Chapter 1 — Chemical Calculation

1.1 Why we need a counting unit

• Atoms, molecules and ions are so tiny that we can never count them one by one, and we cannot weigh a single one on a laboratory balance.
• Chemists therefore group them into a very large "package", exactly as a shop sells eggs by the dozen or paper by the ream.
• The package used in chemistry is the mole.

• The amount of a chemical substance is expressed in the International System of Units (SI) in the unit mole (symbol: mol).
• One mole of any substance contains the same fixed number of particles (molecules, atoms, ions, formula units or electrons).
• That fixed number is called Avogadro's number:

Avogadro's number = 6.02 × 10²³ particles per mole

The central relation is therefore:

Number of particles = number of moles × Avogadro's number

or, written with symbols, N = n × 6.02 × 10²³. Rearranged, n = N / (6.02 × 10²³).

Important notes:

• The word "particle" must always be specified. "1 mol of hydrogen" is ambiguous: it may mean 1 mol of H atoms or 1 mol of H₂ molecules. One mole of H₂ molecules contains 6.02 × 10²³ molecules but twice as many atoms, because each molecule has two atoms.
• The mole of a formula unit (an ionic compound such as NaCl) contains 6.02 × 10²³ formula units, and these release a known number of ions when dissolved or melted.

1.2 Moles of atoms inside a molecule or formula unit

• The subscripts in a formula tell you how many moles of each atom are present in one mole of the compound.
• For example, in one mole of methane CH₄ there is 1 mol of C atoms and 4 mol of H atoms.

1.3 Molar mass

• The molar mass is the mass of one mole of a substance.
• It is the gram-relative atomic mass of an atom, or the gram molecular (formula) mass of a molecule, and is measured in g/mol.
• It equals the sum of the gram atomic masses of all the atoms forming the molecule or formula unit.

Worked example 1.3 — simple compounds. [H = 1, O = 16, C = 12, Ca = 40]

• Water, H₂O = (2 × 1) + 16 = 18 g/mol.
• Carbon dioxide, CO₂ = 12 + (2 × 16) = 44 g/mol.
• Calcium carbonate, CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol.

Worked example 1.4 — compounds with brackets. Ammonium sulphate (NH₄)₂SO₄ [N = 14, H = 1, S = 32, O = 16]

• The bracket (NH₄) occurs twice: 2 × (14 + 4 × 1) = 36.
• Add S and O₄: 32 + 64 = 96.
• Molar mass = 36 + 96 = 132 g/mol.

Worked example 1.5 — hydrated salts. In a hydrated crystal the water of crystallisation is counted as part of the formula. Magnesium sulphate heptahydrate MgSO₄·7H₂O [Mg = 24, S = 32, O = 16, H = 1]:

• MgSO₄ = 24 + 32 + 64 = 120 g/mol.
• 7H₂O = 7 × 18 = 126 g/mol.
• Molar mass = 120 + 126 = 246 g/mol.

Likewise Na₂CO₃·10H₂O [Na = 23, C = 12, O = 16, H = 1] = 106 + 180 = 286 g/mol. (Compare the book's CuSO₄·5H₂O = 249.5 g/mol, calculated in the same way.)

1.4 The mass–mole relation

Because the molar mass is the mass of one mole, the number of moles in any sample follows from its mass:

Number of moles (mol) = mass of the substance (g) / molar mass (g/mol)

and the reverse, mass = n × molar mass.

Combining the two relations lets us move between mass, moles and number of particles:

mass (g) ⟷ (÷ or × molar mass) ⟷ moles ⟷ (× or ÷ 6.02 × 10²³) ⟷ number of particles

Worked example 1.6 — a full chain (sodium carbonate pattern). A sample of sodium sulphate Na₂SO₄ has a mass of 71 g [Na = 23, S = 32, O = 16]. Calculate (a) the number of moles, (b) the number of moles of sodium ions, (c) the number of sodium ions released when the sample dissolves in water.

(a) Molar mass = 2 × 23 + 32 + 64 = 142 g/mol. n = 71 / 142 = 0.5 mol.

