CHEMICAL CALCULATIONS
Relative Atomic Mass (Aᵣ) and Relative Molecular Mass (Mᵣ)
1. Relative Atomic Mass (Aᵣ): The weighted average mass of the isotopes of an element
compared to 1/12 of the mass of a carbon-12 atom.
2. Relative Formula Mass / Relative Molecular Mass (Mᵣ): Sum of the relative atomic
masses of the atoms in a formula / molecule.
3. Example: Mᵣ of H₂O = (2 x Aᵣ of H) + (Aᵣ of O) = (2 x 1) + 16 = 18
4. Key points:
Use the periodic table to find Aᵣ values.
Mᵣ applies to molecules and compounds, Aᵣ applies to atoms.
Question 1
Calculate the relative molecular mass (Mᵣ) of the following compounds:
a) CO₂
Aᵣ of C = 12, Aᵣ of O = 16
Mᵣ of CO₂ = (1 × 12) + (2 × 16) = 12 + 32 = 44
b) H₂SO₄
Aᵣ of H = 1, S = 32, O = 16
Mᵣ of H₂SO₄ = (2 × 1) + (1 × 32) + (4 × 16) = 2 + 32 + 64 = 98
c) NH₃
Aᵣ of N = 14, H = 1
Mᵣ of NH₃ = (1 × 14) + (3 × 1) = 14 + 3 = 17
d) Na₂CO₃
Aᵣ of Na = 23, C = 12, O = 16
Mᵣ of Na₂CO₃ = (2 × 23) + (1 × 12) + (3 × 16) = 46 + 12 + 48 = 106
Percentage composition refers to the percentage by mass of an element in a compound.
Steps to Calculate Percentage Composition:
1. Find the molar mass of the compound.
2. Calculate the total mass of the element in the compound.
3. Divide the total mass of the element by the molar mass of the compound.
4. Multiply by 100 to get the percentage.
Question 3
Find the percentage composition of:
i. hydrogen in (NH4)2SO4
molar mass of (NH4)2SO4 = (2 × 14) + (4 × 2 × 1) + (1 × 32) + (4 × 16) = 132
mass of hydrogen in (NH4)2SO4 = 4 × 2 × 1 = 8
percentage of hydrogen in (NH4)2SO4 = (8 ÷ 132) × 100 = 6.06 %
ii. oxygen in CaCO3
molar mass of CaCO3 = (1 × 40) + (1 × 12) + (3 × 16) = 100
mass of oxygen in CaCO3 = 3 × 16 = 48
percentage of oxygen in CaCO3 = (48 ÷ 100) × 100 = 48 %
iii. carbon in C6H12O6
molar mass of C6H12O6 = (6 × 12) + (12 × 1) + (6 × 16) = 180
mass of carbon in C6H12O6 = 6 × 12 = 72
percentage of carbon in C6H12O6 = (72 ÷ 180) × 100 = 40 %
The Mole Concept
1. 1 mole of a substance contains Avogadro's number of particles (6.02 x 10²³ particles).
2. Molar Mass (g/mol): The mass of one mole of a substance. 1 mole of any substance
weighs as much in grams as its relative atomic/molecular mass.
3. Example: Molar mass of NaCl = Mᵣ of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol
4. Formula: Number of moles (mol) = Mass (g) / Molar Mass (g/mol)
5. Key points:
Know how to rearrange the formula based on the given values.
Question 2
a) Calculate the number of moles in 18 g of water (H₂O).
Moles of H₂O = Mass / Molar Mass = 18 g / 18 g/mol = 1 mole
b) Given that the relative atomic mass of magnesium is 24, how many grams of
magnesium are there in 2 moles of magnesium atoms?
Molar mass of Mg = 24 g/mol
Mass = Moles × Molar Mass
Mass = 2 moles × 24 g/mol = 48 g
c) Calculate the number of moles in:
i. 40 g of NaOH
Mᵣ of NaOH = 23 + 16 + 1 = 40 g/mol
Moles = Mass / Molar Mass = 40 g / 40 g/mol = 1 mole
ii. 56 g of N₂
Mᵣ of N₂ = 2 × 14 = 28 g/mol
Moles = 56 g / 28 g/mol = 2 moles
d) How many molecules are there in 2.5 moles of H₂O?
1 mole contains 6.02 × 10²³ molecules.
