise 4.2 (1): information from an equation

What quantitative information do you get from

C X_{3} H X_{8} (g) + 5 O X_{2} (g) 3 CO X_{2} (g) + 4 H X_{2} O (g)

Read the coefficients as moles, molecules, mass and gas volumes at STP (each mole of gas is $22.414 d m^{3}$ and $N = 6.023 \times 1 0^{23}$ ).

	$C X_{3} H X_{8}$	$5 O X_{2}$	$3 CO X_{2}$	$4 H X_{2} O$
Moles	1	5	3	4
Molecules	$6.023 \times 1 0^{23}$	$5 \times 6.023 \times 1 0^{23}$	$3 \times 6.023 \times 1 0^{23}$	$4 \times 6.023 \times 1 0^{23}$
Mass (g)	44	$5 \times 32 = 160$	$3 \times 44 = 132$	$4 \times 18 = 72$
Volume at STP ( $d m^{3}$ )	22.414	112.07	67.242	89.656

The total mass of the reactants, $44 + 160 = 204 g$ , equals that of the products, $132 + 72 = 204 g$ , as the law of conservation of mass requires.

Concept assessment exercise 4.2 (2): molecular mass and molar mass

Compare and contrast the terms molecular mass and molar mass.

	Molecular mass	Molar mass
Meaning	Mass of one molecule, compared on the atomic mass scale	Mass of one mole ( $6.023 \times 1 0^{23}$ particles) of a substance
Unit	atomic mass unit (amu)	$g mo l^{- 1}$
Example, water	$18 amu$ for one $H X_{2} O$ molecule	$18 g mo l^{- 1}$ for one mole of $H X_{2} O$

The two have the same numerical value, because the mass of one mole in grams equals the molecular mass in amu. They differ in the quantity they refer to (one molecule against one mole) and in the unit.

Concept assessment exercise 4.2 (3): zinc for hydrogen

What mass of

Zn

is needed to produce

100 c m^{3}

H X_{2}

at STP? (

Zn (s) + 2 HCl (aq) ZnCl X_{2} (aq) + H X_{2} (g)

)

Given

$V_{H X_{2}} = 100 c m^{3} = 0.100 d m^{3}$
$V_{m} = 22.414 d m^{3} mo l^{- 1}$
$M_{Zn} = 65 g mo l^{- 1}$

To find

$m_{Zn}$

Convert the volume to $d m^{3}$ and then to moles of hydrogen.
$n_{H X_{2}} = \frac{0.100}{22.414} = 4.46 \times 1 0^{- 3} mol$
From the equation 1 mol of Zn gives 1 mol of $H X_{2}$ , so the moles of Zn equal the moles of $H X_{2}$ .
$n_{Zn} = 4.46 \times 1 0^{- 3} mol$
Convert moles of zinc to mass.
$m = n \times M = 4.46 \times 1 0^{- 3} \times 65 = 0.29 g$

m_{Zn} \approx 0.29 g

Worked example: mass and volume of carbon dioxide from burning butane

Butane burns completely:

2 C X_{4} H X_{10} + 13 O X_{2} 8 CO X_{2} + 10 H X_{2} O

. For

5.8 g

of butane find (a) the mass of

CO X_{2}

formed, (b) its volume at STP and (c) the mass of oxygen needed. Check your answers with the law of conservation of mass.

Given

$m_{C X_{4} H X_{10}} = 5.8 g$
$M_{C X_{4} H X_{10}}, M_{CO X_{2}}, M_{O X_{2}}, M_{H X_{2} O} = 58, 44, 32, 18 g mo l^{- 1}$
$V_{m} = 22.4 d m^{3} mo l^{- 1}$ at STP

To find

$m_{CO X_{2}}$
$V_{CO X_{2}}$ at STP
$m_{O X_{2}}$

Moles of butane.
$n = \frac{5.8}{58} = 0.10 mol$
Mole ratio from the equation: 2 mol $C X_{4} H X_{10}$ give 8 mol $CO X_{2}$ , so the ratio is $8/2 = 4$ .
$n_{CO X_{2}} = 0.10 \times \frac{8}{2} = 0.40 mol$
(a) Convert to mass.
$m_{CO X_{2}} = 0.40 \times 44 = 17.6 g$
(b) Convert to volume at STP.
$V = 0.40 \times 22.4 = 8.96 d m^{3}$
(c) Mole ratio for oxygen: 2 mol butane need 13 mol $O X_{2}$ .
$n_{O X_{2}} = 0.10 \times \frac{13}{2} = 0.65 mol, m = 0.65 \times 32 = 20.8 g$
Water formed, for the check: 10 mol per 2 mol butane.
$n_{H X_{2} O} = 0.10 \times \frac{10}{2} = 0.50 mol = 9.0 g$
Mass before and after the reaction must be equal.
$5.8 + 20.8 = 26.6 g, 17.6 + 9.0 = 26.6 g$

The calculation route: mass to moles, mole ratio from the equation, then moles to the quantity asked. The branch at the end is the gas volume at STP.

Common mistake: the coefficients 2 : 13 : 8 : 10 are ratios of moles, not of masses. The masses ( $5.8 : 20.8 : 17.6 : 9.0$ ) only come out after multiplying by the molar masses.

m_{CO X_{2}} = 17.6 g; V = 8.96 d m^{3}; m_{O X_{2}} = 20.8 g

Using the mole as the unit of quantity, we can calculate masses, volumes and numbers of molecules or other particles. The diagram shows how the mole links them.

The mole links mass, number of particles, gas volume at STP and the volume of a solution. Convert to moles first, then out to the quantity asked for.

4.1.4Mole-mole calculationOften examined

In a balanced chemical equation the coefficients are taken as moles, and they show the proportion of each chemical to the others. For example:

\ce{CH4 + 2O2 -> CO2 + 2H2O}

The reaction indicates that the complete burning of one mole of \ce{CH4} requires 2 moles of \ce{O2}, and that 1 mole of methane produces 2 moles of water and 1 mole of \ce{CO2}.

Example 4.5: magnesium burning in oxygen

How much \ce{MgO} is formed when 4 moles of magnesium react with excess of \ce{O2}? Also calculate the moles of \ce{O2} required.

Given

n_{\ce{Mg}} = 4\,\mathrm{mol} oxygen in excess

To find

n_{\ce{MgO}}
n_{\ce{O2}}

Write the balanced equation.
\ce{2Mg + O2 -> 2MgO}
Moles of \ce{MgO}. Two moles of Mg produce 2 moles of \ce{MgO}, so 1 mole of Mg produces 2/2=1 mole of \ce{MgO}.
4\,\mathrm{mol\ Mg}\times\dfrac{2\,\mathrm{mol\ MgO}}{2\,\mathrm{mol\ Mg}}=4\,\mathrm{mol\ MgO}
Moles of \ce{O2}. Two moles of Mg need 1 mole of \ce{O2}, so 1 mole of Mg needs 1/2 mole of \ce{O2}.
4\,\mathrm{mol\ Mg}\times\dfrac{1\,\mathrm{mol\ O_2}}{2\,\mathrm{mol\ Mg}}=2\,\mathrm{mol\ O_2}
State the result: 4 moles of Mg give 4 moles of MgO and need 2 moles of \ce{O2}.

n_{\ce{MgO}}=4\,\mathrm{mol},\quad n_{\ce{O2}}=2\,\mathrm{mol}

4.1.5Mole-mass calculationsOften examined

In mole-mass calculations the mass of a substance is changed into moles (mass divided by molar mass), the mole ratio of the balanced equation is applied, and the moles of the wanted substance are changed back into mass (moles multiplied by molar mass).

Example 4.6: decomposition of potassium chlorate

Potassium chlorate is used in making matches and dyes. On decomposition it produces \ce{KCl} and oxygen gas. If 25\,\mathrm{g} of \ce{KClO3} is decomposed, calculate (a) how many moles of \ce{O2} are produced and (b) how much mass in grams of \ce{O2} is produced.

Given

m_{\ce{KClO3}} = 25\,\mathrm{g}
M_{\ce{KClO3}} = 39+35.5+3\times16=122.5\,\mathrm{g\,mol^{-1}}
M_{\ce{O2}} = 32\,\mathrm{g\,mol^{-1}}

To find

(a) n_{\ce{O2}}
(b) m_{\ce{O2}}

Write the balanced equation.
\ce{2KClO3 -> 2KCl + 3O2}
Convert the mass of \ce{KClO3} to moles.
n_{\ce{KClO3}}=\dfrac{25}{122.5}=0.204\,\mathrm{mol}
(a) Apply the mole ratio: 2 moles of \ce{KClO3} give 3 moles of \ce{O2}.
n_{\ce{O2}}=\dfrac{3}{2}\times0.204=0.306\,\mathrm{mol}
(b) Mass of \ce{O2} is moles multiplied by molar mass.
m_{\ce{O2}}=0.306\times32=9.796\,\mathrm{g}

n_{\ce{O2}}=0.306\,\mathrm{mol},\quad m_{\ce{O2}}=9.796\,\mathrm{g}

Example 4.7: ionization of sulphuric acid

20\,\mathrm{g} of \ce{H2SO4} on dissolving in water ionizes completely. Calculate (a) the number of \ce{H2SO4} molecules, (b) the number of \ce{H+} and \ce{SO4^2-} ions and (c) the mass of the individual ions.

Given

m_{\ce{H2SO4}} = 20\,\mathrm{g}
M_{\ce{H2SO4}} = 98.016\,\mathrm{g\,mol^{-1}}
N_A = 6.02\times10^{23}\,\mathrm{mol^{-1}}

To find

(a) number of \ce{H2SO4} molecules
(b) numbers of \ce{H+} and \ce{SO4^2-}
(c) masses of \ce{H+} and \ce{SO4^2-}

(a) Number of molecules is moles multiplied by Avogadro's number, that is mass divided by molar mass, times 6.02\times10^{23}.
N=\dfrac{20}{98.016}\times6.02\times10^{23}=1.228\times10^{23}\ \text{molecules}
(b) The acid ionizes completely, and each molecule gives 2 \ce{H+} ions and 1 \ce{SO4^2-} ion.
\ce{H2SO4 -> 2H+ + SO4^2-}
So the number of \ce{H+} ions is twice the number of molecules, and the number of \ce{SO4^2-} ions is equal to it.
N_{\ce{H+}}=2\times1.228\times10^{23}=2.456\times10^{23},\qquad N_{\ce{SO4^2-}}=1.228\times10^{23}
(c) Mass of the ions is the number of ions divided by 6.02\times10^{23} (moles), multiplied by the molar mass of the ion. For \ce{H+} (molar mass 1.008):
m_{\ce{H+}}=\dfrac{2.456\times10^{23}}{6.02\times10^{23}}\times1.008=0.411\,\mathrm{g}
For \ce{SO4^2-} (molar mass 96):
m_{\ce{SO4^2-}}=\dfrac{1.228\times10^{23}}{6.02\times10^{23}}\times96=19.58\,\mathrm{g}
Check: the masses of the ions add up to the mass of acid taken.
0.411+19.58=19.99\approx20\,\mathrm{g}

N_{\ce{H2SO4}}=1.228\times10^{23},\ N_{\ce{H+}}=2.456\times10^{23},\ N_{\ce{SO4^2-}}=1.228\times10^{23};\ m_{\ce{H+}}=0.411\,\mathrm{g},\ m_{\ce{SO4^2-}}=19.58\,\mathrm{g}

Concept assessment exercise 4.3 (1): moles of CO to reduce iron(III) oxide

\ce{CO} can reduce iron(III) oxide into iron metal. How many moles of \ce{CO} are needed to reduce 5 moles of \ce{Fe2O3}? (\ce{Fe2O3 + 3CO -> 2Fe + 3CO2})

From the equation, 1 mole of \ce{Fe2O3} needs 3 moles of \ce{CO}.
Multiply by 5 moles of \ce{Fe2O3}.
n_{\ce{CO}}=5\times3=15\,\mathrm{mol}

n_{\ce{CO}}=15\,\mathrm{mol}

Concept assessment exercise 4.3 (2): phosphine from calcium phosphide

Calculate the amount of phosphine (\ce{PH3}) that can be prepared when 1 mole of calcium phosphide (\ce{Ca3P2}) reacts with excess \ce{H2O}. (\ce{Ca3P2 + 6H2O -> 3Ca(OH)2 + 2PH3})

From the equation, 1 mole of \ce{Ca3P2} gives 2 moles of \ce{PH3}.
n_{\ce{PH3}}=1\times2=2\,\mathrm{mol}
In mass, with M_{\ce{PH3}}=31+3=34\,\mathrm{g\,mol^{-1}}:
m_{\ce{PH3}}=2\times34=68\,\mathrm{g}

2\,\mathrm{mol}\ \ce{PH3}=68\,\mathrm{g}

Concept assessment exercise 4.3 (3): burning methanol

Methanol burns according to \ce{2CH3OH(l) + 3O2(g) -> 2CO2(g) + 4H2O(g)}. If 3.50 moles of methanol are burnt in oxygen, calculate (a) how many moles of oxygen are used and (b) how many moles of water are produced.

(a) Two moles of \ce{CH3OH} use 3 moles of \ce{O2}, so the ratio is 3/2.
n_{\ce{O2}}=3.50\times\dfrac{3}{2}=5.25\,\mathrm{mol}
(b) Two moles of \ce{CH3OH} give 4 moles of \ce{H2O}, so the ratio is 4/2=2.
n_{\ce{H2O}}=3.50\times\dfrac{4}{2}=7.00\,\mathrm{mol}

n_{\ce{O2}}=5.25\,\mathrm{mol},\quad n_{\ce{H2O}}=7.00\,\mathrm{mol}

Concept assessment exercise 4.3 (4): oxygen from potassium chlorate

Calculate the number of molecules and the volume of \ce{O2} at STP produced by the thermal decomposition of 490\,\mathrm{g} of \ce{KClO3}.

