Chemistry - Stoichiometry and Mole Concept

Relative Atomic and Molecular Mass

Relative Atomic Mass ($A_r$)

The relative atomic mass of an element is the weighted average mass of one atom of the element relative to $1/12$ the mass of one atom of carbon-12.

Relative Molecular Mass ($M_r$)

The relative molecular mass of a compound is the sum of the relative atomic masses of all atoms in one molecule.

$M_r = \sum A_r \mathrm{ of all atoms in the formula}$

Worked Example 1

Calculate the relative molecular mass of sulfuric acid ($\mathrm{H_2SO_4}$).

Solution

$M_r(\mathrm{H_2SO_4}) = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98$

---

The Mole Concept

A mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, or other particles) as there are atoms in exactly $12 \mathrm{ g}$ of carbon-12.

$n = \frac{N}{N_A}$

where:

• $n$ = number of moles
• $N$ = number of particles
• $N_A = 6.02 \times 10^{23} \mathrm{ mol^{-1}}$ (Avogadro's number)

Molar Mass ($M$)

The molar mass is the mass of one mole of a substance. It is numerically equal to $A_r$ or $M_r$ but has units of $\mathrm{g/mol}$.

$n = \frac{m}{M}$

where $m$ is the mass in grams and $M$ is the molar mass in $\mathrm{g/mol}$.

Worked Example 2

How many moles are there in $24.5 \mathrm{ g}$ of $\mathrm{H_2SO_4}$?

Solution

$M(\mathrm{H_2SO_4}) = 98 \mathrm{ g/mol}$

$n = \frac{m}{M} = \frac{24.5}{98} = 0.250 \mathrm{ mol}$

How many molecules does this correspond to?

$N = n \times N_A = 0.250 \times 6.02 \times 10^{23} = 1.51 \times 10^{23} \mathrm{ molecules}$

---

Molar Volume of Gas

At a given temperature and pressure, one mole of any ideal gas occupies the same volume.

Condition	Temperature	Pressure	Molar Volume
STP	$273 \mathrm{ K}$ ($0^\circ\mathrm{C}$)	$1.01 \times 10^5 \mathrm{ Pa}$	$22.4 \mathrm{ dm^3/mol}$
RTP	$298 \mathrm{ K}$ ($25^\circ\mathrm{C}$)	$1.01 \times 10^5 \mathrm{ Pa}$	$24.0 \mathrm{ dm^3/mol}$

$n = \frac{V}{V_m}$

where $V$ is the volume of gas and $V_m$ is the molar volume.

Worked Example 3

Calculate the volume occupied by $5.00 \mathrm{ g}$ of carbon dioxide at RTP.

Solution

$n = \frac{m}{M} = \frac{5.00}{44} = 0.1136 \mathrm{ mol}$

$V = n \times V_m = 0.1136 \times 24.0 = 2.73 \mathrm{ dm^3}$

---

Empirical and Molecular Formulae

Empirical Formula

The empirical formula gives the simplest whole-number ratio of atoms in a compound.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in one molecule. It is a whole-number multiple of the empirical formula:

$\mathrm{Molecular formula} = (\mathrm{Empirical formula})_n$

Determining Empirical Formula from Percentage Composition

1. Assume $100 \mathrm{ g}$ of compound.
2. Convert each percentage to mass in grams.
3. Convert mass to moles (divide by $A_r$).
4. Divide all mole values by the smallest mole value.
5. Round to the nearest whole numbers (multiply if necessary).

Worked Example 4

A compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. Determine the empirical formula.

Solution

Element	Mass (g)	Moles ($m/A_r$)	Ratio (divide by smallest)
C	40.0	$40.0/12 = 3.33$	$3.33/3.33 = 1$
H	6.7	$6.7/1 = 6.7$	$6.7/3.33 \approx 2$
O	53.3	$53.3/16 = 3.33$	$3.33/3.33 = 1$

Empirical formula: $\mathrm{CH_2O}$

If the molar mass of the compound is $180 \mathrm{ g/mol}$, find the molecular formula:

$M_r(\mathrm{empirical}) = 12 + 2 + 16 = 30$

$n = \frac{180}{30} = 6$

Molecular formula: $\mathrm{C_6H_{12}O_6}$ (glucose).

---

Percentage Composition

$\%\mathrm{ element} = \frac{n \times A_r \mathrm{ of element}}{M_r \mathrm{ of compound}} \times 100$

Worked Example 5

Calculate the percentage by mass of nitrogen in ammonium nitrate ($\mathrm{NH_4NO_3}$).