Molecules = 2.5 moles × 6.02 × 10²³ molecules/mol = 1.505 × 10²⁴ molecules
e) What is the mass of 0.75 moles of calcium carbonate (CaCO₃)?
Mᵣ of CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol
Mass = Moles × Molar Mass = 0.75 mol × 100 g/mol = 75 g
Empirical Formula is the simplest whole number ratio of atoms in a compound.
Steps to calculate:
1. Find the percentage and mass of each element.
2. Find the molar mass of each element.
3. Find the moles of each element.
4. Divide each by the smallest number of moles to find the ratio.
Question 4 (a)
i. A compound contains 40% sulfur and 60% oxygen. Find the empirical formula.
Assuming 100 g, mass of S = 40 g, mass of O = 60 g.
Moles of S = 40 g / 32 g/mol = 1.25 mol, Moles of O = 60 g / 16 g/mol = 3.75 mol
Ratio of S = 1.25 / 1.25 = 1, Ratio of O = 3.75 / 1.25 = 3
Ratio: S:O = 1:3 → Empirical formula: SO₃
ii. Compound contains 44% phosphorus. Only other element is oxygen. Find empirical
formula.
Percentage of oxygen = 100 – 44 = 56%
Assuming 100g, mass of P = 44 g, mass of O = 56 g
Molar mass of P = 31 g/mol, molar mass of O = 16 g/mol
Moles of P = 44 / 31 = 1.42 mol, moles of O = 56 / 16 = 3.5 mol
Ratio of P = 1.42 / 1.42 = 1, ratio of O = 3.5 / 1.42 = 2.5
(Multiply both by 2 to get a whole number ratio)
P = 1 × 2 = 2, O = 2.5 × 2 = 5
Ratio (P:O) = 2:5 → Empirical formula: P2O5
iii. A sample of antifreeze has the composition by mass: 38.7% carbon, 9.7% hydrogen,
51.6% oxygen. Calculate its empirical formula.
1. In 100g, mass of C = 38.7 g, mass of H = 9.7 g mass of O = 51.6 g
2. Molar mass: C = 12 g/mol, H = 1 g/mol O = 16 g/mol
3. Moles: C = 38.7 / 12 = 3.225 mol, H = 9.7 / 1 = 9.7 mol O = 51.6 / 16 = 3.225 mol
4. Ratios: C = 3.225 / 3.225 = 1, H = 9.7 / 3.225 = 2, O = 3.225 / 3.225 = 1
5. Ratio (C:H:O) = 1:2:1 → Empirical formula: CH2O
iv. If 5 g of hydrated copper (II) sulphate crystals are heated to drive off water of
crystallization, the remaining solid has mass of 3.2 g. Calculate formula of the crystals.
CuSO4 H2O
Mass 3.2 g 5 – 3.2 = 1.8 g
Molar mass (1 × 64) + (1 × 32) + (4 × 16) = 160 (2 × 1) + (1 × 16) = 18
Number of moles 3.2 / 160 = 0.02 mol 1.8 / 18 = 0.1 mol
Simplest ratio 0.02 / 0.02 = 1 1.8 / 0.02 = 5
Formula CuSO4•5H2O
v. One of the ores of copper is the mineral chalcopyrite. A laboratory analysis showed that
15.15 g of chalcopyrite had the following composition: copper 5.27 g and iron 4.61 g.
sulphur is the only other element present. Calculate the empirical formula of
chalcopyrite.
Cu Fe S
Mass 5.27 g 4.61 g 15.15 – 5.27 – 4.61 = 5.27 g
Molar mass 64 56 32
Moles 5.27 / 64 = 0.082 4.61 / 56 = 0.082 5.27 / 32 = 0.165
Ratio 0.082 / 0.082 = 1 0.082 / 0.082 = 1 0.165 / 0.082 = 2
Formula CuFeS2
Molecular Formula
1. It is the actual number of atoms of each element in a molecule.
2. Formula: Molecular Formula = n x (Empirical Formula)
3. Find n by dividing Mᵣ (molecular mass) by the empirical formula mass.
Question 3 (b)
The molecular mass of the compound is 160 g/mol. Determine the molecular formula of the
compound.
Aᵣ of S = 32; Aᵣ of O = 16
Empirical Formula Mass (E.F.M.) of SO₃ = 32 + (3 × 16) = 80 g/mol
n = Molecular mass / Empirical formula mass = 160 g/mol / 80 g/mol = 2
Molecular formula = (SO₃) × 2 = S₂O₆
Question 4
a) A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen by mass.