(b) Na₂SO₄ → 2Na⁺ + SO₄²⁻ (in water). One mole gives 2 mol Na⁺, so moles of Na⁺ = 2 × 0.5 = 1 mol.

(c) Number of Na⁺ ions = 1 × 6.02 × 10²³ = 6.02 × 10²³ ions.

Worked example 1.7 — ions in sodium chloride. How many ions are produced when 5.85 g of NaCl dissolve in water? [Na = 23, Cl = 35.5]

• Molar mass NaCl = 58.5 g/mol; n = 5.85 / 58.5 = 0.1 mol.
• NaCl → Na⁺ + Cl⁻: 1 mol gives 2 mol of ions, so 0.2 mol of ions.
• Number of ions = 0.2 × 6.02 × 10²³ = 1.204 × 10²³ ions.

2.1 Avogadro's hypothesis

• Solids and liquids of one mole have different volumes, because their particles are packed close together and differ in size.
• Gas particles are very far apart compared to their own size, so the volume of a gas depends almost only on the pressure and temperature, not on what the gas is.
• This is the content of Avogadro's hypothesis:

Equal volumes of different gases, under the same conditions of temperature and pressure, contain the same number of molecules.

Consequently, one mole of any gas occupies the same volume under the same conditions. Measurements showed:

• At STP (standard temperature and pressure: 0 °C = 273 K and 1 atm = 760 mmHg), one mole of any gas occupies 22.4 L.
• At rtp (room temperature and pressure: 25 °C = 298 K and 1 atm), one mole of any gas occupies 24 L.

The two values differ because a gas expands when warmed: at the higher temperature of rtp the same number of molecules needs more space.

2.2 Comparing gases

• The molecules differ in mass, but a mole of each takes the same volume.
• Hence 22.4 L of hydrogen weighs 2 g whereas 22.4 L of oxygen weighs 32 g (both at STP).

2.3 The working relations

Volume of gas (L) = number of moles × molar volume (22.4 L at STP; 24 L at rtp)

Number of moles = volume (L) / 22.4 (at STP) or volume (L) / 24 (at rtp)

Number of molecules = (V / 22.4) × 6.02 × 10²³.

Worked example 2.1. Volume of 8.8 g of carbon dioxide at (a) STP and (b) rtp. [C = 12, O = 16]

• M(CO₂) = 44 g/mol; n = 8.8 / 44 = 0.2 mol.
• (a) V = 0.2 × 22.4 = 4.48 L at STP.
• (b) V = 0.2 × 24 = 4.8 L at rtp.

Worked example 2.2. Mass of 11.2 L of oxygen measured at STP. [O = 16]

• n = 11.2 / 22.4 = 0.5 mol; M(O₂) = 32 g/mol.
• Mass = 0.5 × 32 = 16 g.

2.4 Notes

• The molar volume 22.4 L applies only to gases. Never use it for liquid water or solid sodium.
• Always check the condition in the question: STP gives 22.4, rtp gives 24.
• Volumes of gases that react or form can be compared directly in the ratio of the coefficients, as we see in Lesson 3.

3.1 What a balanced equation tells us

• A balanced symbolic equation fixes the proportions in which substances react and form.
• The coefficients give moles (and, for gases, volumes in the same ratio, and particles in the same ratio).
• From it we can calculate:

1. masses of substances,
2. numbers of moles,
3. numbers of particles,
4. volumes of gases,
5. the limiting reactant (Lesson 4).

For example 2H₂O(l) → 2H₂(g) + O₂(g) (electrolysis of acidified water) states: 2 mol of water give 2 mol of hydrogen and 1 mol of oxygen; in terms of mass, 36 g of water give 4 g of H₂ and 32 g of O₂ (conservation of mass: 36 = 4 + 32); in terms of particles, 2 molecules of water give 2 molecules of H₂ and 1 of O₂.

3.2 Method

1. Write and balance the equation.
2. Under the substances in question write the amount from the equation (moles, or mass = coefficient × molar mass, or volume = coefficient × 22.4 L).
3. Write the given amount above the known substance and "?" above the required one.
4. Cross-multiply. Convert units at the end.

Conservation of mass guarantees that the sum of reactant masses equals the sum of product masses; use this as a check.