Given

m_{\ce{KClO3}} = 490\,\mathrm{g}
M_{\ce{KClO3}} = 122.5\,\mathrm{g\,mol^{-1}}

To find

number of \ce{O2} molecules
V_{\ce{O2}} at STP

Write the balanced equation.
\ce{2KClO3 -> 2KCl + 3O2}
Moles of \ce{KClO3}.
n_{\ce{KClO3}}=\dfrac{490}{122.5}=4.00\,\mathrm{mol}
Moles of \ce{O2}: 2 moles of \ce{KClO3} give 3 moles of \ce{O2}.
n_{\ce{O2}}=4.00\times\dfrac{3}{2}=6.00\,\mathrm{mol}
Number of molecules is moles times Avogadro's number.
N=6.00\times6.023\times10^{23}=3.61\times10^{24}\ \text{molecules}
Volume at STP is moles times the molar volume.
V=6.00\times22.414=134.5\,\mathrm{dm^3}

N=3.61\times10^{24}\ \text{molecules},\quad V=134.5\,\mathrm{dm^3}\ \text{at STP}

Worked example: iron from iron(III) oxide and a mass check

Iron(III) oxide is reduced by excess carbon monoxide: \ce{Fe2O3 + 3CO -> 2Fe + 3CO2}. When 80\,\mathrm{g} of \ce{Fe2O3} reacts, find the mass of iron produced, the mass of \ce{CO2} produced and the mass of \ce{CO} used. Show that mass is conserved.

Given

m_{\ce{Fe2O3}} = 80\,\mathrm{g}
M_{\ce{Fe2O3}},\;M_{\ce{Fe}},\;M_{\ce{CO}},\;M_{\ce{CO2}} = 160,\;56,\;28,\;44\,\mathrm{g\,mol^{-1}}

To find

m_{\ce{Fe}}
m_{\ce{CO2}}
m_{\ce{CO}}

Molar mass of \ce{Fe2O3}.
2\times56+3\times16=160\,\mathrm{g\,mol^{-1}}
Moles of \ce{Fe2O3}.
n=\dfrac{80}{160}=0.50\,\mathrm{mol}
Mole ratios: 1 mol \ce{Fe2O3} gives 2 mol Fe and 3 mol \ce{CO2} and needs 3 mol CO.
n_{\ce{Fe}}=1.00\,\mathrm{mol},\quad n_{\ce{CO2}}=1.50\,\mathrm{mol},\quad n_{\ce{CO}}=1.50\,\mathrm{mol}
Convert each to mass.
m_{\ce{Fe}}=56\,\mathrm{g},\quad m_{\ce{CO2}}=1.50\times44=66\,\mathrm{g},\quad m_{\ce{CO}}=1.50\times28=42\,\mathrm{g}
Check: mass of reactants equals mass of products.
80+42=122\,\mathrm{g},\qquad 56+66=122\,\mathrm{g}

m_{\ce{Fe}}=56\,\mathrm{g};\; m_{\ce{CO2}}=66\,\mathrm{g};\; m_{\ce{CO}}=42\,\mathrm{g}

4.1.6Solution stoichiometryOften examined

In solutions the most common concentration unit is molarity. The molarity of a solution is simply the numerical value of its concentration in \mathrm{mol\,dm^{-3}}.

If you read the label on a bottle of concentrated \ce{H2SO4} you will notice 98% \ce{H2SO4} by mass and also 18 M \ce{H2SO4}. What does 18 M stand for? It means that there are 18 moles of \ce{H2SO4} in each \mathrm{dm^3} of solution. Similarly, concentrated \ce{HCl} is 37% and 12.1 M \ce{HCl}: there are 12.1 moles of \ce{HCl} in each \mathrm{dm^3} of solution. We can express the concentration in terms of the moles of solute in a given volume of solution.

Molarity (M): The number of moles of solute dissolved per \mathrm{dm^3} of solution. It is the concentration unit in which the amount of solute is expressed in moles and the quantity of solution in \mathrm{dm^3}.

M=\dfrac{\text{moles of solute}}{\mathrm{dm^3}\ \text{of solution}}

Example 4.8: molarity of a urea solution

Urea (\ce{NH2CONH2}) is a white solid used as fertilizer and as a starting material for synthetic plastic. A solution contains 40\,\mathrm{g} urea dissolved in 500\,\mathrm{cm^3} of solution. Calculate the molarity of this solution.

Given

m = 40\,\mathrm{g} urea
V = 500\,\mathrm{cm^3} of solution

To find

molarity M

Molar mass of urea.
M_r=14+1\times2+12+16+14+1\times2=60\,\mathrm{g\,mol^{-1}}
Moles of urea.
n=\dfrac{40\,\mathrm{g}}{60\,\mathrm{g\,mol^{-1}}}=0.667\,\mathrm{mol}
Volume of solution in \mathrm{dm^3}.
V=\dfrac{500}{1000}=0.5\,\mathrm{dm^3}
Apply the definition of molarity.
M=\dfrac{0.667\,\mathrm{mol}}{0.5\,\mathrm{dm^3}}=1.334\,\mathrm{M}

M=1.334\,\mathrm{mol\,dm^{-3}}

In solution stoichiometry we use the volumes of solutions of known concentration instead of the masses of the reactants and products, and their molarity is used to determine their moles.

Example 4.9: silver chloride from silver nitrate

When 250\,\mathrm{cm^3} of 0.1\,\mathrm{M} \ce{AgNO3} solution is added to an excess of \ce{NaCl} solution, what mass of \ce{AgCl} will be formed?

Given

V_{\ce{AgNO3}} = 250\,\mathrm{cm^3}
M_{\ce{AgNO3}} = 0.1\,\mathrm{M}
M_{\ce{AgCl}} = 143.5\,\mathrm{g\,mol^{-1}}

To find

m_{\ce{AgCl}}

Write the balanced equation.
\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}
Find the moles of silver nitrate. A 0.1\,\mathrm{M} solution has 0.1 mol in 1000\,\mathrm{cm^3}.
n=\dfrac{0.1\times250}{1000}=0.025\,\mathrm{mol\ AgNO_3}
From the equation 1 mol of \ce{AgNO3} produces 1 mol of \ce{AgCl}, so 0.025 mol of \ce{AgNO3} produces 0.025 mol of \ce{AgCl}.
Convert to mass: moles times molar mass.
m=0.025\times143.5=3.59\,\mathrm{g}

m_{\ce{AgCl}}=3.59\,\mathrm{g}

When two solutions are mixed, the same idea gives a direct relation between their volumes and molarities.

Derivation Relation between the volumes and molarities of two reacting solutions

Aim. Obtain the relation between M_1,V_1 and M_2,V_2 for two solutions that react in the ratio n_1:n_2.

Assumptions.

The solutions react exactly, in the ratio of the coefficients of the balanced equation.
Both volumes are in the same unit.

For a solution, moles of solute are molarity times volume, since M=\text{moles}/V. Write this for each reactant.
\text{moles of reactant 1}=M_1V_1,\qquad \text{moles of reactant 2}=M_2V_2
The ratio between the moles of the reactants is the ratio given by the balanced chemical equation, where n_1 and n_2 are the coefficients (moles) of the two reactants.
M_1V_1:M_2V_2=n_1:n_2
Write the ratio as a fraction.
\dfrac{M_1V_1}{M_2V_2}=\dfrac{n_1}{n_2}
Cross-multiply to put each solution on its own side.
\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}

Result\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}

In this expression the ratio of the two volumes stays the same whether \mathrm{dm^3} or \mathrm{cm^3} is used, so volumes may be left in \mathrm{cm^3} provided both are in the same unit.

Example 4.10: sodium sulphate and barium chloride

What volume of 0.5\,\mathrm{M} sodium sulphate will react with 275\,\mathrm{cm^3} of 0.25\,\mathrm{M} barium chloride solution to completely precipitate \ce{Ba^2+} in solution?

Given

M_1 = 0.5\,\mathrm{M} \ce{Na2SO4}
n_1 = 1
V_1 = ?
M_2 = 0.25\,\mathrm{M} \ce{BaCl2}
n_2 = 1
V_2 = 275\,\mathrm{cm^3}

To find

Write the balanced equation. The coefficients of both reactants are 1.
\ce{Na2SO4(aq) + BaCl2(aq) -> BaSO4(s) + 2NaCl(aq)}
Write the relation and substitute.
\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}\ \Rightarrow\ \dfrac{0.5\times V_1}{1}=\dfrac{0.25\times275}{1}
Solve for V_1.
V_1=\dfrac{0.25\times275}{0.5}=137.5\,\mathrm{cm^3}
Check with moles: \ce{BaCl2} is 0.25\times0.275=0.06875 mol, so the same moles of \ce{Na2SO4} are needed, in 0.06875/0.5=0.1375\,\mathrm{dm^3}.

V_1=137.5\,\mathrm{cm^3}

Worked example: neutralizing sulphuric acid with sodium hydroxide

What volume of 0.25\,\mathrm{M} \ce{NaOH} is needed to neutralize 25.0\,\mathrm{cm^3} of 0.10\,\mathrm{M} \ce{H2SO4}, and what mass of \ce{Na2SO4} is formed? \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}

Given

V_{\ce{H2SO4}} = 25.0\,\mathrm{cm^3}
M_{\ce{H2SO4}} = 0.10\,\mathrm{M}
M_{\ce{NaOH}} = 0.25\,\mathrm{M}
M_r(\ce{Na2SO4}) = 142\,\mathrm{g\,mol^{-1}}

To find

V_{\ce{NaOH}}
m_{\ce{Na2SO4}}

Moles of acid, with the volume in \mathrm{dm^3} (25.0\,\mathrm{cm^3}=0.0250\,\mathrm{dm^3}).
n_{\ce{H2SO4}}=0.10\times0.0250=2.50\times10^{-3}\,\mathrm{mol}
Mole ratio: 1 mol acid needs 2 mol \ce{NaOH}.
n_{\ce{NaOH}}=2\times2.50\times10^{-3}=5.00\times10^{-3}\,\mathrm{mol}
Volume of base from V=n/M.
V=\dfrac{5.00\times10^{-3}}{0.25}=0.0200\,\mathrm{dm^3}=20.0\,\mathrm{cm^3}
Check with the relation between volumes and molarities: the ratio must equal the ratio of moles in the equation.
\dfrac{M_1V_1}{M_2V_2}=\dfrac{0.10\times25.0}{0.25\times20.0}=\dfrac12=\dfrac{n_{\ce{H2SO4}}}{n_{\ce{NaOH}}}
Moles and mass of salt: 1 mol acid gives 1 mol \ce{Na2SO4}.
m=2.50\times10^{-3}\times142=0.355\,\mathrm{g}

Common mistake: treating the acid and base as 1 : 1. Sulphuric acid has two replaceable hydrogens, so the base needed is twice the moles of acid.

V_{\ce{NaOH}}=20.0\,\mathrm{cm^3};\; m_{\ce{Na2SO4}}=0.355\,\mathrm{g}

Concept assessment exercise 4.4 (1): molarity of potassium chlorate

Potassium chlorate (\ce{KClO3}) is a white solid used in making matches and dyes. Calculate the molarity of the solution that contains (a) 1.5 moles of this compound dissolved in 250\,\mathrm{cm^3} of solution, (b) 75\,\mathrm{g} dissolved to produce 1.25\,\mathrm{dm^3} of solution, (c) What is the molarity of a 50\,\mathrm{cm^3} sample of potassium chlorate solution that yields 0.25\,\mathrm{g} residue after evaporation of the water?

(a) Convert the volume to \mathrm{dm^3} and divide the moles by it.
M=\dfrac{1.5\,\mathrm{mol}}{0.250\,\mathrm{dm^3}}=6.0\,\mathrm{M}
(b) Molar mass of \ce{KClO3} is 122.5\,\mathrm{g\,mol^{-1}}, so convert 75\,\mathrm{g} to moles first.
n=\dfrac{75}{122.5}=0.612\,\mathrm{mol},\qquad M=\dfrac{0.612}{1.25}=0.490\,\mathrm{M}
(c) The residue is the dissolved solute, so 0.25\,\mathrm{g} is the solute in 50\,\mathrm{cm^3}=0.050\,\mathrm{dm^3}.
n=\dfrac{0.25}{122.5}=2.04\times10^{-3}\,\mathrm{mol},\qquad M=\dfrac{2.04\times10^{-3}}{0.050}=0.0408\,\mathrm{M}

\text{(a) } 6.0\,\mathrm{M},\ \text{(b) } 0.490\,\mathrm{M},\ \text{(c) } 0.0408\,\mathrm{M}

Concept assessment exercise 4.4 (2): lithium hydroxide and sulphuric acid

What volume of 0.25\,\mathrm{M} \ce{LiOH} will completely react with 0.500\,\mathrm{dm^3} of 0.25\,\mathrm{M} \ce{H2SO4}? (\ce{2LiOH(aq) + H2SO4(aq) -> Li2SO4(aq) + 2H2O(l)})

Given

M_1 = 0.25\,\mathrm{M} \ce{LiOH}
n_1 = 2
V_1 = ?
M_2 = 0.25\,\mathrm{M} \ce{H2SO4}
n_2 = 1
V_2 = 0.500\,\mathrm{dm^3}

To find

Use the relation with n_1=2 for \ce{LiOH} and n_2=1 for \ce{H2SO4}.
\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}\ \Rightarrow\ \dfrac{0.25\times V_1}{2}=\dfrac{0.25\times0.500}{1}
Solve for V_1.
V_1=\dfrac{2\times0.25\times0.500}{0.25}=1.00\,\mathrm{dm^3}

V_{\ce{LiOH}}=1.00\,\mathrm{dm^3}

Concept assessment exercise 4.4 (3): sodium hydroxide to neutralize hydrochloric acid

How many grams of \ce{NaOH} are required to neutralize 40\,\mathrm{cm^3} of 0.5\,\mathrm{M} \ce{HCl} solution? (\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)})

Moles of \ce{HCl} in the solution.
n_{\ce{HCl}}=\dfrac{0.5\times40}{1000}=0.020\,\mathrm{mol}
The equation gives a 1 : 1 ratio, so 0.020 mol of \ce{NaOH} is needed.
Convert to mass with M_{\ce{NaOH}}=23+16+1=40\,\mathrm{g\,mol^{-1}}.
m=0.020\times40=0.80\,\mathrm{g}

m_{\ce{NaOH}}=0.80\,\mathrm{g}

4.2Limiting and non-limiting reactantsOften examined

When two reactants are mixed for a reaction, usually one of them reacts completely and the other does not. The reactant that is completely consumed in the reaction and limits the amount of product formed is called the limiting reactant. The reactant that is left unreacted after the completion of the reaction is called the non-limiting or excess reactant.

