Stoichiometry

Stoichiometry is the arithmetic of chemistry: given a balanced equation, how much of one substance reacts with, or produces, how much of another. It rests on a single idea — atoms are neither created nor destroyed, only rearranged — and on a single counting unit, the mole. This is the longest unit in the physical chemistry area and the one that carries most of the numerical marks in the board paper, so every step here is worth practising until it is automatic.

2.1 Dalton's atomic theory and its postulates

John Dalton (1808) explained the mass laws by assuming matter is made of indivisible particles. His postulates are:

1. All matter is made up of extremely small, indivisible particles called atoms.
2. Atoms can neither be created nor destroyed in a chemical reaction.
3. Atoms of the same element are identical in mass, size and all other properties.
4. Atoms of different elements differ in mass, size and properties.
5. Atoms combine in small whole-number ratios to form compound atoms (molecules).
6. The relative number and kind of atoms in a given compound are fixed.

Postulate	Law it explains	Modern status
Atoms are indivisible	Conservation of mass	Wrong — atoms contain e⁻, p⁺, n
Atoms are neither created nor destroyed	Conservation of mass	Valid for chemical change; fails in nuclear reactions
Atoms of one element are identical in mass	Definite proportions	Wrong — isotopes (¹H, ²H, ³H) differ in mass
Atoms combine in small whole-number ratios	Multiple proportions	Valid for simple compounds; fails for polymers and non-stoichiometric solids such as Fe₀.₉₅O

2.2 Laws of stoichiometry

Law	Statement	Example
Conservation of mass (Lavoisier, 1789)	Matter is neither created nor destroyed in a chemical reaction; total mass of reactants = total mass of products	2H₂ + O₂ → 2H₂O : 4 g + 32 g = 36 g
Definite (constant) proportions (Proust, 1799)	A pure compound always contains the same elements combined in the same fixed ratio by mass, whatever its source	H₂O from a Kathmandu tap or from a lab always has H : O = 1 : 8 by mass
Multiple proportions (Dalton, 1803)	When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio	Cu₂O and CuO: O per 63.5 g Cu = 8 g and 16 g, ratio 1 : 2
Reciprocal proportions (Richter, 1792)	The masses of two elements that combine separately with a fixed mass of a third element are either in the same ratio, or a simple multiple of it, as when they combine together	H and O each combine with S; H : O in H₂S and SO₂ relate simply to H : O in H₂O
Gaseous volumes (Gay-Lussac, 1808)	Gases react, and the gaseous products form, in volumes that bear a simple whole-number ratio to one another at the same temperature and pressure	2 vol H₂ + 1 vol O₂ → 2 vol steam

2.3 Avogadro's law and its deductions

Avogadro's law rescued Gay-Lussac's volume ratios, which contradicted Dalton's assumption that elementary gases were monoatomic. Writing hydrogen and oxygen as H₂ and O₂ makes 2 volumes + 1 volume → 2 volumes come out exactly right:

2H₂ + O₂ → 2H₂O (2 molecules + 1 molecule → 2 molecules)

Deduction 1 — molecular mass and vapour density

Deduction 2 — molecular mass and volume of a gas

One mole of any gas occupies 22.4 litres at STP (273 K, 1 atm). This volume is called the gram molecular volume. Hence the molecular mass in grams of any gas occupies 22.4 L at STP, and

Deduction 3 — molecular mass and number of particles

The gram molecular mass of any substance contains the same number of molecules, the Avogadro number N_A = 6.022\times10^{23}\ \text{mol}^{-1}.

2.4 Mole and its relation with mass, volume and number of particles

Starting from	To get moles	From moles
Mass w (g)	n = w/M	w = n\times M
Number of particles N	n = N/N_A	N = n\times N_A
Volume of gas at STP V (L)	n = V/22.4	V = n\times22.4

2.5 Calculations based on mole concept

A balanced equation is read in moles. For

CaCO₃ --Δ--> CaO + CO₂↑

the coefficients 1 : 1 : 1 mean 1 mol (100 g) of calcium carbonate gives 1 mol (56 g) of calcium oxide and 1 mol (22.4 L at STP) of carbon dioxide.