Worked example 3.1 — mass from mass. Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide formed from 6 g of magnesium. [Mg = 24, O = 16]

• n(Mg) = 6 / 24 = 0.25 mol.
• 2 mol Mg → 2 mol MgO, so n(MgO) = 0.25 mol.
• M(MgO) = 40 g/mol; mass = 0.25 × 40 = 10 g.
• Check: oxygen that combined = 10 − 6 = 4 g = 0.125 mol O₂, which is half of 0.25 mol, as the equation requires.

Worked example 3.2 — gas volume from mass. Thermal decomposition of limestone: CaCO₃(s) → CaO(s) + CO₂(g) (on heating). Volume of CO₂ at STP from 50 g of CaCO₃? [Ca = 40, C = 12, O = 16]

• M(CaCO₃) = 100; n = 0.5 mol.
• 1 mol CaCO₃ → 1 mol CO₂, so n(CO₂) = 0.5 mol.
• V = 0.5 × 22.4 = 11.2 L.

Worked example 3.3 — several quantities at once. Ammonia is made from N₂ + 3H₂ → 2NH₃. For 3.4 g of NH₃ calculate (a) moles of H₂ needed, (b) volume of H₂ at STP, (c) number of NH₃ molecules. [N = 14, H = 1]

• M(NH₃) = 17; n(NH₃) = 3.4 / 17 = 0.2 mol.
• (a) 3 mol H₂ → 2 mol NH₃, so n(H₂) = 0.2 × 3/2 = 0.3 mol.
• (b) V = 0.3 × 22.4 = 6.72 L.
• (c) Molecules = 0.2 × 6.02 × 10²³ = 1.204 × 10²³.

Worked example 3.4 — gas at rtp and a second reactant. Zinc and hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. 13 g of zinc react completely. Find (a) the volume of hydrogen at rtp and (b) the mass of HCl used. [Zn = 65, H = 1, Cl = 35.5]

• n(Zn) = 13 / 65 = 0.2 mol.
• (a) n(H₂) = 0.2 mol; V = 0.2 × 24 = 4.8 L.
• (b) n(HCl) = 2 × 0.2 = 0.4 mol; mass = 0.4 × 36.5 = 14.6 g.

Worked example 3.5 — from moles of the first substance (book pattern: KNO₃). 2KNO₃(s) →(Δ) 2KNO₂(s) + O₂(g). Heating 3 mol KNO₃ gives n(O₂) = 3 × 1/2 = 1.5 mol, i.e. 33.6 L at STP or 36 L at rtp.

Worked example 3.6 — volume ratios (gas–gas). 2CO(g) + O₂(g) → 2CO₂(g). By Avogadro's hypothesis volumes react in the ratio of coefficients (at the same T and P): 10 L of CO need 5 L of O₂ and give 10 L of CO₂, with no need to convert to moles.

Worked example 3.7 — balancing first. Iron(III) oxide is reduced by carbon monoxide: Fe₂O₃ + CO → Fe + CO₂. Balanced: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Mass of iron from 80 g of Fe₂O₃? [Fe = 56, O = 16] M(Fe₂O₃) = 160; n = 0.5 mol; n(Fe) = 1 mol; mass = 56 g.

4.1 Concept

• In a laboratory or factory the reactants are rarely mixed in exactly the ratio of the equation.
• One of them is consumed completely first and the reaction stops, even though some of the other reactant is still present.

• The limiting reactant is the substance completely consumed in the reaction; it determines the amount of product formed.
• The other reactant is the excess reactant; part of it remains unreacted.

Picture from the book: in H₂ + F₂ → 2HF (1 : 1 : 2), mixing 6 mol H₂ with 4 mol F₂ consumes all 4 mol of fluorine with 4 mol of hydrogen to form 8 mol HF. Two mol H₂ stay unreacted. Fluorine is therefore the limiting reactant.

4.2 How to find it

For each reactant calculate

ratio = number of moles present (given) / number of moles in the balanced equation (coefficient)

• The reactant with the smaller ratio is the limiting reactant.
• Then do every product calculation from the limiting reactant, never from the excess one.
• To find the excess mass, compute how much of the excess reactant is consumed (from the limiting one), then subtract from the amount given.