Limiting reactant: The reactant that is completely consumed in the reaction and limits the amount of product formed.
Non-limiting (excess) reactant: The reactant that is left unreacted after the reaction is complete.

To understand the concept, consider the reaction:

\ce{2H2 + O2 -> 2H2O}

When 1 mole of \ce{O2} and 1 mole of \ce{H2} are mixed, all the \ce{H2} reacts completely and \ce{O2} is left unreacted. For 1 mole of \ce{H2}, only \tfrac12 mole of \ce{O2} is required, which is available. But for the complete reaction of 1 mole of \ce{O2}, 2 moles of \ce{H2} would be required, which are not available. That is why \ce{H2} is the limiting reactant, and \ce{O2} is the non-limiting or excess reactant.

Two H2 react with one O2. Hydrogen runs out first, so it limits the water formed, and the second O2 is left over in excess (the same ratio as 1 mol H2 with 1 mol O2).

Example 4.11: calcium nitride

Calculate the mass of calcium nitride (\ce{Ca3N2}) prepared from 54.9\,\mathrm{g} of \ce{Ca} and 43.2\,\mathrm{g} of \ce{N2}: \ce{3Ca + N2 -> Ca3N2}.

Given

m_{\ce{Ca}} = 54.9\,\mathrm{g}
M_{\ce{Ca}} = 40\,\mathrm{g\,mol^{-1}}
m_{\ce{N2}} = 43.2\,\mathrm{g}
M_{\ce{N2}} = 28\,\mathrm{g\,mol^{-1}}

To find

limiting reactant
m_{\ce{Ca3N2}}

Moles of each reactant.
n_{\ce{Ca}}=\dfrac{54.9}{40}=1.37\,\mathrm{mol},\qquad n_{\ce{N2}}=\dfrac{43.2}{28}=1.54\,\mathrm{mol}
Moles of product from Ca: 3 moles of Ca give 1 mole of \ce{Ca3N2}.
\dfrac13\times1.37=0.457\,\mathrm{mol\ Ca_3N_2}
Moles of product from \ce{N2}: 1 mole of \ce{N2} gives 1 mole of \ce{Ca3N2}.
1.54\,\mathrm{mol\ Ca_3N_2}
Ca gives the least product (0.457 mol), so Ca is the limiting reactant and \ce{N2} is in excess. The product formed is 0.457 mol.
The book stops at this point. To finish the question, convert the moles of product to mass with M_{\ce{Ca3N2}}=3\times40+2\times14=148\,\mathrm{g\,mol^{-1}}.
m=0.457\times148=67.6\,\mathrm{g}
Unreacted \ce{N2}: 0.457 mol of \ce{N2} is used (1 mol per mol of product), so the excess is
1.54-0.457=1.08\,\mathrm{mol}\ \ce{N2}\ (\approx30\,\mathrm{g})

\ce{Ca}\ \text{is limiting};\quad m_{\ce{Ca3N2}}\approx67.6\,\mathrm{g}

Worked example: limiting reactant in the burning of magnesium

A mixture of 4.8\,\mathrm{g} of magnesium and 4.0\,\mathrm{g} of oxygen reacts: \ce{2Mg + O2 -> 2MgO}. Find the limiting reactant, the mass of \ce{MgO} formed and the mass of the excess reactant left over.

Given

m_{\ce{Mg}} = 4.8\,\mathrm{g}
m_{\ce{O2}} = 4.0\,\mathrm{g}
M_{\ce{Mg}},\;M_{\ce{O2}},\;M_{\ce{MgO}} = 24,\;32,\;40\,\mathrm{g\,mol^{-1}}

To find

limiting reactant
m_{\ce{MgO}}
mass of excess reactant left

Moles of each reactant.
n_{\ce{Mg}}=\dfrac{4.8}{24}=0.200\,\mathrm{mol},\qquad n_{\ce{O2}}=\dfrac{4.0}{32}=0.125\,\mathrm{mol}
Moles of \ce{MgO} each reactant could give. The equation gives 2 mol \ce{MgO} per 2 mol Mg and 2 mol \ce{MgO} per 1 mol \ce{O2}.
\text{from Mg: }0.200\times\dfrac22=0.200\,\mathrm{mol},\qquad \text{from }\ce{O2}\text{: }0.125\times\dfrac21=0.250\,\mathrm{mol}
The reactant giving the least product is limiting. Magnesium gives 0.200 mol, so Mg is the limiting reactant and the product formed is 0.200 mol.
Mass of \ce{MgO} formed.
m=0.200\times40=8.0\,\mathrm{g}
Oxygen used up: 1 mol \ce{O2} per 2 mol Mg, so
n_{\ce{O2}}\text{ used}=0.200\times\dfrac12=0.100\,\mathrm{mol}=3.2\,\mathrm{g}
Oxygen left over.
4.0-3.2=0.8\,\mathrm{g}\ (0.025\,\mathrm{mol})
Check by conservation of mass: reactants used 4.8+3.2 equal the product.
4.8+3.2=8.0\,\mathrm{g}

Each reactant could make a different amount of MgO. The shorter bar belongs to the limiting reactant (Mg) and sets the product actually formed.

\ce{Mg}\text{ is limiting};\; m_{\ce{MgO}}=8.0\,\mathrm{g};\; 0.8\,\mathrm{g\ O_2\ left}

Concept assessment exercise 4.5 (1): volume of marsh gas

How much volume is occupied by marsh gas (\ce{CH4}) at STP containing 4.8\times10^{24} molecules?

Convert the number of molecules to moles by dividing by Avogadro's number.
n=\dfrac{4.8\times10^{24}}{6.023\times10^{23}}=7.97\,\mathrm{mol}
Multiply by the molar volume.
V=7.97\times22.414=178.6\,\mathrm{dm^3}

V\approx178.6\,\mathrm{dm^3}\ \ (\approx 8\ \text{mol})

Concept assessment exercise 4.5 (2): molecular weight from density

Calculate the molecular weight of a gas with a density of 1.25\,\mathrm{g\,dm^{-3}}.

Use M=\rho\times22.414 (density at STP times molar volume).
M=1.25\times22.414=28.0\,\mathrm{g\,mol^{-1}}

M=28.0\,\mathrm{g\,mol^{-1}}

Concept assessment exercise 4.5 (3): 500 molecules each of hydrogen and oxygen

Calculate the molecules of \ce{H2O} produced by 500 molecules each of \ce{H2} and \ce{O2}. Which one is the limiting reactant? How much excess reactant is left unreacted? What would happen if the molecules of \ce{O2} were doubled?

Write the balanced equation. Two molecules of \ce{H2} react with one molecule of \ce{O2} to give two molecules of \ce{H2O}.
\ce{2H2 + O2 -> 2H2O}
Product from each reactant. 500 molecules of \ce{H2} would give 500 molecules of \ce{H2O}. 500 molecules of \ce{O2} would give 2\times500=1000 molecules.
The smaller amount of product is from \ce{H2}, so \ce{H2} is the limiting reactant and 500 molecules of \ce{H2O} are formed.
Excess reactant. 500 molecules of \ce{H2} use 500/2=250 molecules of \ce{O2}.
500-250=250\ \text{molecules of}\ \ce{O2}\ \text{left unreacted}
If the molecules of \ce{O2} are doubled (1000), \ce{H2} is still the limiting reactant, so still only 500 molecules of \ce{H2} react and 500 molecules of \ce{H2O} form. The excess \ce{O2} simply rises to 1000-250=750 molecules.

500\ \ce{H2O}\ \text{molecules};\ \ce{H2}\ \text{limiting};\ 250\ \ce{O2}\ \text{left}

Concept assessment exercise 4.5 (4): iron(III) chloride and silver nitrate

Excess \ce{AgNO3} was added to \ce{FeCl3} and 2.02\,\mathrm{g} of \ce{AgCl} was produced. Calculate the amount of \ce{FeCl3} in the reaction.

Given

m_{\ce{AgCl}} = 2.02\,\mathrm{g}
M_{\ce{AgCl}} = 143.5\,\mathrm{g\,mol^{-1}}
M_{\ce{FeCl3}} = 56+3\times35.5=162.5\,\mathrm{g\,mol^{-1}}

To find

m_{\ce{FeCl3}}

Write the balanced equation. \ce{AgNO3} is in excess, so \ce{FeCl3} is the limiting reactant and all of its chloride ends up in \ce{AgCl}.
\ce{FeCl3 + 3AgNO3 -> 3AgCl + Fe(NO3)3}
Moles of \ce{AgCl}.
n_{\ce{AgCl}}=\dfrac{2.02}{143.5}=0.01408\,\mathrm{mol}
Three moles of \ce{AgCl} come from 1 mole of \ce{FeCl3}.
n_{\ce{FeCl3}}=\dfrac{0.01408}{3}=4.69\times10^{-3}\,\mathrm{mol}
Convert to mass.
m=4.69\times10^{-3}\times162.5=0.762\,\mathrm{g}

m_{\ce{FeCl3}}\approx0.762\,\mathrm{g}

4.3Theoretical yield, actual yield and percent yieldOften examined

So far we have taken an optimistic approach to the amount of product from a chemical reaction. We assumed that 100% of the limiting reactant becomes product. The amount of product calculated from the balanced chemical equation is called the theoretical yield.

In reality, the amount of product in a reaction does not match what the equation shows, because some side reactions produce alternative products, some of the product may remain in solution, the reaction may be stopped before completion, and so on. Whatever the reason, the reaction produces less product than the calculated amount. The quantity of product produced experimentally in a chemical reaction is called the actual yield, and the percent yield is 100 times the ratio of the actual yield to the theoretical yield.

Theoretical yield: The amount of product calculated from the balanced chemical equation.
Actual yield: The quantity of product produced experimentally in a chemical reaction.
Percent yield: 100 times the ratio of the actual yield to the theoretical yield.

\text{Percent yield}=\dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times100

Example 4.12: percent yield of ozone

Calculate the percent yield of ozone that could be produced by 10\,\mathrm{g} of \ce{O2}. The actual yield of \ce{O3} is 1.05\,\mathrm{g}.

Given

m_{\ce{O2}} = 10\,\mathrm{g}
M_{\ce{O2}} = 32\,\mathrm{g\,mol^{-1}}
M_{\ce{O3}} = 48\,\mathrm{g\,mol^{-1}}
\text{actual yield} = 1.05\,\mathrm{g} \ce{O3}

To find

percent yield

Write the balanced equation.
\ce{3O2 -> 2O3}
Moles of \ce{O2}.
n_{\ce{O2}}=\dfrac{10}{32}=0.3125\,\mathrm{mol}
Three moles of \ce{O2} give 2 moles of \ce{O3}, so the moles of \ce{O3} are \tfrac23 of the moles of \ce{O2}.
n_{\ce{O3}}=\dfrac23\times0.3125=0.208\,\mathrm{mol}
Theoretical yield: convert to mass.
m_{\ce{O3}}=0.208\times48=10\,\mathrm{g}
Percent yield from the actual yield of 1.05\,\mathrm{g}.
\text{percent yield}=\dfrac{1.05}{10}\times100=10.5\%

\text{percent yield}=10.5\%

Worked example: percent yield when one reactant is in excess

Zinc (6.5\,\mathrm{g}) is added to 14.6\,\mathrm{g} of hydrochloric acid: \ce{Zn + 2HCl -> ZnCl2 + H2}. The mass of \ce{ZnCl2} actually obtained is 11.9\,\mathrm{g}. Find the limiting reactant, the theoretical yield and the percent yield of \ce{ZnCl2}.