Solution

$M_r(\mathrm{NH_4NO_3}) = 14 + 4(1) + 14 + 3(16) = 14 + 4 + 14 + 48 = 80$

$\%\mathrm{N} = \frac{2 \times 14}{80} \times 100 = \frac{28}{80} \times 100 = 35.0\%$

---

Reacting Mass Calculations

The stoichiometric coefficients in a balanced chemical equation give the molar ratio of reactants and products.

Steps

1. Write the balanced chemical equation.
2. Convert the given mass to moles.
3. Use the molar ratio from the equation to find moles of the unknown.
4. Convert moles back to mass.

Worked Example 6

What mass of magnesium oxide is produced when $12.0 \mathrm{ g}$ of magnesium reacts completely with oxygen?

Solution

$2\mathrm{Mg} + \mathrm{O_2} \to 2\mathrm{MgO}$

Step 1: Moles of Mg:

$n(\mathrm{Mg}) = \frac{12.0}{24.3} = 0.494 \mathrm{ mol}$

Step 2: Molar ratio $\mathrm{Mg} : \mathrm{MgO} = 2 : 2 = 1 : 1$

$n(\mathrm{MgO}) = 0.494 \mathrm{ mol}$

Step 3: Mass of MgO:

$m(\mathrm{MgO}) = 0.494 \times (24.3 + 16) = 0.494 \times 40.3 = 19.9 \mathrm{ g}$

---

Gas Volume Calculations

When one or more reactants/products are gases, use the molar volume at the relevant conditions.

Worked Example 7

Calculate the volume of carbon dioxide produced (at RTP) when $25.0 \mathrm{ g}$ of calcium carbonate decomposes on heating.

Solution

$\mathrm{CaCO_3} \xrightarrow{\Delta} \mathrm{CaO} + \mathrm{CO_2}$

$n(\mathrm{CaCO_3}) = \frac{25.0}{100} = 0.250 \mathrm{ mol}$

Molar ratio: $1 : 1$, so $n(\mathrm{CO_2}) = 0.250 \mathrm{ mol}$

$V(\mathrm{CO_2}) = 0.250 \times 24.0 = 6.00 \mathrm{ dm^3}$

---

Concentration (Molarity)

Molar Concentration

$c = \frac{n}{V}$

where:

• $c$ = concentration ($\mathrm{mol/dm^3}$)
• $n$ = number of moles (mol)
• $V$ = volume ($\mathrm{dm^3}$)

Worked Example 8

What is the concentration of a solution prepared by dissolving $10.6 \mathrm{ g}$ of sodium carbonate ($\mathrm{Na_2CO_3}$) in water and making up to $250 \mathrm{ cm^3}$?

Solution

$M(\mathrm{Na_2CO_3}) = 2(23) + 12 + 3(16) = 106 \mathrm{ g/mol}$

$n = \frac{10.6}{106} = 0.100 \mathrm{ mol}$

$V = 250 \mathrm{ cm^3} = 0.250 \mathrm{ dm^3}$

$c = \frac{0.100}{0.250} = 0.400 \mathrm{ mol/dm^3}$

Mass Concentration

$\rho = \frac{m}{V}$

where $\rho$ is the mass concentration ($\mathrm{g/dm^3}$).

Converting Between Concentrations

$c = \frac{\rho}{M}$

---

Titration Calculations

A titration is a technique used to determine the concentration of a solution by reacting it with a solution of known concentration.

Steps

1. Write the balanced equation for the reaction.
2. Calculate moles of the known solution from its concentration and volume.
3. Use the molar ratio to find moles of the unknown.
4. Calculate the concentration of the unknown.

Worked Example 9

$25.0 \mathrm{ cm^3}$ of sodium hydroxide solution is titrated with $0.100 \mathrm{ mol/dm^3}$ hydrochloric acid. The average titre is $21.5 \mathrm{ cm^3}$. Calculate the concentration of the sodium hydroxide solution.