Warning: the larger number of moles is not necessarily the excess; always compare ratios.

4.3 Worked examples

Worked example 4.1 — moles. N₂ + 3H₂ → 2NH₃. A mixture of 2 mol N₂ and 4 mol H₂ reacts. Find the limiting reactant, the moles of NH₃ and the moles of excess left.

• Ratio N₂ = 2 / 1 = 2. Ratio H₂ = 4 / 3 = 1.33.
• H₂ has the smaller ratio, so H₂ is limiting.
• n(NH₃) = 4 × 2/3 = 2.67 mol.
• N₂ consumed = 4 × 1/3 = 1.33 mol; N₂ left = 2 − 1.33 = 0.67 mol.

Worked example 4.2 — masses. 2Al + 3Cl₂ → 2AlCl₃. A mixture of 5.4 g Al and 10.65 g Cl₂ reacts. Find the limiting reactant, the mass of AlCl₃ and the mass of excess left. [Al = 27, Cl = 35.5]

• n(Al) = 5.4 / 27 = 0.2 mol; n(Cl₂) = 10.65 / 71 = 0.15 mol.
• Ratio Al = 0.2 / 2 = 0.1. Ratio Cl₂ = 0.15 / 3 = 0.05.
• Cl₂ is limiting (smaller ratio).
• n(AlCl₃) = 0.15 × 2/3 = 0.1 mol; M(AlCl₃) = 133.5; mass = 13.35 g.
• Al consumed = 0.15 × 2/3 = 0.1 mol = 2.7 g; Al left = 5.4 − 2.7 = 2.7 g.
• Check: mass before 5.4 + 10.65 = 16.05 g; after 13.35 + 2.7 = 16.05 g. ✓.

Worked example 4.3 — gas volumes. CH₄ + 2O₂ → CO₂ + 2H₂O. 6 L of methane are burned in 8 L of oxygen (both at STP). Volumes are in the ratio of moles: ratio CH₄ = 6/1 = 6; ratio O₂ = 8/2 = 4. O₂ is limiting. CO₂ formed = 8 × 1/2 = 4 L; CH₄ consumed = 4 L; CH₄ left = 2 L.

Worked example 4.4 — pattern of the book's H₂/O₂ example. 2H₂ + O₂ → 2H₂O. 12 L of hydrogen are mixed with 9.6 L of oxygen at rtp. n(H₂) = 12 / 24 = 0.5 mol; n(O₂) = 9.6 / 24 = 0.4 mol. Ratios: H₂ 0.5/2 = 0.25; O₂ 0.4/1 = 0.4. H₂ is limiting. O₂ consumed = 0.25 mol (= 8 g); O₂ given = 0.4 × 32 = 12.8 g; excess O₂ = 12.8 − 8 = 4.8 g. Water formed = 0.5 mol = 9 g.

Worked example 4.5 — mixture of gases after reaction. C₂H₄ + 3O₂ → 2CO₂ + 2H₂O. 3 mol C₂H₄ with 6 mol O₂: ratios 3/1 = 3 and 6/3 = 2, so O₂ is limiting. CO₂ = 6 × 2/3 = 4 mol; H₂O = 4 mol; C₂H₄ consumed 2 mol, left 1 mol.

Why does it matter? In industry the cheaper reactant is used in excess so that the expensive one is completely converted.

5.1 Types of formula

Formulas of compounds are classified by the information they give:

• The empirical formula gives the simplest whole-number ratio of the atoms (or ions) of the elements in the compound.
• The molecular formula gives the actual types and numbers of atoms in one molecule.
• The structural formula gives the types and numbers of atoms and also how they are bonded to each other.

The molecular formula is a whole-number multiple of the empirical formula: molecular formula = (empirical formula)ₙ. For some compounds (e.g. water H₂O, methane CH₄) the two are identical. For ionic compounds (NaCl) only the empirical (formula unit) ratio is meaningful.