Given

m_{\ce{Zn}},\;m_{\ce{HCl}} = 6.5\,\mathrm{g},\;14.6\,\mathrm{g}
M_{\ce{Zn}},\;M_{\ce{HCl}},\;M_{\ce{ZnCl2}} = 65,\;36.5,\;136\,\mathrm{g\,mol^{-1}}
\text{actual yield} = 11.9\,\mathrm{g}

To find

limiting reactant
theoretical yield
percent yield

Moles of each reactant.
n_{\ce{Zn}}=\dfrac{6.5}{65}=0.10\,\mathrm{mol},\qquad n_{\ce{HCl}}=\dfrac{14.6}{36.5}=0.40\,\mathrm{mol}
Moles of \ce{ZnCl2} each could give: 1 mol Zn gives 1 mol; 2 mol HCl give 1 mol.
\text{from Zn: }0.10\,\mathrm{mol},\qquad\text{from HCl: }\dfrac{0.40}{2}=0.20\,\mathrm{mol}
Zinc gives the least product, so Zn is the limiting reactant. The theoretical yield is 0.10 mol.
m_{\text{theoretical}}=0.10\times136=13.6\,\mathrm{g}
Percent yield, with both masses in grams.
\text{percent yield}=\dfrac{11.9}{13.6}\times100=87.5\%

The percent yield is below 100\%, as it must be for a real reaction. Never use the excess reactant (here HCl, 14.6\,\mathrm{g}) to find the theoretical yield.

\text{Zn limiting};\; 13.6\,\mathrm{g};\; 87.5\%

Concept assessment exercise 4.6 (1): iron(III) chloride from FeCl2

Determine the amount of ferric chloride produced by the reaction of \ce{KMnO4}, 10 moles of \ce{FeCl2} and 22 moles of \ce{HCl}: \ce{KMnO4 + 5FeCl2 + 8HCl -> MnCl2 + KCl + 5FeCl3 + 4H2O}.

The amount of \ce{KMnO4} is not given, so it is taken as sufficient. The limiting reactant is then decided between \ce{FeCl2} and \ce{HCl}.

Product from \ce{FeCl2}: 5 moles of \ce{FeCl2} give 5 moles of \ce{FeCl3} (ratio 5/5=1).
n_{\ce{FeCl3}}=10\times\dfrac55=10\,\mathrm{mol}
Product from \ce{HCl}: 8 moles of \ce{HCl} give 5 moles of \ce{FeCl3}.
n_{\ce{FeCl3}}=22\times\dfrac58=13.75\,\mathrm{mol}
\ce{FeCl2} gives the smaller amount, so it is the limiting reactant, and the \ce{FeCl3} produced is 10 mol. (\ce{HCl} used is 10\times8/5=16 mol, so 6 mol of \ce{HCl} is left.)
In mass, with M_{\ce{FeCl3}}=56+3\times35.5=162.5\,\mathrm{g\,mol^{-1}}:
m=10\times162.5=1625\,\mathrm{g}

n_{\ce{FeCl3}}=10\,\mathrm{mol}\ (1625\,\mathrm{g})

Concept assessment exercise 4.6 (2): percent yield of baking soda

Baking soda is commercially prepared by passing ammonia and carbon dioxide through saturated solution of \ce{NaCl}. If 20\,\mathrm{g} \ce{NH3} and 30\,\mathrm{g} \ce{CO2} produced 40\,\mathrm{g} baking soda, calculate the percentage yield of baking soda taking \ce{NaCl} and \ce{H2O} in excess. (\ce{NaCl + H2O + CO2 + NH3 -> NaHCO3 + NH4Cl})

Given

m_{\ce{NH3}} = 20\,\mathrm{g} M=17
m_{\ce{CO2}} = 30\,\mathrm{g} M=44
\text{actual yield} = 40\,\mathrm{g} \ce{NaHCO3}, M=84

To find

percent yield

Moles of each reactant.
n_{\ce{NH3}}=\dfrac{20}{17}=1.18\,\mathrm{mol},\qquad n_{\ce{CO2}}=\dfrac{30}{44}=0.682\,\mathrm{mol}
The equation has 1 : 1 : 1 ratios, so each mole of reactant gives at most 1 mole of \ce{NaHCO3}. \ce{CO2} gives the smaller amount, so it is the limiting reactant.
n_{\ce{NaHCO3}}=0.682\,\mathrm{mol}
Theoretical yield: convert to mass with M_{\ce{NaHCO3}}=23+1+12+48=84\,\mathrm{g\,mol^{-1}}.
m=0.682\times84=57.3\,\mathrm{g}
Percent yield from the actual yield of 40\,\mathrm{g}.
\dfrac{40}{57.3}\times100=69.8\%

\text{percent yield}\approx69.8\%

4.4Importance of stoichiometry in the production and dosage of medicine

In the production of medicines, the amount of active ingredient is essential to produce the desired effects. Stoichiometry ensures the accuracy of drug synthesis. Any deviation can result in an incomplete reaction or contamination with unreacted reactants or by-products. Stoichiometry allows chemists to precisely control chemical reactions to produce drugs, and so to ensure their efficiency, effectiveness and safe use. For example:

In the preparation of antibiotics, stoichiometry ensures that each dose matches the active ingredient and the target bacteria.
The use of insulin relies on stoichiometry for the precise control of blood sugar levels.
Stoichiometry determines the concentration of viral antigens in the preparation of vaccines, for effective results.

Activity: synthesis of aspirin

Materials required: salicylic acid, acetic anhydride, sulphuric acid, sodium hydrogen carbonate, ice bath, water, filtration set-up.

\ce{C7H6O3 + (CH3CO)2O -> C9H8O4 + CH3COOH}

salicylic acid + acetic anhydride gives aspirin + acetic acid

Salicylic acid

Acetic anhydride

Aspirin

Acetic acid

Steps

Mix the reactants in stoichiometrically calculated amounts (one mole of each reactant) in a flask and add a few drops of sulphuric acid to catalyse the reaction.
Place the flask in an ice bath.
Stir the mixture with a glass rod.
The flask will be filled with insoluble aspirin.
When the reaction is over, add sodium hydrogen carbonate solution to neutralize the excess acid.
Filter the aspirin, wash and dry the aspirin crystals.
Calculate the amount of aspirin formed.
Calculate the theoretical yield using stoichiometry.
Compare the actual yield obtained experimentally with the theoretical yield and assess the efficiency of the synthesis.

This activity helps you to understand how stoichiometry influences the production and dosage of medicine.

Formula sheet

Number of particles

N=n\times6.023\times10^{23}

N is the number of representative particles (atoms, molecules or ions) and n the number of moles.

no unit

Moles from mass

n=\dfrac{m}{M}

m is the mass of the substance and M its molar mass.

Unit: mole (\mathrm{mol})

Volume of a gas at STP

V=n\times22.414\,\mathrm{dm^3\,mol^{-1}}

n is the number of moles of the gas and 22.414\,\mathrm{dm^3\,mol^{-1}} is the molar volume.

Unit: \mathrm{dm^3}

Molar mass from density at STP

M=\rho\times22.414

\rho is the density of the gas at STP in \mathrm{g\,dm^{-3}}.

Unit: \mathrm{g\,mol^{-1}}

Molarity

M=\dfrac{\text{moles of solute}}{\mathrm{dm^3}\ \text{of solution}}

The number of moles of solute dissolved in one \mathrm{dm^3} of solution.

Unit: \mathrm{mol\,dm^{-3}} (M)

Reacting solutions

\dfrac{M_1V_1}{n_1}=\dfrac{M_2V_2}{n_2}

M and V are the molarity and volume of each solution, and n_1, n_2 the coefficients of the two reactants in the balanced equation. The ratio of volumes is the same in \mathrm{dm^3} or \mathrm{cm^3}.

no unit for the ratio

Percent yield

\text{Percent yield}=\dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times100

The actual yield is the product obtained experimentally, the theoretical yield the product calculated from the balanced equation.

Unit: per cent (%)

Key terms

Stoichiometry: The study of the relationships between the amounts of the reactants and the products in a chemical reaction.
Mole: The amount of substance that contains 6.023\times10^{23} representative particles.
Avogadro's number: The experimentally determined number 6.023\times10^{23} of representative particles in one mole.
Representative particles: The atoms, molecules or ions of which a substance is made.
Molar volume: The volume of 22.414\,\mathrm{dm^3} occupied by one mole of any gas at STP.
Gram molecular mass: The mass in grams of one mole of a gas or molecular substance, found at STP as density times 22.414.
Molarity: The number of moles of solute dissolved per \mathrm{dm^3} of solution.
Limiting reactant: The reactant that is completely consumed in a reaction and limits the amount of product formed.
Non-limiting (excess) reactant: The reactant that is left unreacted after the reaction is complete.
Theoretical yield: The amount of product calculated from the balanced chemical equation.
Actual yield: The quantity of product produced experimentally in a chemical reaction.
Percent yield: 100 times the ratio of the actual yield to the theoretical yield.

Multiple choice questionsp. 92–93

Choose the correct option. Each answer uses the mole, the molar volume at STP and the mole ratios of the balanced equation.

Which sample produces most hydrogen by reaction with excess of HCl?

A0.25\,\mathrm{mol} \ce{Ca}
B0.25\,\mathrm{mol} \ce{Al}Answer
C0.25\,\mathrm{mol} \ce{Zn}
D0.25\,\mathrm{mol} \ce{Na}

Why

Calcium and zinc give 1 mol of \ce{H2} per mol of metal: \ce{Ca + 2HCl -> CaCl2 + H2} and \ce{Zn + 2HCl -> ZnCl2 + H2}. So 0.25 mol of either gives 0.25 mol of \ce{H2}.

Aluminium gives 3 mol of \ce{H2} for every 2 mol of Al: \ce{2Al + 6HCl -> 2AlCl3 + 3H2}. So 0.25 mol Al gives 0.25\times\dfrac{3}{2}=0.375 mol of \ce{H2}, the most. Sodium gives only 0.125 mol, because \ce{2Na + 2HCl -> 2NaCl + H2}.

Note. The book prints no key for this question. Key B worked out here. Option (d) is printed as '0.25 Na' and the unit is added here.

A flask contains 500\,\mathrm{cm^3} of \ce{SO2} at STP. The mass of \ce{SO2} in the flask is:

A40\,\mathrm{g}
B100\,\mathrm{g}
C50\,\mathrm{g}
D1.42\,\mathrm{g}Answer

Why

At STP, 1 mol of any gas occupies 22.414\,\mathrm{dm^3}. The volume is 500\,\mathrm{cm^3}=0.500\,\mathrm{dm^3}, so n=\dfrac{0.500}{22.414}=0.0223 mol of \ce{SO2}.

Mass =0.0223\times64=1.43\,\mathrm{g}, which is closest to option D (1.42 g). Options A, B and C are many times larger.

Note. The question's second sentence is printed as 'The flask contains SO2' (cleaned here to 'The mass of SO2 in the flask is'). The exact value is 1.43 g; the printed option 1.42 g is the nearest.

iii

When 1 mole of each of the following is completely burnt in oxygen, which will give the greater mass of \ce{CO2}?

A\ce{CO}
BDiamond
CEthaneAnswer
DMethane

Why

Each carbon atom in the fuel gives one \ce{CO2}: \ce{C + O2 -> CO2} (diamond), \ce{2CO + O2 -> 2CO2} (CO), and \ce{CH4 + 2O2 -> CO2 + 2H2O} (methane). Each of these gives 1 mol of \ce{CO2} (44 g) from 1 mol of fuel.

Ethane has two carbon atoms: \ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}. So 1 mol of ethane gives 2 mol of \ce{CO2}, that is 88 g, the greatest mass.

Which one occupies more volume at STP?

A1\,\mathrm{g} \ce{O2}
B1\,\mathrm{g} \ce{H2}Answer
C1\,\mathrm{g} \ce{CH4}
D1\,\mathrm{g} \ce{NO2}

Why

At STP, equal numbers of moles occupy equal volumes, so the sample with the most moles in 1 g occupies the most volume.

Moles in 1 g: \ce{O2} =\dfrac{1}{32}=0.031; \ce{H2} =\dfrac{1}{2}=0.50; \ce{CH4} =\dfrac{1}{16}=0.063; \ce{NO2} =\dfrac{1}{46}=0.022. Hydrogen has the most moles, so V=0.50\times22.414=11.2\,\mathrm{dm^3}, which is more than the others (0.70, 1.4 and 0.49 dm^3).

0.2 moles of \ce{Na2SO4}, when completely ionized, produce \ce{Na+} ions:

A2.4\times10^{22}
B2.4\times10^{23}Answer
C1.204\times10^{23}
D0.12\times10^{23}

Why

\ce{Na2SO4 -> 2Na+ + SO4^2-}, so each formula unit gives 2 \ce{Na+} ions. Moles of \ce{Na+} =2\times0.2=0.4 mol.

Number of ions =0.4\times6.023\times10^{23}=2.4\times10^{23}. Option C is 0.2\times6.02\times10^{23}: it counts the formula units and forgets the two sodium ions in each.

How much volume of \ce{NH3} gas is produced when 3\,\mathrm{g} of \ce{H2} react with excess of \ce{N2} at STP?

A24\,\mathrm{dm^3}Answer
B2.24\,\mathrm{dm^3}
C2.4\,\mathrm{dm^3}
D1.2\,\mathrm{dm^3}

Why

\ce{N2 + 3H2 -> 2NH3}. Moles of \ce{H2} =\dfrac{3}{2}=1.5 mol. Three moles of \ce{H2} give two moles of \ce{NH3}, so 1.5 mol of \ce{H2} give 1.5\times\dfrac{2}{3}=1.0 mol of \ce{NH3}.

Volume at STP =1.0\times22.414=22.4\,\mathrm{dm^3}. Option A (24 dm^3) is the nearest value.