Solution

$\mathrm{NaOH} + \mathrm{HCl} \to \mathrm{NaCl} + \mathrm{H_2O}$

Molar ratio: $1 : 1$

$n(\mathrm{HCl}) = c \times V = 0.100 \times \frac{21.5}{1000} = 2.15 \times 10^{-3} \mathrm{ mol}$

$n(\mathrm{NaOH}) = 2.15 \times 10^{-3} \mathrm{ mol}$

$c(\mathrm{NaOH}) = \frac{n}{V} = \frac{2.15 \times 10^{-3}}{25.0/1000} = \frac{2.15 \times 10^{-3}}{0.0250} = 0.0860 \mathrm{ mol/dm^3}$

Worked Example 10

$20.0 \mathrm{ cm^3}$ of sulphuric acid is titrated with $0.0500 \mathrm{ mol/dm^3}$ sodium hydroxide. The average titre is $30.0 \mathrm{ cm^3}$. Find the concentration of the sulphuric acid.

Solution

$\mathrm{H_2SO_4} + 2\mathrm{NaOH} \to \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$

Molar ratio: $\mathrm{H_2SO_4} : \mathrm{NaOH} = 1 : 2$

$n(\mathrm{NaOH}) = 0.0500 \times \frac{30.0}{1000} = 1.50 \times 10^{-3} \mathrm{ mol}$

$n(\mathrm{H_2SO_4}) = \frac{1.50 \times 10^{-3}}{2} = 7.50 \times 10^{-4} \mathrm{ mol}$

$c(\mathrm{H_2SO_4}) = \frac{7.50 \times 10^{-4}}{20.0/1000} = \frac{7.50 \times 10^{-4}}{0.0200} = 0.0375 \mathrm{ mol/dm^3}$

---

Limiting Reactants

The limiting reactant is the reactant that is completely consumed first and determines the maximum amount of product that can be formed.

Steps

1. Calculate the moles of each reactant.
2. Divide each by its stoichiometric coefficient.
3. The reactant with the smallest value is the limiting reactant.

Worked Example 11

$5.0 \mathrm{ g}$ of zinc reacts with $25.0 \mathrm{ cm^3}$ of $2.0 \mathrm{ mol/dm^3}$ hydrochloric acid. Identify the limiting reactant and calculate the volume of hydrogen produced at RTP.

Solution

$\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnCl_2} + \mathrm{H_2}$

Moles of Zn: $n = 5.0/65.4 = 0.0765 \mathrm{ mol}$

Moles of HCl: $n = 2.0 \times 25.0/1000 = 0.0500 \mathrm{ mol}$

Divide by coefficients:

• Zn: $0.0765 / 1 = 0.0765$
• HCl: $0.0500 / 2 = 0.0250$

HCl is the limiting reactant.

$n(\mathrm{H_2}) = \frac{n(\mathrm{HCl})}{2} = \frac{0.0500}{2} = 0.0250 \mathrm{ mol}$

$V(\mathrm{H_2}) = 0.0250 \times 24.0 = 0.600 \mathrm{ dm^3} = 600 \mathrm{ cm^3}$

---

Percentage Yield

$\%\mathrm{ yield} = \frac{\mathrm{actual yield}}{\mathrm{theoretical yield}} \times 100$

The theoretical yield is the maximum amount of product calculated from the limiting reactant.

Worked Example 12

$10.0 \mathrm{ g}$ of calcium carbonate is heated, and $4.80 \mathrm{ g}$ of calcium oxide is collected. Calculate the percentage yield.

Solution

$\mathrm{CaCO_3} \to \mathrm{CaO} + \mathrm{CO_2}$

$n(\mathrm{CaCO_3}) = 10.0/100 = 0.100 \mathrm{ mol}$

Theoretical $n(\mathrm{CaO}) = 0.100 \mathrm{ mol}$

Theoretical $m(\mathrm{CaO}) = 0.100 \times 56 = 5.60 \mathrm{ g}$

$\%\mathrm{ yield} = \frac{4.80}{5.60} \times 100 = 85.7\%$

---

Water of Crystallisation

Some ionic compounds contain water molecules as part of their crystal structure. These are called hydrated salts.

$\mathrm{CuSO_4 \cdot 5H_2O}$

The formula indicates 5 moles of water per mole of $\mathrm{CuSO_4}$.

Worked Example 13

$12.5 \mathrm{ g}$ of hydrated copper(II) sulphate ($\mathrm{CuSO_4 \cdot xH_2O}$) is heated to constant mass, leaving $8.00 \mathrm{ g}$ of anhydrous $\mathrm{CuSO_4}$. Find the value of $x$.