5.2 Finding the empirical formula

Percentage composition means mass of each element in 100 g of the compound. Steps:

1. Write the mass of each element (percentage → grams in a 100 g sample). If one percentage is missing, subtract the others from 100.
2. Divide each mass by its atomic mass to get moles of atoms.
3. Divide all by the smallest number of moles.
4. If a ratio is not a whole number, multiply all by the smallest integer that makes them whole: 0.5 → ×2; 0.33 or 0.67 → ×3; 0.25 or 0.75 → ×4; 0.2 → ×5. Do not just round ratios such as 1.5 or 2.33.

Worked example 5.1 (pattern of compound X). Compound with 85.7 % C and 14.3 % H [C = 12, H = 1]. C: 85.7/12 = 7.14; H: 14.3/1 = 14.3. Divide by 7.14: C = 1, H = 2.0. Empirical formula CH₂.

Worked example 5.2 — three elements. A compound is 40 % C, 6.67 % H and 53.33 % O [C = 12, H = 1, O = 16]. C: 3.33; H: 6.67; O: 3.33. Ratio 1 : 2 : 1 → CH₂O. If M = 180 g/mol, n = 180/30 = 6, molecular formula C₆H₁₂O₆ (glucose).

Worked example 5.3 — multiplier needed. Aluminium oxide: 52.94 % Al, 47.06 % O [Al = 27, O = 16]. Al: 1.96; O: 2.94. Ratio O/Al = 1.5, so Al : O = 1 : 1.5 → ×2 → Al₂O₃.

Worked example 5.4 — phosphorus oxide. 43.7 % P, 56.3 % O [P = 31, O = 16]. P: 1.41; O: 3.52; ratio 1 : 2.5 → ×2 → P₂O₅. If M = 284 g/mol, the empirical mass = 142, n = 2, molecular formula P₄O₁₀ (this matches the P₄O₁₀ in the chapter exercises).

Worked example 5.5 — masses given. 1.2 g of carbon is combined with 0.3 g of hydrogen: C = 0.1 mol, H = 0.3 mol → CH₃. With M = 30 g/mol: empirical mass 15, n = 2 → C₂H₆.

5.3 From empirical to molecular formula

n = molar mass of the compound / molar mass of the empirical formula

Molecular formula = (empirical formula)ₙ.

Worked example 5.6. Compound X: empirical formula CH₂ (mass 14), molar mass 28 → n = 2 → C₂H₄.

Worked example 5.7 — hydrocarbon from carbon percentage (pattern). A hydrocarbon has 82.76 % carbon and a molar mass of 58 g/mol. [C = 12, H = 1] H % = 17.24. C: 6.897; H: 17.24; ratio H/C = 2.5 → ×2 → C₂H₅ (mass 29); n = 58/29 = 2 → C₄H₁₀.

5.4 Combustion analysis

When a compound is burned completely, all its carbon appears as CO₂ and all its hydrogen as H₂O. Therefore

mass of C = mass of CO₂ × (12/44), mass of H = mass of H₂O × (2/18).

• If the compound contains only C and H, these two masses add up to the sample mass.
• If they add up to less, the difference is oxygen (found by subtraction).

Worked example 5.8 — hydrocarbon (pattern of the book). A hydrocarbon of molar mass 78 g/mol, mass 3.9 g, burns to give 13.2 g CO₂ and 2.7 g H₂O [CO₂ = 44, H₂O = 18]. Mass of C = 13.2 × 12/44 = 3.6 g; mass of H = 2.7 × 2/18 = 0.3 g (total 3.9 g, so no oxygen). Moles: C = 0.3; H = 0.3 → CH (mass 13); n = 78/13 = 6 → C₆H₆.

Worked example 5.9 — compound with oxygen. 2.3 g of an organic compound C, H, O gives 4.4 g CO₂ and 2.7 g H₂O. C = 4.4 × 12/44 = 1.2 g; H = 2.7 × 2/18 = 0.3 g; O = 2.3 − 1.5 = 0.8 g. Moles: C 0.1, H 0.3, O 0.05. Divide by 0.05: C₂H₆O (empirical). With molar mass 46 it is also the molecular formula.

Worked example 5.10 — finding x and y. A hydrocarbon CₓH_y (M = 44) yields 13.2 g CO₂ and 7.2 g H₂O from 4.4 g. C = 13.2 × 12/44 = 3.6 g, H = 7.2 × 2/18 = 0.8 g. Moles C 0.3, H 0.8 → ratio 1 : 2.67 → ×3 → C₃H₈ (44). So x = 3, y = 8.