Note. The exact answer is 22.4\,\mathrm{dm^3} (3\,\mathrm g of \ce{H2} is 1.5\,\mathrm{mol}, which gives 1\,\mathrm{mol} of \ce{NH3}). None of the printed options matches it, so this is a misprint in the book. Option A (24\,\mathrm{dm^3}) is the closest. In the exam, write 22.4\,\mathrm{dm^3} if the options let you.

vii

When equal volumes of \ce{SO2} and \ce{O2} are taken for the formation of \ce{SO3}, which one will be left unreacted?

A\ce{SO2}
B\ce{O2}Answer
CBoth
DNot possible

Why

\ce{2SO2 + O2 -> 2SO3}. Equal volumes at the same conditions contain equal moles, n each. Each mole of \ce{O2} needs 2 mol of \ce{SO2}, so n mol of \ce{SO2} need only \dfrac{n}{2} mol of \ce{O2}.

\ce{SO2} is used up completely and is the limiting reactant. n-\dfrac{n}{2}=\dfrac{n}{2} mol of \ce{O2} is left over, so the answer is B.

viii

0.1 moles of laughing gas (\ce{N2O}) consist of:

A6.022\times10^{22} molecules
B1.806\times10^{23} atoms
C1.204\times10^{23} atoms of N
DAllAnswer

Why

Molecules: 0.1\times6.023\times10^{23}=6.02\times10^{22}, so A is true.

Atoms: each \ce{N2O} molecule has 3 atoms, so 3\times6.02\times10^{22}=1.806\times10^{23} atoms, so B is true. N atoms: each molecule has 2 N, so 2\times6.02\times10^{22}=1.204\times10^{23}, so C is true.

All three statements are true, so the answer is D.

Which pair contains equal quantities?

AVolume of 28\,\mathrm{g} \ce{N2} and 8\,\mathrm{g} \ce{CH4} at STP
BMolecules in 0.1 mole \ce{NH3} and 2.2414\,\mathrm{dm^3} \ce{O2} at STPAnswer
CMass of 1.204\times10^{24} molecules of \ce{CO2} and 4.8\times10^{24} atoms of \ce{NH3}
DBonds in 56\,\mathrm{g} \ce{N2} and 2 mole \ce{CH4}

Why

A: 28\,\mathrm{g} \ce{N2} is 1 mol, which occupies 22.4\,\mathrm{dm^3}. 8\,\mathrm{g} \ce{CH4} is 0.5 mol, which occupies 11.2\,\mathrm{dm^3}. Not equal.

B: 0.1 mol \ce{NH3} has 0.1\times6.023\times10^{23}=6.02\times10^{22} molecules. 2.2414\,\mathrm{dm^3} of \ce{O2} at STP is 0.1 mol, which also has 6.02\times10^{22} molecules. Equal, so B is correct.

C: 1.204\times10^{24} molecules of \ce{CO2} is 2 mol, that is 88 g. 4.8\times10^{24} atoms of \ce{NH3} is \dfrac{4.8\times10^{24}}{4}=1.2\times10^{24} molecules, or about 2 mol, that is about 34 g. Not equal.

D: 56\,\mathrm{g} \ce{N2} is 2 mol with a triple bond each, so 6 mol of bonds. 2 mol \ce{CH4} has 8 mol of C–H bonds. Not equal.

For 2\ce{X} + 3\ce{Y} \rightarrow \ce{Z}: when 12 moles of \ce{Y} react with excess of \ce{X} and give 3 moles of \ce{Z}, the percentage yield is:

A25\%
B33.33\%
C66\%
D75\%Answer

Why

Theoretical yield: 3 mol of \ce{Y} give 1 mol of \ce{Z}, so 12 mol of \ce{Y} give 4 mol of \ce{Z}.

Percent yield =\dfrac{\text{actual yield}}{\text{theoretical yield}}\times100=\dfrac{3}{4}\times100=75\%. Option A comes from dividing the actual yield by the starting moles (3/12).

Note. The printed equation has '1Z' and '3y' (cleaned to \ce{Z} and \ce{Y}).

Short questionsp. 93

Each item is a statement: say whether it is true and explain why.

i3 marks

49 g each of \ce{H2SO4} and \ce{H3PO4} have the same number of molecules but a different number of atoms.

Solution

True. Both acids have molar mass 98\,\mathrm{g\,mol^{-1}}, so 49 g of each is 0.5 mol. Each contains 0.5\times6.023\times10^{23}=3.01\times10^{23} molecules, the same number.

The number of atoms differs because the formulas differ. \ce{H2SO4} has 2+1+4=7 atoms per molecule, giving 7\times3.01\times10^{23}=2.11\times10^{24} atoms. \ce{H3PO4} has 3+1+4=8 atoms per molecule, giving 8\times3.01\times10^{23}=2.41\times10^{24} atoms.

ii3 marks

Different gases having different masses occupy equal volume at STP.

Solution

True. At STP every gas has the molar volume 22.414\,\mathrm{dm^3} per mole, so equal volumes contain equal numbers of moles and therefore equal numbers of molecules (Avogadro's law).

The mass of one mole is the molar mass, and molar masses differ. For example, 22.414\,\mathrm{dm^3} of \ce{H2} weighs 2 g, of \ce{NH3} 17 g and of \ce{O2} 32 g.

iii3 marks

One mole of Na and Al has equal number of atoms but the size of atoms are different.

Solution

True. One mole of any element contains 6.023\times10^{23} atoms, so 1 mol of Na and 1 mol of Al contain the same number of atoms.

The sizes differ because the atoms of the two elements are different: each element has its own atomic structure, and sodium atoms are larger than aluminium atoms. The same number of atoms does not mean the same size.

iv3 marks

Limiting reactant is always in lesser quantity in reaction mixture or not.

Solution

No. The limiting reactant is the one that is used up first, and it is found from the mole ratio of the balanced equation, not from the amount present alone.

Example: in \ce{2H2 + O2 -> 2H2O}, 2 mol of \ce{O2} needs 4 mol of \ce{H2}. If the mixture has 3 mol of \ce{H2} and 2 mol of \ce{O2}, the \ce{H2} runs out first, so \ce{H2} is limiting even though it is the larger amount.

v3 marks

Amount of product obtained through balance chemical equation is greater than the amount obtained experimentally.

Solution

True. The theoretical yield is calculated from the balanced equation on the assumption that all of the limiting reactant becomes product. In practice the actual yield is less, because side reactions give other products, some product remains in solution or is lost during filtering and transfer, and the reaction may stop before it is complete.

So \text{percent yield}=\dfrac{\text{actual yield}}{\text{theoretical yield}}\times100 is always less than 100% in a real experiment.

vi3 marks

What are the basic assumption in Stoichiometric calculations?

Solution

The basic assumptions are:

The chemical equation is balanced, so its coefficients give the mole ratio of reactants and products.
The limiting reactant is completely used up (100% conversion), so the calculated amount is the theoretical yield.
Masses, volumes of gases at STP (22.414 dm^3 per mole) and numbers of particles are all converted through the mole, using molar mass, molar volume and Avogadro's number.

vii3 marks

18g of steam has Avogadro's No of molecules but 58.5g of NaCl has not.

Solution

The statement is correct as far as molecules go. 18 g of \ce{H2O} is 1 mol, which contains 6.023\times10^{23} molecules, Avogadro's number of molecules.

58.5 g of \ce{NaCl} is also 1 mol, and it contains 6.023\times10^{23} formula units, that is 6.023\times10^{23} \ce{Na+} ions and 6.023\times10^{23} \ce{Cl-} ions. However, \ce{NaCl} is an ionic solid with no discrete molecules, so it has formula units rather than molecules.

viii3 marks

Why 2moles of Na react with 1mole of chlorine gas to produce 1mole of NaCl?

Solution

The question's premise does not match the balanced equation. \ce{2Na + Cl2 -> 2NaCl} shows that 2 mol of Na react with 1 mol of \ce{Cl2} to produce 2 mol of NaCl, not 1 mol.

The mole ratio is Na : \ce{Cl2} : NaCl =2:1:2. To make 1 mol of NaCl, 1 mol of Na and 0.5 mol of \ce{Cl2} are needed.

Note. Misprint or unclear premise: the balanced equation gives 2 mol of NaCl from 2 mol of Na, so the question's '1 mole of NaCl' cannot be right as printed.

Long questionsp. 95

Conceptual questions, answered in full with headings. The book gives no answers to these.

166 marks

Critically evaluate the importance of the mole concept in understanding chemical reactions.

Solution

The mole is the bridge between the invisible world of particles and the measurable world of grams, litres and solution volumes. Its importance can be judged by what it allows chemists to do and by where it stops being enough.

1. The mole links mass and particles

One mole is the amount containing 6.023\times10^{23} representative particles, and its mass in grams equals the molar mass. So 1 mol of \ce{O} atoms is 16 g and 1 mol of \ce{H2O} molecules is 18 g. Chemists can weigh out an exact number of particles without counting them.

2. The balanced equation is a mole ratio

The coefficients of a balanced equation are the mole ratios. In \ce{2Mg + O2 -> 2MgO}, 4 mol of Mg give 4 mol of MgO and need 2 mol of \ce{O2} (book Example 4.5). Without moles, the equation would only describe the reaction in words.

3. Gases, solutions and volumes

At STP, 1 mol of any gas occupies 22.414\,\mathrm{dm^3}, so mass and volume of a gas convert through moles (Example 4.1). For solutions, molarity gives moles from volume, and M_1V_1/n_1=M_2V_2/n_2 links two solutions that react (Example 4.10).

4. Limiting reactant and yield

The mole decides which reactant runs out first and how much product can form. The theoretical yield is found from moles, and the percent yield compares it with the actual yield (Example 4.11 and section 4.3).

Critical evaluation

Strength: one unit converts between mass, volume and number of particles, so every quantitative question about a reaction follows the same route.
Strength: it lets chemists predict yields and check experimental results, which matters in making medicines (section 4.4).
Limit: the calculation assumes that the equation is balanced and that the limiting reactant reacts completely. Real reactions give less product because of side reactions, losses and incomplete reaction, so mole calculations give the maximum amount, not the actual amount.
Limit: the mole relates amounts only. It needs correct formulas and balanced equations, so an error in either spoils every later step.

Overall, the mole concept is essential. It turns a chemical equation into numbers that can be checked against experiment, and its value depends on remembering that it gives the maximum product, not the actual yield.

176 marks

Analyze the relationship between molar volume and Avogadro's number.

Solution

Definitions

Molar volume (V_m) is the volume occupied by 1 mol of any gas at STP: V_m=22.414\,\mathrm{dm^3\,mol^{-1}}. Avogadro's number (N_A) is the number of particles in 1 mol: N_A=6.023\times10^{23}\,\mathrm{mol^{-1}}.

Derivation Volume of one molecule of gas

Aim. Find the volume that one molecule of gas occupies at STP and show how it links V_m and N_A.

Assumptions.

The gas is at STP, where 1 mol of any gas occupies V_m.
1 mol contains N_A particles (definition of the mole).

Volume of one mole of gas.
V_m=22.414\,\mathrm{dm^3\,mol^{-1}}
Number of molecules in one mole.
N_A=6.023\times10^{23}\,\mathrm{mol^{-1}}
Divide the volume of one mole by the number of molecules in it to get the volume per molecule.
v_{\text{molecule}}=\dfrac{V_m}{N_A}=\dfrac{22.414}{6.023\times10^{23}}
Evaluate.
v_{\text{molecule}}=3.72\times10^{-23}\,\mathrm{dm^3}
For any volume V of gas at STP, the number of molecules is V/V_m moles, multiplied by N_A.
N=\dfrac{V}{V_m}\,N_A

ResultV_m=N_A\times v_{\text{molecule}}

The molar volume is the space taken up by N_A molecules of gas, including the empty space between them. Because N_A is fixed and STP is fixed, V_m is the same for every gas.

Check with the book: 2.5 mol of \ce{Cl2} occupies 2.5\times22.414=56.035\,\mathrm{dm^3} at STP (Example 4.1) and contains 2.5\times6.023\times10^{23}=1.51\times10^{24} molecules. The ratio 56.035/1.51\times10^{24} matches the volume per molecule found above.

What the relationship shows

Avogadro's number fixes how many molecules are in 1 mol; the molar volume fixes how much space they take at STP. The two are linked by the volume of a single molecule.
Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (Avogadro's law), so one mole of any gas fills the same volume.
If one of the two were changed, the other would have to change: a different V_m would give a different N_A for the same molecular volume.

186 marks

Compare and contrast the molar volumes of different gases under the same condition of temperature and pressure.

Solution

Introduction

Under the same temperature and pressure, 1 mol of any gas occupies the same volume: the molar volume. At STP this is 22.414\,\mathrm{dm^3}. The molar volume does not depend on what the gas is, but the mass of that volume does.

Table: 1 mol of gas at STP

The volume is the same for every gas; the mass and the size of the molecules are not.
Gas	Molar mass (g mol^{-1})	Mass of 22.414\,\mathrm{dm^3} at STP (g)	Molecules in 22.414\,\mathrm{dm^3}
\ce{H2}	2	2	6.023\times10^{23}
\ce{CH4}	16	16	6.023\times10^{23}
\ce{NH3}	17	17	6.023\times10^{23}
\ce{O2}	32	32	6.023\times10^{23}
\ce{CO2}	44	44	6.023\times10^{23}

Similarities

The volume of one mole is the same, 22.414\,\mathrm{dm^3} at STP, so the molar volume is the same for all gases.
Each mole contains the same number of molecules, 6.023\times10^{23} (Avogadro's number).
Equal volumes under the same conditions contain equal numbers of molecules, so the gases behave alike in volume.