Solution

$M(\mathrm{CuSO_4}) = 64 + 32 + 64 = 160 \mathrm{ g/mol}$

Mass of water lost: $12.5 - 8.00 = 4.50 \mathrm{ g}$

$n(\mathrm{H_2O}) = 4.50/18 = 0.250 \mathrm{ mol}$

$n(\mathrm{CuSO_4}) = 8.00/160 = 0.0500 \mathrm{ mol}$

$x = \frac{n(\mathrm{H_2O})}{n(\mathrm{CuSO_4})} = \frac{0.250}{0.0500} = 5$

The formula is $\mathrm{CuSO_4 \cdot 5H_2O}$.

---

Common Pitfalls

• Forgetting to convert $\mathrm{cm^3}$ to $\mathrm{dm^3}$ when calculating concentration or moles from volume ($1 \mathrm{ dm^3} = 1000 \mathrm{ cm^3}$).
• Using the wrong molar volume (22.4 for STP, 24.0 for RTP). Check the conditions carefully.
• Failing to identify the limiting reactant correctly. Always compare the mole ratio, not just the mass.
• Confusing empirical and molecular formulae. The empirical formula is the simplest ratio; the molecular formula is the actual formula.
• Forgetting to use the molar ratio from the balanced equation in stoichiometry calculations.
• Dividing mass by molar mass but using the wrong molar mass (e.g., using atomic mass instead of molecular mass).

---

Summary Table

Concept	Key Equation
Moles from mass	$n = m/M$
Moles from particles	$n = N/N_A$
Moles from gas volume	$n = V/V_m$
Concentration	$c = n/V$
Ideal gas (STP)	$V_m = 22.4 \mathrm{ dm^3/mol}$
Ideal gas (RTP)	$V_m = 24.0 \mathrm{ dm^3/mol}$
Percentage yield	$(\mathrm{actual} / \mathrm{theoretical}) \times 100$

---

Problem Set

Problem 1: Calculate the number of molecules in $3.20 \mathrm{ g}$ of oxygen gas ($\mathrm{O_2}$).

If you get this wrong, revise: The Mole Concept

Solution

$n = \frac{3.20}{32} = 0.100 \mathrm{ mol}$

$N = 0.100 \times 6.02 \times 10^{23} = 6.02 \times 10^{22} \mathrm{ molecules}$

Problem 2: A compound contains $36.5\%$ Na, $25.4\%$ S, and $38.1\%$ O. Find its empirical formula.

If you get this wrong, revise: Empirical and Molecular Formulae

Solution

Element	Mass (g)	Moles	Ratio
Na	36.5	$36.5/23 = 1.587$	$1.587/0.793 = 2$
S	25.4	$25.4/32 = 0.794$	$0.794/0.793 = 1$
O	38.1	$38.1/16 = 2.381$	$2.381/0.793 = 3$

Empirical formula: $\mathrm{Na_2SO_3}$

Problem 3: $2.40 \mathrm{ dm^3}$ of ammonia gas at RTP is dissolved in water to make $500 \mathrm{ cm^3}$ of solution. Calculate the concentration of the ammonia solution.

If you get this wrong, revise: Concentration (Molarity) and Molar Volume of Gas

Solution

$n(\mathrm{NH_3}) = 2.40/24.0 = 0.100 \mathrm{ mol}$

$c = 0.100/0.500 = 0.200 \mathrm{ mol/dm^3}$

Problem 4: $15.0 \mathrm{ g}$ of $\mathrm{NaOH}$ is reacted with excess $\mathrm{H_2SO_4}$. Calculate the mass of $\mathrm{Na_2SO_4}$ produced.

If you get this wrong, revise: Reacting Mass Calculations

Solution

$2\mathrm{NaOH} + \mathrm{H_2SO_4} \to \mathrm{Na_2SO_4} + 2\mathrm{H_2O}$

$n(\mathrm{NaOH}) = 15.0/40 = 0.375 \mathrm{ mol}$

$n(\mathrm{Na_2SO_4}) = 0.375/2 = 0.1875 \mathrm{ mol}$

$m(\mathrm{Na_2SO_4}) = 0.1875 \times 142 = 26.6 \mathrm{ g}$

Problem 5: In a titration, $25.0 \mathrm{ cm^3}$ of $\mathrm{H_2SO_4}$ reacts with $20.0 \mathrm{ cm^3}$ of $0.200 \mathrm{ mol/dm^3}$ $\mathrm{KOH}$. Find the concentration of the acid.