2.6 Limiting reactant and excess reactant

When reactants are not mixed in exactly the ratio required by the equation, one runs out first and stops the reaction.

To find it, divide the moles available of each reactant by its coefficient in the balanced equation. The smallest quotient marks the limiting reactant.

2.7 Theoretical yield, experimental yield and percentage yield

Real reactions rarely give everything the equation promises: they may be reversible, side reactions occur, or product is lost on filtering and drying.

Term	Meaning
Theoretical yield	Maximum mass of product calculated from the balanced equation, assuming the limiting reactant is fully converted
Experimental (actual) yield	Mass of pure product actually obtained in the laboratory
Percentage yield	Actual yield expressed as a percentage of the theoretical yield

2.8 Calculation of empirical and molecular formula from percentage composition

Chapter summary

• Dalton's postulates: atoms are indivisible, conserved, identical within an element and combine in small whole-number ratios. Isotopes and subatomic particles have since disproved the first and third.
• The mass laws: conservation of mass, definite proportions, multiple proportions, reciprocal proportions, and Gay-Lussac's law of gaseous volumes.
• Avogadro's law: equal volumes of gases at the same T and P contain equal numbers of molecules. Deductions: molecular mass = 2 × vapour density; 1 mol of gas occupies 22.4 L at STP; 1 mol contains N_A = 6.022\times10^{23} particles.
• Mole conversions: n = w/M = N/N_A = V/22.4 (V in litres, at STP).
• Stoichiometry route: mass → moles → mole ratio from the balanced equation → moles → mass or volume.
• The limiting reactant is the one with the smallest value of (moles ÷ coefficient); it fixes the amount of product. The rest is excess.
• \%\ \text{yield} = (\text{actual}/\text{theoretical})\times100. Purity is applied to the reactant, yield to the product.
• Empirical formula: % → ÷ atomic mass → ÷ smallest → clear fractions; molecular formula = n\times empirical formula with n = \text{molecular mass}/\text{EF mass}.

Practice questions

Group A — Multiple choice (1 mark each)

1. The number of molecules in 11.2 L of any gas at STP is [1]a6.022 × 10²³b3.011 × 10²³c1.204 × 10²⁴d12.044 × 10²³
2. The vapour density of a gas is 32. Its molecular mass is [1]a16b32c64d128
3. In the reaction 2H₂ + O₂ → 2H₂O, 4 g of H₂ is mixed with 16 g of O₂. The limiting reactant is [1]aH₂bO₂cneitherdboth equally
4. Which law is explained by the existence of CO and CO₂? [1]aconservation of massbdefinite proportionscmultiple proportionsdreciprocal proportions
5. The empirical formula of a compound is CH₂O and its molecular mass is 60. Its molecular formula is [1]aCH₂ObC₂H₄O₂cC₃H₆O₃dC₆H₁₂O₆

Answers to Group A

1. (b) — 11.2 L is half a mole, so 0.5 × 6.022 × 10²³ = 3.011 × 10²³.

2. (c) — molecular mass = 2 × VD = 64.

3. (b) — n(H₂) = 2, n(O₂) = 0.5; dividing by coefficients gives 1.0 and 0.5, so oxygen runs out first.

4. (c) — for a fixed 12 g of carbon, the oxygen masses are 16 g and 32 g, a ratio of 1 : 2.

5. (b) — EF mass = 30, n = 60/30 = 2.