Differences

Molar mass: \ce{H2} has 2 g mol^{-1} and \ce{CO2} has 44 g mol^{-1}, so equal volumes have very different masses.
Density: density is mass per unit volume, \rho=\dfrac{M}{V_m}, so a gas with a higher molar mass has a higher density (book section 4.1.2).
Molecules: the molecules differ in size and mass, although their number in a given volume is the same.

Conclusion

The gases are alike in volume and in the number of molecules per mole, and different in molar mass, mass per volume and density. The molar volume is therefore a property of the number of molecules, not of the gas itself.

Numerical problemsp. 93–94

Items 3 and 4 to 15 of the exercise. Each working uses the mole, molar mass, the balanced equation and, for gases, the molar volume at STP.

Calculate each of the following quantities.

Solution

(i) Mass in gram of 0.74 mol \ce{KMnO4}.

Molar mass of \ce{KMnO4} from the atomic masses K 39, Mn 55, O 16.
M=39+55+4\times16=158\,\mathrm{g\,mol^{-1}}
Mass = moles \times molar mass.
m=0.74\times158=116.9\,\mathrm{g}

m=116.9\,\mathrm{g}

(ii) Moles of O atoms in 9.22 g \ce{Mg(NO3)2}.

Molar mass of \ce{Mg(NO3)2}.
M=24+2(14+3\times16)=148\,\mathrm{g\,mol^{-1}}
Moles of formula units.
n=\dfrac{9.22}{148}=0.0623\,\mathrm{mol}
Each formula unit has 2\times3=6 O atoms, so multiply by 6.
n_{\mathrm{O}}=6\times0.0623=0.374\,\mathrm{mol}

n_{\mathrm{O}}=0.374\,\mathrm{mol}

(iii) Number of O atoms in 0.037 g \ce{CuSO4.5H2O}.

Molar mass of \ce{CuSO4.5H2O}.
M=63.5+32+4\times16+5\times18=249.5\,\mathrm{g\,mol^{-1}}
Moles of the hydrate.
n=\dfrac{0.037}{249.5}=1.48\times10^{-4}\,\mathrm{mol}
O atoms per formula unit: 4 in the sulphate and 5 in the water of crystallisation, 9 in all.
N=1.48\times10^{-4}\times9\times6.023\times10^{23}=8.04\times10^{20}

N_{\mathrm{O}}=8.04\times10^{20}\ \text{atoms}

(iv) Mass in kg of 2.6\times10^{20} molecules of \ce{SO3}.

Moles of \ce{SO3}.
n=\dfrac{2.6\times10^{20}}{6.023\times10^{23}}=4.32\times10^{-4}\,\mathrm{mol}
Mass with M(\ce{SO3})=80\,\mathrm{g\,mol^{-1}}.
m=4.32\times10^{-4}\times80=0.0345\,\mathrm{g}
Convert grams to kilograms.
m=3.45\times10^{-5}\,\mathrm{kg}

m=3.45\times10^{-5}\,\mathrm{kg}

(v) Total number of ions in 14.3 g \ce{CaBr2}.

Molar mass of \ce{CaBr2} (Ca 40, Br 80).
M=40+2\times80=200\,\mathrm{g\,mol^{-1}}
Moles of \ce{CaBr2}.
n=\dfrac{14.3}{200}=0.0715\,\mathrm{mol}
Each formula unit gives one \ce{Ca^2+} and two \ce{Br-}, which is 3 ions.
N=0.0715\times3\times6.023\times10^{23}=1.29\times10^{23}

N_{\text{ions}}=1.29\times10^{23}

(vi) Mass in mg of 0.45 mole of \ce{CuCl2.2H2O}.

Molar mass of \ce{CuCl2.2H2O} (Cu 63.5, Cl 35.5).
M=63.5+2\times35.5+2\times18=170.5\,\mathrm{g\,mol^{-1}}
Mass = moles \times molar mass.
m=0.45\times170.5=76.7\,\mathrm{g}
Convert to milligrams.
m=7.67\times10^{4}\,\mathrm{mg}

m=7.67\times10^{4}\,\mathrm{mg}

(vii) Mass in grams of 2.78\times10^{21} molecules of \ce{N2O4}.

Molar mass of \ce{N2O4}.
M=2\times14+4\times16=92\,\mathrm{g\,mol^{-1}}
Moles of molecules.
n=\dfrac{2.78\times10^{21}}{6.023\times10^{23}}=4.62\times10^{-3}\,\mathrm{mol}
Mass.
m=4.62\times10^{-3}\times92=0.425\,\mathrm{g}

m=0.425\,\mathrm{g}

(viii) Volume of \ce{SO2} at STP of 4.8\times10^{23} molecules of \ce{SO2}.

Moles of \ce{SO2}.
n=\dfrac{4.8\times10^{23}}{6.023\times10^{23}}=0.797\,\mathrm{mol}
Volume at STP, with 1 mol occupying 22.414 dm^3.
V=0.797\times22.414=17.86\,\mathrm{dm^3}

V=17.86\,\mathrm{dm^3}

(ix) Mass in gram of \ce{Ca(NO3)2} having 2\times10^{21} ions of nitrate.

Each formula unit contains 2 nitrate ions, so find the moles of \ce{NO3-} first.
n_{\mathrm{NO_3^-}}=\dfrac{2\times10^{21}}{6.023\times10^{23}}=3.32\times10^{-3}\,\mathrm{mol}
Moles of \ce{Ca(NO3)2} are half of this.
n=\dfrac{3.32\times10^{-3}}{2}=1.66\times10^{-3}\,\mathrm{mol}
Molar mass M=40+2\times62=164\,\mathrm{g\,mol^{-1}}, so the mass is:
m=1.66\times10^{-3}\times164=0.272\,\mathrm{g}

m=0.272\,\mathrm{g}

(x) Covalent bonds in 22 gram of dry ice.

Dry ice is solid \ce{CO2}. Moles: M=44\,\mathrm{g\,mol^{-1}}.
n=\dfrac{22}{44}=0.5\,\mathrm{mol}
Each \ce{O=C=O} molecule has two C=O double bonds, so 4 covalent bonds (each double bond counted as two).
N_{\text{bonds}}=0.5\times4\times6.023\times10^{23}=1.20\times10^{24}

N_{\text{bonds}}=1.20\times10^{24}

Note. Part (x): if each double bond is counted as one bond, the answer is 0.5\times2\times6.023\times10^{23}=6.02\times10^{23} bonds. The book prints no answers for this part.

42026 Annual I, Set D, Q.3 OR

Calcium ion can be precipitated from solution by sodium oxalate. Is 15 g \ce{Ca^2+} completely precipitated by 15 g of sodium oxalate? If \ce{Ca^2+} ions are left in the solution, calculate how much \ce{Ca^2+} ions are left in the solution?

\ce{Ca^2+ + Na2C2O4 -> CaC2O4 + 2Na+}

Solution

Given

m_{\ce{Ca^2+}} = 15\,\mathrm{g}
M_{\ce{Ca}} = 40\,\mathrm{g\,mol^{-1}}
m_{\ce{Na2C2O4}} = 15\,\mathrm{g}
M_{\ce{Na2C2O4}} = 134\,\mathrm{g\,mol^{-1}}

To find

Whether all \ce{Ca^2+} is precipitated
mass of \ce{Ca^2+} left

Moles of \ce{Ca^2+} available.
n_{\mathrm{Ca}}=\dfrac{15}{40}=0.375\,\mathrm{mol}
Moles of sodium oxalate.
n_{\mathrm{ox}}=\dfrac{15}{134}=0.112\,\mathrm{mol}
The ratio is 1:1, so the oxalate is the limiting reactant (0.112 mol is less than 0.375 mol). Only 0.112 mol of \ce{Ca^2+} is precipitated, so the answer to the first part is no.
Moles of \ce{Ca^2+} left in solution.
n_{\text{left}}=0.375-0.112=0.263\,\mathrm{mol}
Mass left.
m_{\text{left}}=0.263\times40=10.5\,\mathrm{g}

m_{\ce{Ca^2+}\,\text{left}}=0.263\,\mathrm{mol}\times40=10.5\,\mathrm{g}

0.05 mol of potassium chlorate heated for a time and its 0.015 mol left. How much KCl produced, also calculate molecules of \ce{O2} produced?

2\ce{KClO3} \longrightarrow 2\ce{KCl} + 3\ce{O2}

Solution

Given

n_{\text{initial}} = 0.05\,\mathrm{mol}
n_{\text{left}} = 0.015\,\mathrm{mol}
M_{\ce{KCl}} = 74.5\,\mathrm{g\,mol^{-1}}

To find

mass of KCl
molecules of \ce{O2}

Moles of \ce{KClO3} decomposed.
n_{\text{dec}}=0.05-0.015=0.035\,\mathrm{mol}
From the equation, 2 mol \ce{KClO3} give 2 mol \ce{KCl}, so 0.035 mol of KCl is formed.
n_{\ce{KCl}}=0.035\,\mathrm{mol}
Mass of KCl.
m=0.035\times74.5=2.61\,\mathrm{g}
From the equation, 2 mol \ce{KClO3} give 3 mol \ce{O2}, so \dfrac{3}{2}\times0.035=0.0525 mol of \ce{O2} is formed.
n_{\ce{O2}}=0.0525\,\mathrm{mol}
Molecules of \ce{O2}.
N=0.0525\times6.023\times10^{23}=3.16\times10^{22}

m_{\ce{KCl}}=2.61\,\mathrm{g},\qquad N_{\ce{O2}}=3.16\times10^{22}\ \text{molecules}

Calculate No. of moles of water produce by 5\times10^{24} molecules of \ce{H2SO4} and 20 g of \ce{NaOH}.

\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}

Solution

Given

N_{\ce{H2SO4}} = 5\times10^{24}
m_{\ce{NaOH}} = 20\,\mathrm{g}
M_{\ce{NaOH}} = 40\,\mathrm{g\,mol^{-1}}

To find

moles of \ce{H2O} produced

Moles of \ce{H2SO4}.
n=\dfrac{5\times10^{24}}{6.023\times10^{23}}=8.30\,\mathrm{mol}
Moles of \ce{NaOH}.
n=\dfrac{20}{40}=0.5\,\mathrm{mol}
Mole ratio: 2 mol \ce{NaOH} need 1 mol \ce{H2SO4}. So 0.5 mol NaOH needs only 0.25 mol \ce{H2SO4}. The \ce{H2SO4} is in excess and NaOH is limiting.
Water is formed in the ratio 2 mol \ce{H2O} : 2 mol \ce{NaOH}, so 0.5 mol of water is formed.
n_{\ce{H2O}}=0.5\,\mathrm{mol}

n_{\ce{H2O}}=0.5\,\mathrm{mol}

Note. The question gives no equation. The neutralisation equation \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O} is assumed here.

Formalin is an aqueous solution of formaldehyde (\ce{HCHO}), used as a preservative for biological specimens. A biologist wants to prepare 1\,\mathrm{dm^3} of 11.5 M formalin. What mass of formaldehyde he requires?

Solution

Given

V = 1\,\mathrm{dm^3}
M = 11.5\,\mathrm{mol\,dm^{-3}}
M_{\ce{HCHO}} = 30\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{HCHO}

Molar mass of \ce{HCHO} (C 12, H 1, O 16).
M=12+2+16=30\,\mathrm{g\,mol^{-1}}
Moles needed from molarity = mol/dm^3 \times dm^3.
n=11.5\times1=11.5\,\mathrm{mol}
Mass = moles \times molar mass.
m=11.5\times30=345\,\mathrm{g}

m=345\,\mathrm{g}

What mass of \ce{CaCO3} would you use to add to 100\,\mathrm{cm^3} of 0.5 M HCl to completely neutralize acid?

\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}

Solution

Given

V_{\ce{HCl}} = 100\,\mathrm{cm^3}=0.100\,\mathrm{dm^3}
M_{\ce{HCl}} = 0.5\,\mathrm{mol\,dm^{-3}}
M_{\ce{CaCO3}} = 100\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{CaCO3}

Moles of HCl.
n_{\ce{HCl}}=0.100\times0.5=0.050\,\mathrm{mol}
From the equation, 1 mol \ce{CaCO3} reacts with 2 mol HCl, so the moles of \ce{CaCO3} are half.
n_{\ce{CaCO3}}=\dfrac{0.050}{2}=0.025\,\mathrm{mol}
Mass.
m=0.025\times100=2.5\,\mathrm{g}

m_{\ce{CaCO3}}=2.5\,\mathrm{g}

Calculate mass of oxygen required for complete combustion of 1 mole of gasoline (\ce{C8H18}).

2\ce{C8H18} + 25\ce{O2} \longrightarrow 16\ce{CO2} + 18\ce{H2O}

Solution

Given

n_{\ce{C8H18}} = 1\,\mathrm{mol}
M_{\ce{O2}} = 32\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{O2}

Balanced equation for one mole of \ce{C8H18}: \ce{C8H18 + 12.5O2 -> 8CO2 + 9H2O}, which is half of the equation above.
Moles of \ce{O2} needed.
n_{\ce{O2}}=12.5\,\mathrm{mol}
Mass.
m=12.5\times32=400\,\mathrm{g}

m_{\ce{O2}}=400\,\mathrm{g}

Graphite is the crystalline form of carbon used in 'lead' pencil. (a) How many moles are present in 315 mg graphite? (b) How many carbon atoms are in it?