If you get this wrong, revise: Titration Calculations

Solution

$\mathrm{H_2SO_4} + 2\mathrm{KOH} \to \mathrm{K_2SO_4} + 2\mathrm{H_2O}$

$n(\mathrm{KOH}) = 0.200 \times 0.0200 = 4.00 \times 10^{-3} \mathrm{ mol}$

$n(\mathrm{H_2SO_4}) = 4.00 \times 10^{-3}/2 = 2.00 \times 10^{-3} \mathrm{ mol}$

$c(\mathrm{H_2SO_4}) = 2.00 \times 10^{-3}/0.0250 = 0.0800 \mathrm{ mol/dm^3}$

Problem 6: $6.50 \mathrm{ g}$ of $\mathrm{Zn}$ reacts with $50.0 \mathrm{ cm^3}$ of $2.00 \mathrm{ mol/dm^3}$ $\mathrm{HCl}$. Calculate the maximum volume of $\mathrm{H_2}$ produced at RTP.

If you get this wrong, revise: Limiting Reactants and Gas Volume Calculations

Solution

$\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnCl_2} + \mathrm{H_2}$

$n(\mathrm{Zn}) = 6.50/65.4 = 0.0994 \mathrm{ mol}$

$n(\mathrm{HCl}) = 2.00 \times 0.0500 = 0.100 \mathrm{ mol}$

Divide by coefficients: Zn: $0.0994/1 = 0.0994$; HCl: $0.100/2 = 0.0500$

HCl is limiting. $n(\mathrm{H_2}) = 0.0500 \mathrm{ mol}$

$V(\mathrm{H_2}) = 0.0500 \times 24.0 = 1.20 \mathrm{ dm^3}$

Problem 7: Calculate the percentage of oxygen by mass in $\mathrm{CaCO_3}$.

If you get this wrong, revise: Percentage Composition

Solution

$M_r(\mathrm{CaCO_3}) = 40 + 12 + 3(16) = 100$

$\%\mathrm{O} = \frac{3 \times 16}{100} \times 100 = \frac{48}{100} \times 100 = 48.0\%$

Problem 8: A compound has the empirical formula $\mathrm{CH_2O}$ and a molar mass of $150 \mathrm{ g/mol}$. Determine the molecular formula.

If you get this wrong, revise: Empirical and Molecular Formulae

Solution

$M_r(\mathrm{empirical}) = 12 + 2(1) + 16 = 30$

$n = \frac{150}{30} = 5$

Molecular formula: $(\mathrm{CH_2O})_5 = \mathrm{C_5H_{10}O_5}$

Problem 9: $8.00 \mathrm{ g}$ of $\mathrm{Fe_2O_3}$ is reduced by excess carbon monoxide. Calculate the mass of iron produced and the volume of $\mathrm{CO_2}$ formed at RTP.

$\mathrm{Fe_2O_3} + 3\mathrm{CO} \to 2\mathrm{Fe} + 3\mathrm{CO_2}$

If you get this wrong, revise: Reacting Mass Calculations and Gas Volume Calculations

Solution

$n(\mathrm{Fe_2O_3}) = \frac{8.00}{160} = 0.0500 \mathrm{ mol}$

Molar ratio $\mathrm{Fe_2O_3} : \mathrm{Fe} : \mathrm{CO_2} = 1 : 2 : 3$

$n(\mathrm{Fe}) = 2 \times 0.0500 = 0.100 \mathrm{ mol}$

$m(\mathrm{Fe}) = 0.100 \times 55.8 = 5.58 \mathrm{ g}$

$n(\mathrm{CO_2}) = 3 \times 0.0500 = 0.150 \mathrm{ mol}$

$V(\mathrm{CO_2}) = 0.150 \times 24.0 = 3.60 \mathrm{ dm^3}$

Problem 10: $16.2 \mathrm{ g}$ of hydrated magnesium sulphate ($\mathrm{MgSO_4 \cdot xH_2O}$) is heated to constant mass, leaving $7.8 \mathrm{ g}$ of anhydrous $\mathrm{MgSO_4}$. Find $x$.

If you get this wrong, revise: Water of Crystallisation

Solution

$M(\mathrm{MgSO_4}) = 24 + 32 + 4(16) = 120 \mathrm{ g/mol}$

Mass of water lost: $16.2 - 7.8 = 8.4 \mathrm{ g}$

$n(\mathrm{H_2O}) = \frac{8.4}{18} = 0.467 \mathrm{ mol}$

$n(\mathrm{MgSO_4}) = \frac{7.8}{120} = 0.0650 \mathrm{ mol}$

$x = \frac{0.467}{0.0650} = 7.18 \approx 7$

The formula is $\mathrm{MgSO_4 \cdot 7H_2O}$ (Epsom salt).