Determine its empirical formula.
Assume 100 g of compound: C = 40 g, H = 6.67 g, O = 53.33 g
Moles of C = 40 g / 12 g/mol = 3.33 moles
Moles of H = 6.67 g / 1 g/mol = 6.67 moles
Moles of O = 53.33 g / 16 g/mol = 3.33 moles
Ratio: C:H:O (divide all by 3.33) = 3.33:6.67:3.33 = 1:2:1
Empirical formula = CH₂O
b) The empirical formula of a compound is CH₂. Its molecular mass is 56 g/mol. Determine
the molecular formula.
E.F.M. of CH₂ = (1 × 12) + (2 × 1) = 14 g/mol
n = Molecular mass / Empirical formula mass
n = 56 g/mol / 14 g/mol = 4
Molecular formula = 4 × (CH₂)₄ = C₄H₈
c) A compound with an empirical formula of P₂O₅ has a molecular mass of 284 g/mol. Find
the molecular formula.
E.F.M. of P₂O₅ = (2 × 31) + (5 × 16) = 142 g/mol
n = (Molecular mass / Empirical formula mass)
n = 284 g/mol / 142 g/mol = 2
Molecular formula = 2 × (P₂O₅) = P₄O₁₀
Chemical Equations and Stoichiometry
1. Write the balanced chemical equation.
2. Calculate moles of known substance.
3. Use mole ratios from the equation to find moles of other substances.
Question 5
a) 2H₂ + O₂ → 2H₂O
How many moles of H₂O are produced when 4 moles of H₂ react?
According to the balanced equation, 2 moles of H₂ produce 2 moles of H₂O.
Therefore, 4 moles of H₂ produce 4 moles of H₂O.
b) In the reaction 2Al + 3Cl₂ → 2AlCl₃, calculate the mass of AlCl₃ produced when 54 g of
aluminium reacts completely.
Moles of Al = 54 g / 27 g/mol = 2 moles
According to the balanced equation, 2 moles of Al produce 2 moles of AlCl₃.
Molar mass of AlCl₃ = 27 + (3 × 35.5) = 133.5 g/mol
Mass of AlCl₃ = 2 moles × 133.5 g/mol = 267 g
c) How many moles of oxygen gas are required to completely react with 8 moles of
hydrogen gas in the reaction 2H₂ + O₂ → 2H₂O?
According to the equation, 2 moles of H₂ react with 1 mole of O₂.
Therefore, 8 moles of H₂ react with 8/2 = 4 moles of O₂.
Mass Calculations in Reactions
1. Calculate moles of the given substance.
2. Use mole ratios to find moles of the required substance.
3. Convert moles to mass.
4. Formula: Mass = Moles × Molar Mass
Question 6
a) Calculate the mass of magnesium oxide produced when 12 g of magnesium reacts with
oxygen. (2Mg + O₂ → 2MgO)
Moles of Mg = 12 g / 24 g/mol = 0.5 mol
Mole ratio (Mg: MgO) = 1:1 → Moles of MgO = 0.5 mol
Mass of MgO = 0.5 mol × 40 g/mol = 20 g
b) Calculate the mass of calcium oxide (CaO) produced when 20 g of calcium carbonate
(CaCO₃) decomposes. CaCO₃ → CaO + CO₂
Mᵣ of CaCO₃ = 40 + 12 + 48 = 100 g/mol
Moles of CaCO₃ = 20 g / 100 g/mol = 0.2 moles
According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CaO.
Molar mass of CaO = 40 + 16 = 56 g/mol
Mass of CaO = 0.2 moles × 56 g/mol = 11.2 g
c) How many grams of iron are needed to react with 16 g of oxygen in the reaction:
4Fe + 3O2→ 2Fe2O3
Mᵣ of O₂ = 2 × 16 = 32 g/mol
Moles of O₂ = 16 g / 32 g/mol = 0.5 moles
According to the equation, 3 moles of O₂ react with 4 moles of Fe.
Moles of Fe required = (4/3) × 0.5 moles = 0.67 moles
Molar mass of Fe = 56 g/mol
Mass of Fe = 0.67 moles × 56 g/mol = 37.5 g
d) If 5.0 g of zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas, how
many grams of hydrogen are produced? (Zn + 2HCl → ZnCl₂ + H₂)
Mᵣ of Zn = 65.4 g/mol
Moles of Zn = 5.0 g / 65.4 g/mol = 0.076 moles
According to the balanced equation, 1 mole of Zn produces 1 mole of H₂.