Solution

Given

m = 315\,\mathrm{mg}=0.315\,\mathrm{g}
M_{\text{C}} = 12\,\mathrm{g\,mol^{-1}}

To find

moles of graphite
number of carbon atoms

Part (a): moles = mass / molar mass.
n=\dfrac{0.315}{12}=2.63\times10^{-2}\,\mathrm{mol}
Part (b): atoms = moles \times Avogadro's number.
N=0.02625\times6.023\times10^{23}=1.58\times10^{22}

\text{(a) }n=2.63\times10^{-2}\,\mathrm{mol};\qquad \text{(b) }N=1.58\times10^{22}\text{ atoms}

Manganese is transition metal essential for the growth of strong bones. Calculate mass of 3.22\times10^{20} atoms of manganese found in 1 kilogram of bone?

Solution

Given

N = 3.22\times10^{20}\,\text{atoms}
M_{\text{Mn}} = 55\,\mathrm{g\,mol^{-1}}

To find

mass of manganese

Moles of Mn atoms.
n=\dfrac{3.22\times10^{20}}{6.023\times10^{23}}=5.35\times10^{-4}\,\mathrm{mol}
Mass.
m=5.35\times10^{-4}\times55=2.94\times10^{-2}\,\mathrm{g}

m_{\text{Mn}}=2.94\times10^{-2}\,\mathrm{g}=29.4\,\mathrm{mg}\ \text{per kilogram of bone}

122026 Annual I, Set B, Q.3 OR

How much mass of excess reactant left after 40.5 g of aluminium metal reacts with 196 g of \ce{H2SO4}?

2\ce{Al} + 3\ce{H2SO4} \longrightarrow \ce{Al2(SO4)3} + 3\ce{H2}

Solution

Given

m_{\ce{Al}} = 40.5\,\mathrm{g}
m_{\ce{H2SO4}} = 196\,\mathrm{g}
M_{\ce{Al}} = 27\,\mathrm{g\,mol^{-1}}
M_{\ce{H2SO4}} = 98\,\mathrm{g\,mol^{-1}}

To find

mass of the excess reactant left

Moles of each reactant.
n_{\ce{Al}}=\dfrac{40.5}{27}=1.5\,\mathrm{mol};\qquad n_{\ce{H2SO4}}=\dfrac{196}{98}=2\,\mathrm{mol}
Al needed to react with all 2 mol of \ce{H2SO4} (ratio 2 Al : 3 \ce{H2SO4}).
n_{\ce{Al}}=2\times\dfrac{2}{3}=1.33\,\mathrm{mol}
Al available (1.5 mol) is more than 1.33 mol, so \ce{H2SO4} is limiting and Al is in excess.
Excess Al left.
n_{\text{left}}=1.5-1.33=0.167\,\mathrm{mol}
Mass of Al left.
m=0.167\times27=4.5\,\mathrm{g}

m_{\text{Al left}}=4.5\,\mathrm{g}

132026 Annual I, Set B, Q.2 (iv) OR2026 Annual I, Set D, Q.2 (iv) OR

Calculate mass of \ce{SO2} that will be produced with 155 g of Cu from the roasting of \ce{CuS}.

\ce{CuS} + \ce{O2} \longrightarrow \ce{Cu} + \ce{SO2}

Solution

Given

m_{\ce{Cu}} = 155\,\mathrm{g}
M_{\ce{Cu}} = 63.5\,\mathrm{g\,mol^{-1}}
M_{\ce{SO2}} = 64\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{SO2}

Moles of Cu.
n_{\ce{Cu}}=\dfrac{155}{63.5}=2.44\,\mathrm{mol}
The equation is already balanced (1 Cu, 1 S and 2 O on each side): 1 mol Cu is formed with 1 mol \ce{SO2}, so the mole ratio Cu : \ce{SO2} is 1 : 1.
n_{\ce{SO2}}=2.44\,\mathrm{mol}
Mass of \ce{SO2}.
m=2.44\times64=156.2\,\mathrm{g}

m_{\ce{SO2}}=156.2\,\mathrm{g}

\ce{SO2} is air pollutant, it contributes to acid rain. Its emission is controlled by absorbing it into a base (NaOH). Calculate mass of \ce{SO2} absorbed by 33 g of NaOH.

\ce{SO2} + 2\ce{NaOH} \longrightarrow \ce{Na2SO3} + \ce{H2O}

Solution

Given

m_{\ce{NaOH}} = 33\,\mathrm{g}
M_{\ce{NaOH}} = 40\,\mathrm{g\,mol^{-1}}
M_{\ce{SO2}} = 64\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{SO2} absorbed

Moles of NaOH.
n_{\ce{NaOH}}=\dfrac{33}{40}=0.825\,\mathrm{mol}
From the equation, 2 mol NaOH absorb 1 mol \ce{SO2}.
n_{\ce{SO2}}=\dfrac{0.825}{2}=0.4125\,\mathrm{mol}
Mass of \ce{SO2}.
m=0.4125\times64=26.4\,\mathrm{g}

m_{\ce{SO2}}=26.4\,\mathrm{g}

Potassium super oxide (\ce{KO2}) is used as source of oxygen in re-breathing mask. Identify limiting reactant in each of following reactant mixtures. (i) 6.4 moles \ce{KO2} and 2.1 moles of \ce{H2O}. (ii) 8.4 moles of \ce{KO2} and 1.5 moles of \ce{H2O}.

4\ce{KO2} + 2\ce{H2O} \longrightarrow 4\ce{KOH} + 3\ce{O2}

Solution

Given

n_{\ce{KO2}} = 6.4\,\mathrm{mol}\text{ or }8.4\,\mathrm{mol}
n_{\ce{H2O}} = 2.1\,\mathrm{mol}\text{ or }1.5\,\mathrm{mol}

To find

limiting reactant in each mixture

(i) Divide each amount by its coefficient: \ce{KO2} gives 6.4/4=1.6; \ce{H2O} gives 2.1/2=1.05.
The smaller value is for \ce{H2O}, so water is limiting. Check: 2.1 mol \ce{H2O} needs 4.2 mol \ce{KO2}; 6.4 mol is available, leaving 2.2 mol \ce{KO2} unreacted.
(ii) \ce{KO2} gives 8.4/4=2.1; \ce{H2O} gives 1.5/2=0.75.
Water is limiting again. Check: 1.5 mol \ce{H2O} needs 3.0 mol \ce{KO2}; 8.4 mol is available, leaving 5.4 mol \ce{KO2} unreacted.

\text{(i) and (ii): }\ce{H2O}\text{ is the limiting reactant}

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 11 mark

Number of atoms in 8\,\mathrm{g} \ce{He} are:

A6.02\times10^{22}
B1.204\times10^{24}Answer
C6.02\times10^{21}
D1.204\times10^{23}

Why

Helium is monatomic, so its molar mass is 4\,\mathrm{g\,mol^{-1}} and 8\,\mathrm{g} is 8/4=2 mol. Number of atoms =2\times6.02\times10^{23}=1.204\times10^{24}, which is B.

Option D (1.204\times10^{23}) is the count for only 0.2 mol, ten times too few, which comes from forgetting to divide by 4\,\mathrm{g\,mol^{-1}} correctly (using 40). Options A and C are smaller still.

2025 Annual I, Set B, MCQ 41 mark

The number of molecules present in 67.2\ \mathrm{dm^{3}} of \ce{He} gas at STP is:

A6.02\times10^{24}
B1.204\times10^{24}
C1.806\times10^{24}Answer
D2.4\times10^{24}

Why

At STP one mole of any gas occupies 22.4\,\mathrm{dm^3}, so 67.2\,\mathrm{dm^3} is 67.2/22.4=3 mol. Number of molecules =3\times6.02\times10^{23}=1.806\times10^{24}, which is C.

Option A (6.02\times10^{24}) would be 10 mol and B (1.204\times10^{24}) would be 2 mol. Helium is monatomic, so molecules and atoms are the same here.

2026 Annual I, Set B, MCQ 61 mark

Molar volume concept is applicable mainly to:

AGasesAnswer
BLiquids
CSolids
DPlasma

Why

One mole of any gas occupies 22.414\,\mathrm{dm^3} at STP, because the volume of a gas depends on the number of molecules and not on their kind. This is the molar volume, so the answer is A.

Liquids and solids do not follow this rule: the volume of one mole of a liquid or solid is different for each substance, since it depends on the density of the substance.

2025 Annual I, Set B, Q.2 (iv)3 marks

Write down three differences between limiting and non-limiting reactants.

Solution

The limiting reactant is the reactant that is completely consumed and so limits the amount of product. The non-limiting (excess) reactant is the one left unreacted after the reaction is complete.

Point	Limiting reactant	Non-limiting (excess) reactant
Fate in the reaction	Completely used up	Left unreacted when the reaction stops
Effect on the product	Decides how much product is formed (it gives the least product)	Does not decide the product, and could have given more product than is formed
Amount compared with the equation	Present in less than the mole ratio of the equation needs for the other reactant	Present in more than the mole ratio of the equation needs
Remaining at the end	None	Some is left, found by subtracting the amount used from the amount taken

Example: for \ce{2H2 + O2 -> 2H2O}, with 1 mol of \ce{H2} and 1 mol of \ce{O2}, hydrogen is limiting and \tfrac12 mol of oxygen is left over.

2026 Annual I, Set B, Q.2 (iv) OR3 marks

Calculate the mass of \ce{SO2} that will be produced with 155\,\mathrm{g} of Cu from the roasting of CuS. \ce{CuS + O2 -> Cu + SO2} (\mathrm{Cu}=63.5\ \mathrm{g\,mol^{-1}}, \mathrm{S}=32\ \mathrm{g\,mol^{-1}})

Solution

Given

m_{\ce{Cu}} = 155\,\mathrm{g}
M_{\ce{Cu}} = 63.5\,\mathrm{g\,mol^{-1}}
M_{\ce{SO2}} = 32+2\times16=64\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{SO2} produced

Check the equation. Each side has 1 Cu, 1 S and 2 O, so it is already balanced and the mole ratio Cu : \ce{SO2} is 1 : 1.
\ce{CuS + O2 -> Cu + SO2}
Moles of Cu produced.
n_{\ce{Cu}}=\dfrac{155}{63.5}=2.44\,\mathrm{mol}
By the mole ratio 1 : 1, the same number of moles of \ce{SO2} is produced.
n_{\ce{SO2}}=2.44\,\mathrm{mol}
Mass of \ce{SO2} = moles \times molar mass.
m_{\ce{SO2}}=2.44\times64=156.2\,\mathrm{g}

m_{\ce{SO2}}=156.2\,\mathrm{g}

2025 Annual I, Set B, Q.2 (xi)3 marks

Calculate the volume occupied by 3.011\times10^{24} molecules of \ce{NH3} gas enclosed in a container at STP.

Solution

Given

N = 3.011\times10^{24}\ \text{molecules}
N_A = 6.023\times10^{23}\,\mathrm{mol^{-1}}
V_m = 22.4\,\mathrm{dm^3\,mol^{-1}} at STP

To find

volume of \ce{NH3} at STP

Convert the number of molecules to moles by dividing by Avogadro's number.
n=\dfrac{3.011\times10^{24}}{6.023\times10^{23}}=5.0\,\mathrm{mol}
Multiply the moles by the molar volume.
V=5.0\times22.4=112\,\mathrm{dm^3}

V=112\,\mathrm{dm^3}

2026 Annual I, Set B, Q.2 (xii) OR3 marks

Calculate the volume of 177.5\,\mathrm{g} of chlorine gas at STP.

Solution

Given

m_{\ce{Cl2}} = 177.5\,\mathrm{g}
M_{\ce{Cl2}} = 2\times35.5=71\,\mathrm{g\,mol^{-1}}
V_m = 22.4\,\mathrm{dm^3\,mol^{-1}} at STP

To find

volume V at STP

Moles of chlorine molecules.
n=\dfrac{177.5}{71}=2.5\,\mathrm{mol}
One mole of any gas occupies 22.4\,\mathrm{dm^3} at STP, so multiply by the molar volume.
V=n\times V_m=2.5\times22.4=56\,\mathrm{dm^3}

Using 22.414\,\mathrm{dm^3\,mol^{-1}} gives 56.04\,\mathrm{dm^3}, the same to three significant figures.