6.1 Molarity

• The concentration of a solution can be expressed in several ways.
• The most important is molar concentration (molarity): the number of moles of solute dissolved in one litre of solution.

Molarity M (mol/L) = number of moles of solute n (mol) / volume of solution V (L)

Units: mol/L, written M. If the volume is given in mL or cm³, divide by 1000. Also n = M × V (L), mass of solute = M × V (L) × molar mass. Note: V is the volume of the solution, not of the solvent.

Worked example 6.1. Molarity of a solution made by dissolving 4.9 g of sulphuric acid in water to make 250 mL. [H = 1, S = 32, O = 16]

• M(H₂SO₄) = 98; n = 4.9 / 98 = 0.05 mol; V = 0.25 L.
• M = 0.05 / 0.25 = 0.2 mol/L.

Worked example 6.2 — mass needed. Mass of Na₂CO₃ needed to prepare 500 mL of 0.2 M solution [Na = 23, C = 12, O = 16]. n = 0.2 × 0.5 = 0.1 mol; mass = 0.1 × 106 = 10.6 g.

Worked example 6.3 — pattern of the NaOH example. 2 g of NaOH (40) in 500 mL: n = 0.05 mol; M = 0.05 / 0.5 = 0.1 M.

6.2 Concentration of ions

A strong ionic solute dissociates completely, so ion concentrations are the solute molarity times the number of that ion in the formula.

Worked example 6.4. 0.3 M Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. [Al³⁺] = 2 × 0.3 = 0.6 M; [SO₄²⁻] = 3 × 0.3 = 0.9 M.

Worked example 6.5 — volume containing a number of moles. What volume of 0.2 M CaCl₂ contains 1 mol of chloride ions? Each CaCl₂ gives 2 Cl⁻, so [Cl⁻] = 0.4 M. V = 1 / 0.4 = 2.5 L.

Worked example 6.6 — concentration from the ions asked (pattern of the chapter exercise). What volume of 0.3 M Na₂SO₄ must be taken to prepare 2 L of a solution with [Na⁺] = 0.4 M? [Na₂SO₄] needed = 0.4/2 = 0.2 M. M₁V₁ = M₂V₂: 0.3 × V₁ = 0.2 × 2, V₁ = 1.33 L.

6.3 Dilution

When water is added to a solution, its volume increases but the number of moles of solute is unchanged, so the concentration decreases. Since n = M × V is constant:

M₁ × V₁ = M₂ × V₂

(1 = concentrated/before dilution; 2 = diluted/after). Volume of water added = V₂ − V₁. (V₁ and V₂ may both be in mL or both in L.)

Worked example 6.7. 20 mL of 5 M HCl are diluted to 250 mL. M₂ = 5 × 20 / 250 = 0.4 M. Water added = 250 − 20 = 230 mL.

Worked example 6.8. Volume of 2 M NaOH needed to prepare 500 mL of 0.1 M solution: V₁ = 0.1 × 500 / 2 = 25 mL; water to add = 475 mL.

Worked example 6.9 — volume of water added (pattern of the book's Na₂CO₃ example). To what volume must 250 mL of 0.6 M NaCl be diluted to give 0.2 M? V₂ = 0.6 × 250 / 0.2 = 750 mL; water added = 750 − 250 = 500 mL.

Worked example 6.10 — concentration given as mass per volume. A solution contains 0.4 g of NaOH in each 10 mL. In 1 L: 40 g = 1 mol, so M = 1 M. If 1 L of it is diluted with 3 L of water, V₂ = 4 L and M₂ = 1 × 1 / 4 = 0.25 M.

6.4 Final tips for the chapter

• Always write down the molar masses first.
• Check units (g vs mg, mL vs L).
• Balance the equation before any stoichiometric calculation.
• Limiting reactant: compute ratios, then base everything on the smaller one.
• Empirical formula: moles → divide by smallest → multiply to get integers.
• Gas volumes: 22.4 L at STP, 24 L at rtp.