Moles of H₂ produced = 0.076 moles
Molar mass of H₂ = 2 g/mol
Mass of H₂ = 0.076 moles × 2 g/mol = 0.152 g
Gas Volumes and Molar Volume (at r.t.p.)
1. Molar Volume: 1 mole of any gas occupies 24 dm³ at room temperature and pressure
(r.t.p.).
2. Volume of gas = Moles of gas × 24 dm³
Question 7
1. Calculate the volume of CO₂ produced when 2 moles of CaCO₃ decompose
(CaCO₃ → CaO + CO₂).
1 mole of CaCO₃ produces 1 mole of CO₂.
Therefore, 2 moles of CaCO₃ produce 2 moles of CO₂.
Volume of CO₂ = 2 mol × 24 dm³ = 48 dm³
2. Calculate the volume of carbon dioxide produced when 1 mole of CaCO₃ decomposes at
room temperature and pressure (CaCO₃ → CaO + CO₂).
1 mole of CaCO₃ produces 1 mole of CO₂.
At r.t.p., 1 mole of gas occupies 24 dm³.
Volume of CO₂ = 24 dm³
3. What volume of oxygen gas (in dm³) is required to react with 4 moles of hydrogen gas
(H₂) at room temperature and pressure?
2H₂ + O₂ → 2H₂O
2 moles of H₂ react with 1 mole of O₂.
Therefore, 4 moles of H₂ react with 2 moles of O₂.
Volume of O₂ = 2 moles × 24 dm³/mol = 48 dm³
4. A sample of nitrogen gas occupies 48 dm³ at room temperature and pressure. How
many moles of nitrogen gas are present?
At r.t.p., 1 mole of gas occupies 24 dm³.
Moles of N₂ = Volume / 24 dm³/mol = 48 dm³ / 24 dm³/mol = 2 moles
End of Chapter Questions
1. A compound contains 70% iron and 30% oxygen by mass. Determine its empirical
formula.
Assume 100 g of the compound: Fe = 70 g, O = 30 g
Moles of Fe = 70 g / 56 g/mol = 1.25 moles
Moles of O = 30 g / 16 g/mol = 1.875 moles
Simplify ratio: Fe : O (divide both by 1.25) = 1.25:1.875 = 2:3
Empirical formula = Fe₂O₃
2. In the reaction:
If 14 g of nitrogen gas reacts, what mass of ammonia (NH₃) is formed?
Mᵣ of N₂ = 28 g/mol
Moles of N₂ = 14 g / 28 g/mol = 0.5 moles
According to the equation, 1 mole of N₂ produces 2 moles of NH₃.
Moles of NH₃ = 0.5 moles × 2 = 1 mole
Mᵣ of NH₃ = 14 + (3 × 1) = 17 g/mol
Mass of NH₃ = 1 mole × 17 g/mol = 17 g
3. The empirical formula of a compound is CH₂. If its molecular mass is 112 g/mol,
calculate its molecular formula.
E.F.M. of CH₂ = 14 g/mol
Molecular mass / Empirical formula mass = 112 g/mol / 14 g/mol = 8
Molecular formula = (CH₂)₈ = C₈H₁₆
4. How many grams of chlorine are needed to react completely with 10 g of sodium in the
reaction:
Mᵣ of Na = 23 g/mol
Moles of Na = 10 g / 23 g/mol = 0.435 moles
According to the equation, 2 moles of Na react with 1 mole of Cl₂.
Moles of Cl₂ = 0.435 moles / 2 = 0.2175 moles
Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol
Mass of Cl₂ = 0.2175 moles × 71 g/mol = 15.44 g
5. Calculate the volume of oxygen gas at room temperature and pressure required to react
with 4 moles of hydrogen gas in the reaction:
2 moles of H₂ react with 1 mole of O₂.
Therefore, 4 moles of H₂ react with 2 moles of O₂.
Volume of O₂ = 2 moles × 24 dm³/mol = 48 dm³
6. What is the mass of 0.5 moles of Na₂SO₄?
Mᵣ of Na₂SO₄ = (2 × 23) + 32 + (4 × 16) = 46 + 32 + 64 = 142 g/mol
Mass = Moles × Molar Mass = 0.5 moles × 142 g/mol = 71 g