V=56\,\mathrm{dm^3}

2026 Annual I, Set B, Q.3 OR6 marks

Hydrogen gas is produced by the following reaction: \ce{2Al + 3H2SO4 -> Al2(SO4)3 + 3H2} When 40.5\,\mathrm{g} of Aluminium reacts with 196\,\mathrm{g} of \ce{H2SO4} then: (a) Calculate the amount of \ce{H2} in gram produced during this reaction. (b) Calculate the amount of excess reactant left unreacted after the completion of chemical reaction. (\mathrm{Al}=27\ \mathrm{g\,mol^{-1}})(\mathrm{S}=32\ \mathrm{g\,mol^{-1}})

Solution

Given

m_{\ce{Al}} = 40.5\,\mathrm{g}
m_{\ce{H2SO4}} = 196\,\mathrm{g}
M_{\ce{Al}} = 27\,\mathrm{g\,mol^{-1}}
M_{\ce{H2SO4}} = 2+32+64=98\,\mathrm{g\,mol^{-1}}
M_{\ce{H2}} = 2\,\mathrm{g\,mol^{-1}}

To find

(a) mass of \ce{H2} produced
(b) mass of excess reactant left

Moles of each reactant

2\ce{Al}+3\ce{H2SO4}\longrightarrow\ce{Al2(SO4)3}+3\ce{H2}

n_{\ce{Al}}=\dfrac{40.5}{27}=1.5\,\mathrm{mol}\qquad n_{\ce{H2SO4}}=\dfrac{196}{98}=2.0\,\mathrm{mol}

(a) Limiting reactant and mass of hydrogen

Moles of \ce{H2} that each reactant could give (ratio 2 Al : 3 \ce{H2} and 3 \ce{H2SO4} : 3 \ce{H2}).
\text{from Al: }1.5\times\dfrac{3}{2}=2.25\,\mathrm{mol}\qquad\text{from }\ce{H2SO4}:\ 2.0\times\dfrac{3}{3}=2.0\,\mathrm{mol}
The reactant that gives the smaller amount of product is the limiting reactant, so \ce{H2SO4} is limiting and 2.0 mol of \ce{H2} is formed.
n_{\ce{H2}}=2.0\,\mathrm{mol}
Mass of hydrogen.
m_{\ce{H2}}=2.0\times2=4.0\,\mathrm{g}

(b) Excess reactant left

Al is in excess. Moles of Al needed to react with all 2.0 mol of \ce{H2SO4}.
n_{\ce{Al}}\text{ used}=2.0\times\dfrac{2}{3}=1.33\,\mathrm{mol}
Moles of Al left.
1.5-1.33=0.167\,\mathrm{mol}
Mass of Al left.
m=0.167\times27=4.5\,\mathrm{g}

m_{\ce{H2}}=4.0\,\mathrm{g};\quad m_{\ce{Al}}\text{ left}=4.5\,\mathrm{g}

2026 Annual I, Set D, MCQ 11 mark

0.2 moles of \ce{NaCl}, when completely ionized produce \ce{Na+} ions:

A1.204\times10^{22}
B1.204\times10^{23}Answer
C2.4\times10^{22}
D2.4\times10^{23}

Why

Sodium chloride gives one \ce{Na+} ion per formula unit (\ce{NaCl -> Na+ + Cl-}), so 0.2 mol of \ce{NaCl} gives 0.2 mol of \ce{Na+}. The number of ions is 0.2\times6.02\times10^{23}=1.204\times10^{23}, which is B.

Option D (2.4\times10^{23}) would need 0.4 mol of \ce{Na+}, which is the case for \ce{Na2SO4} (two \ce{Na+} per formula unit) but not for \ce{NaCl}. Options A and C are ten times too small.

2025 Annual I, Set D, MCQ 31 mark

0.5\ \mathrm{dm^{3}} of \ce{CO2} gas is enclosed in a container at STP. The mass of gas is:

A0.49 g
B0.98 gAnswer
C22.4 g
D48.6 g

Why

Moles of \ce{CO2} at STP: 0.5/22.4=0.0223 mol. Molar mass of \ce{CO2} is 12+32=44\,\mathrm{g\,mol^{-1}}, so the mass is 0.0223\times44=0.98\,\mathrm{g}, which is B.

Option C (22.4\,\mathrm{g}) confuses the molar volume (22.4\,\mathrm{dm^3}) with a mass, and A (0.49\,\mathrm{g}) is half the correct value.

2026 Annual I, Set D, MCQ 61 mark

At STP, 71\,\mathrm{g} of \ce{Cl2} gas occupies:

A11.2\ \mathrm{dm^3}
B22.4\ \mathrm{dm^3}Answer
C44.8\ \mathrm{dm^3}
D2\ \mathrm{dm^3}

Why

The molar mass of \ce{Cl2} is 2\times35.5=71\,\mathrm{g\,mol^{-1}}, so 71\,\mathrm{g} is exactly 1 mol. One mole of any gas occupies 22.4\,\mathrm{dm^3} at STP, so the answer is B.

Option A (11.2\,\mathrm{dm^3}) is the volume of half a mole, which would be 35.5\,\mathrm{g} of \ce{Cl2}, and C (44.8\,\mathrm{dm^3}) is for 2 mol (142\,\mathrm{g}).

2025 Annual I, Set D, Q.2 (iii)1+2 = 3 marks

What is molar volume? Calculate the volume occupied by 2000 g of \ce{Cl2} gas at STP.

Solution

Molar volume (1 mark)

The molar volume is the volume occupied by one mole of any gas at STP. It is 22.414\,\mathrm{dm^3} (about 22.4\,\mathrm{dm^3}).

Volume of 2000 g of \ce{Cl2} (2 marks)

Given

m_{\ce{Cl2}} = 2000\,\mathrm{g}
M_{\ce{Cl2}} = 71\,\mathrm{g\,mol^{-1}}
V_m = 22.4\,\mathrm{dm^3\,mol^{-1}}

To find

volume at STP

Moles of \ce{Cl2}.
n=\dfrac{2000}{71}=28.17\,\mathrm{mol}
Multiply by the molar volume.
V=28.17\times22.4=631\,\mathrm{dm^3}

V=631\,\mathrm{dm^3}\ (\text{at STP})

2026 Annual I, Set D, Q.2 (iv) OR3 marks

Solution

Given

m_{\ce{Cu}} = 155\,\mathrm{g}
M_{\ce{Cu}} = 63.5\,\mathrm{g\,mol^{-1}}
M_{\ce{SO2}} = 32+2\times16=64\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{SO2} produced

Check the equation. Each side has 1 Cu, 1 S and 2 O, so it is already balanced and the mole ratio Cu : \ce{SO2} is 1 : 1.
\ce{CuS + O2 -> Cu + SO2}
Moles of Cu produced.
n_{\ce{Cu}}=\dfrac{155}{63.5}=2.44\,\mathrm{mol}
By the mole ratio 1 : 1, the same number of moles of \ce{SO2} is produced.
n_{\ce{SO2}}=2.44\,\mathrm{mol}
Mass of \ce{SO2} = moles \times molar mass.
m_{\ce{SO2}}=2.44\times64=156.2\,\mathrm{g}

m_{\ce{SO2}}=156.2\,\mathrm{g}

2026 Annual I, Set D, Q.2 (xii) OR3 marks

How many grams of \ce{O2} molecules are there in 20\ \mathrm{dm^3} of oxygen gas at STP?

Solution

Given

V_{\ce{O2}} = 20\,\mathrm{dm^3} at STP
V_m = 22.4\,\mathrm{dm^3\,mol^{-1}}
M_{\ce{O2}} = 32\,\mathrm{g\,mol^{-1}}

To find

mass of \ce{O2}

Moles of oxygen from the volume at STP.
n=\dfrac{V}{V_m}=\dfrac{20}{22.4}=0.893\,\mathrm{mol}
Mass = moles \times molar mass.
m=0.893\times32=28.6\,\mathrm{g}

With 22.414\,\mathrm{dm^3\,mol^{-1}} the answer is 28.55\,\mathrm{g}, which is 28.6\,\mathrm{g} to three significant figures.

m_{\ce{O2}}=28.6\,\mathrm{g}

2026 Annual I, Set D, Q.3 OR6 marks

Calcium ions can be precipitated from solution by sodium oxalate. \ce{Ca^{+2} + Na2C2O4 -> CaC2O4 + 2Na+} When 15\,\mathrm{g} \ce{Ca^{+2}} can be precipitated by 15\,\mathrm{g} of \ce{Na2C2O4} then: (a) Calculate the amount of \ce{CaC2O4} precipitates formed in gram. (b) Calculate the amount of excess reactant left unreacted after the completion of chemical reaction.

Solution

Given

m_{\ce{Ca^2+}} = 15\,\mathrm{g}
M_{\ce{Ca}} = 40\,\mathrm{g\,mol^{-1}}
m_{\ce{Na2C2O4}} = 15\,\mathrm{g}
M_{\ce{Na2C2O4}} = 2\times23+2\times12+4\times16=134\,\mathrm{g\,mol^{-1}}
M_{\ce{CaC2O4}} = 40+24+64=128\,\mathrm{g\,mol^{-1}}

To find

(a) mass of \ce{CaC2O4} formed
(b) mass of excess reactant left

\ce{Ca^2+ + Na2C2O4 -> CaC2O4 + 2Na+}

Moles of each reactant. The equation shows a 1 : 1 : 1 ratio.
n_{\ce{Ca^2+}}=\dfrac{15}{40}=0.375\,\mathrm{mol}\qquad n_{\ce{Na2C2O4}}=\dfrac{15}{134}=0.112\,\mathrm{mol}
With a 1 : 1 ratio, the smaller number of moles gives the smaller amount of product. Sodium oxalate is the limiting reactant and \ce{Ca^2+} is in excess.
0.112<0.375

(a) Mass of the precipitate

Moles of \ce{CaC2O4} equal the moles of the limiting reactant.
n_{\ce{CaC2O4}}=0.112\,\mathrm{mol}
Mass of \ce{CaC2O4}.
m=0.112\times128=14.3\,\mathrm{g}

(b) Excess reactant left

Moles of \ce{Ca^2+} left unreacted.
0.375-0.112=0.263\,\mathrm{mol}
Mass of \ce{Ca^2+} left.
m=0.263\times40=10.5\,\mathrm{g}

m_{\ce{CaC2O4}}=14.3\,\mathrm{g};\quad m_{\ce{Ca^2+}}\text{ left}=10.5\,\mathrm{g}

2025 Annual I, Set D, Q.67 marks

300 g Zn is reacted with 300 g HCl: \ce{Zn + 2HCl -> ZnCl2 + H2}. Calculate the mass of \ce{H2} produced. In an experiment, if 6 g \ce{H2} is obtained, calculate percentage yield. (Atomic masses: Zn = 65.38 g/mol, H = 1 g/mol, Cl = 35.5 g/mol)

Solution

Given

m_{\ce{Zn}} = 300\,\mathrm{g}
M_{\ce{Zn}} = 65.38\,\mathrm{g\,mol^{-1}}
m_{\ce{HCl}} = 300\,\mathrm{g}
M_{\ce{HCl}} = 1+35.5=36.5\,\mathrm{g\,mol^{-1}}
M_{\ce{H2}} = 2\,\mathrm{g\,mol^{-1}}
\text{actual yield} = 6\,\mathrm{g} of \ce{H2}

To find

mass of \ce{H2} produced (theoretical yield)
percentage yield

\ce{Zn + 2HCl -> ZnCl2 + H2}

Step 1: find the limiting reactant

Moles of each reactant.
n_{\ce{Zn}}=\dfrac{300}{65.38}=4.59\,\mathrm{mol}\qquad n_{\ce{HCl}}=\dfrac{300}{36.5}=8.22\,\mathrm{mol}
Moles of \ce{H2} that each reactant could give (1 Zn : 1 \ce{H2} and 2 HCl : 1 \ce{H2}).
\text{from Zn: }4.59\,\mathrm{mol}\qquad\text{from HCl: }\dfrac{8.22}{2}=4.11\,\mathrm{mol}
HCl gives the smaller amount, so HCl is the limiting reactant and Zn is in excess.
\ce{HCl}\ \text{limiting}

Step 2: mass of hydrogen produced (theoretical yield)

Moles of \ce{H2} formed equal those from the limiting reactant.
n_{\ce{H2}}=4.11\,\mathrm{mol}
Mass of \ce{H2}.
m_{\ce{H2}}=4.11\times2=8.22\,\mathrm{g}

Step 3: percentage yield

Percent yield is the actual yield divided by the theoretical yield, times 100. Both are in grams.
\text{Percent yield}=\dfrac{\text{actual yield}}{\text{theoretical yield}}\times100
Substitute the actual yield of 6\,\mathrm{g}.
\dfrac{6}{8.22}\times100=73.0\,\%

The percent yield is below 100%, as it must be for a real experiment.

m_{\ce{H2}}=8.22\,\mathrm{g};\quad \text{percent yield}=73.0\,\%

Exam focus

Limiting reactant numericals: find the limiting reactant, the mass of product and the mass of excess reactant left. (numerical)
A 6-mark question in both 2026 sets (Set B Q.3 OR, Set D Q.3 OR) and a 7-mark question in 2025 Set D Q.6; book numerical questions 4, 12 and 15.
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Percent yield from the theoretical and actual yield, after finding the limiting reactant. (numerical)
2025 Set D Q.6 (7 marks, Zn and HCl); book section 4.3 examples.
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Mass of a product from the mass of another substance using the mole ratio of the balanced equation. (numerical)
Both 2026 sets, Q.2 (iv) OR (mass of \ce{SO2} from 155 g of Cu); same as book numerical question 13.
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Molar volume calculations at STP: volume from mass or number of molecules, and mass from volume. (numerical)
2026 Set B Q.2 (xii) OR and Set D Q.2 (xii) OR, 2025 Set B Q.2 (xi) and Set D Q.2 (iii), and 2025 Set D MCQ 3.
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Number of atoms, molecules or ions from moles, mass or gas volume (including the ions of a salt). (mcq)
2026 Set B MCQ 1, Set D MCQ 1 and MCQ 6, 2025 Set B MCQ 4; book exercise MCQs 5 and 8 and numerical question 3.
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State three differences between limiting and non-limiting reactants, and why the limiting reactant is not always the smaller amount. (short)
2025 Set B Q.2 (iv); book short question iv.
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What is molar volume, and why does it apply to gases? Compare the molar volumes of different gases. (long)
2026 Set B MCQ 6 and Set D MCQ 6, and 2025 Set D Q.2 (iii) (definition); book long questions ii and iii.
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Calculate the molar mass of a gas from its density at STP, and solution stoichiometry using molarity. (numerical)
Not asked in the 2025 or 2026 papers; book sections 4.1.2 and 4.1.6 and numerical question 8.
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