Group B — Short answer (5 marks each)

1. State the postulates of Dalton's atomic theory and mention two of its limitations. [5]
2. State Avogadro's law and derive the relation: molecular mass = 2 × vapour density. [5]
3. Calculate the number of moles, molecules and atoms present in 3.4 g of ammonia gas, and the volume it occupies at STP. (N = 14, H = 1) [5]
4. 6.5 g of zinc is dropped into excess dilute hydrochloric acid. Calculate the volume of hydrogen liberated at STP and the mass of zinc chloride formed. (Zn = 65, Cl = 35.5) [5]
5. Define limiting reactant. 50 g of CaCO₃ is treated with 35 g of HCl. Which is the limiting reactant, and what mass of CO₂ is produced? (Ca = 40, C = 12, O = 16, H = 1, Cl = 35.5) [5]

Answers to Group B

3. M(NH₃) = 17 g mol⁻¹, so n = 3.4/17 = 0.20 mol. Molecules = 0.20 × 6.022 × 10²³ = 1.204 × 10²³. Each NH₃ has 4 atoms, so atoms = 4 × 1.204 × 10²³ = 4.82 × 10²³. Volume at STP = 0.20 × 22.4 = 4.48 L.

4. Zn + 2HCl → ZnCl₂ + H₂↑. n(Zn) = 6.5/65 = 0.10 mol. Mole ratio Zn : H₂ : ZnCl₂ = 1 : 1 : 1, so n(H₂) = n(ZnCl₂) = 0.10 mol. V(H₂) = 0.10 × 22.4 = 2.24 L at STP. M(ZnCl₂) = 65 + 71 = 136, so mass = 0.10 × 136 = 13.6 g.

5. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂↑. n(CaCO₃) = 50/100 = 0.50 mol; n(HCl) = 35/36.5 = 0.959 mol. Dividing by coefficients: 0.50/1 = 0.50 and 0.959/2 = 0.479, so HCl is limiting. n(CO₂) = 0.959/2 = 0.479 mol, giving 0.479 × 44 = 21.1 g of CO₂ (equivalently 10.7 L at STP).

Group C — Long answer (8 marks each)

1. (a) State the law of multiple proportions. Two oxides of nitrogen contain 63.6 % and 46.7 % nitrogen by mass. Show that the data illustrate the law and identify the oxides. (N = 14, O = 16) [5](b) Define theoretical yield and percentage yield, and explain two reasons why the actual yield of an industrial reaction is always less than the theoretical yield. [3]
2. A hydrocarbon contains 85.7 % carbon and 14.3 % hydrogen by mass. Its vapour density is 21. [8](a) Determine its empirical formula.(b) Determine its molecular mass and molecular formula.(c) Calculate the volume of oxygen at STP required for the complete combustion of 4.2 g of this hydrocarbon.

Answers to Group C

1 (a). In 100 g of the first oxide: N = 63.6 g, O = 36.4 g, so oxygen per 1 g of nitrogen = 36.4/63.6 = 0.572. In the second: N = 46.7 g, O = 53.3 g, giving 53.3/46.7 = 1.141. The ratio is 0.572 : 1.141 = 1 : 2, a simple whole-number ratio, so the law holds. The first is N₂O (28 : 16) and the second is NO (14 : 16).

1 (b). Theoretical yield is the maximum product calculable from the balanced equation for the limiting reactant; percentage yield is actual/theoretical × 100. Actual yields fall short because many reactions are reversible and reach equilibrium before completion, side reactions consume reactant, and product is lost during filtration, transfer and purification.

2. (a) C: 85.7/12 = 7.14; H: 14.3/1 = 14.3. Dividing by 7.14 gives 1 : 2, so the empirical formula is CH₂ (EF mass 14).

(b) Molecular mass = 2 × VD = 2 × 21 = 42, so n = 42/14 = 3 and the molecular formula is C₃H₆ (propene).

(c) C₃H₆ + 4.5O₂ → 3CO₂ + 3H₂O, i.e. 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. n(C₃H₆) = 4.2/42 = 0.10 mol, so n(O₂) = 0.10 × 4.5 = 0.45 mol. V(O₂) = 0.45 × 22.4 = 10.08 L at STP.