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

2.3 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

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

(a) Magnesium ribbon burning in air with a dazzling white flame

(b) Limiting reactant: 5.0 g Mg with 2.0 g O2

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.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. State and explain the laws of chemical combination with examples and give the modern explanation by Dalton's atomic theory. [10]
2. Define limiting reactant and percentage yield. Solve a full problem on the formation of ammonia from nitrogen and hydrogen. [10]
3. How are the empirical and molecular formulae of a compound found from its percentage composition? Work out an example. [10]

Model answer D1: laws of chemical combination and Dalton's theory (10 marks)

Introduction. Stoichiometry is the study of the quantitative relationships between reactants and products. It rests on five laws, found by experiment between 1774 and 1811, and on Dalton's atomic theory (1808), which explained them.

1. Law of conservation of mass (Lavoisier, 1774). Matter can neither be created nor destroyed in a chemical reaction, so the total mass of the reactants equals that of the products.

Example 2 g of H₂ burns with 16 g of O₂ to give exactly 18 g of H₂O. (In nuclear reactions mass is converted to energy.)

2. Law of constant (definite) proportions (Proust, 1799). A pure compound always contains the same elements in the same ratio by mass, whatever its source.

Water from the Bagmati and from a Himalayan spring is always 11.2% hydrogen and 88.8% oxygen by mass, a ratio of 1 : 8.

3. Law of multiple proportions (Dalton, 1803). When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.

Example H₂O has 8 g of oxygen per 1 g of hydrogen, and H₂O₂ has 16 g of oxygen per 1 g of hydrogen; the ratio is 8 : 16 = 1 : 2. Similarly CO has 12 : 16 and CO₂ has 12 : 32, with oxygen in the ratio 1 : 2.

4. Gay-Lussac's law of combining volumes (1808). Gases react in volumes that bear a simple whole-number ratio to each other and to the volumes of gaseous products at the same temperature and pressure.

Example 2 volumes of H₂ + 1 volume of O₂ give 2 volumes of steam.

5. Avogadro's law (1811). Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. It explains Gay-Lussac's law, and shows that gases like H₂ and O₂ are diatomic. At STP one mole of any gas occupies 22.4 L.

Dalton's atomic theory.

1. Matter is made of indivisible atoms.
2. Atoms of the same element are identical in mass and properties.
3. Atoms of different elements differ.
4. Atoms combine in simple whole-number ratios to form compounds.
5. Atoms are neither created nor destroyed in a reaction, only rearranged. Postulates 4 and 5 explain the laws of constant proportions, multiple proportions and conservation of mass.

Law	Explained by Dalton's postulate	Example
Conservation of mass	Atoms are not created or destroyed	2 g H₂ + 16 g O₂ = 18 g H₂O
Constant proportions	Fixed ratio of atoms	H : O = 1 : 8 in water
Multiple proportions	Whole-number ratio of atoms	H₂O and H₂O₂

Limitations of Dalton's theory. Atoms are divisible (electron, proton, neutron); isotopes of an element differ in mass; isobars of different elements have the same mass; and some compounds, such as sugar, do not combine in simple ratios.

Numerical. A sample of CuO from the laboratory has 3.18 g Cu with 0.80 g O (ratio 3.975 : 1) and another from a mine has 15.9 g Cu with 4.0 g O (ratio 3.975 : 1). Both ratios are identical, which proves the law of constant proportions.

Conclusion. The five laws fix how the masses and volumes of substances change in a reaction, and Dalton's atomic theory provides a particle-level explanation for them.

Model answer D2: limiting reactant and percentage yield (10 marks)

Introduction. In a real reaction the reactants are rarely taken in the exact ratio of the balanced equation.

The reactant that is used up first and so limits the product is called the limiting reactant (limiting reagent); the other is in excess.

The theoretical yield is the maximum product predicted from the limiting reactant, and the actual yield is what is really obtained.

Problem. 28 g of N₂ is mixed with 10 g of H₂ to make ammonia, \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 (N = 14, H = 1).

Find (a) the limiting reactant, (b) the mass of NH₃ formed, (c) the mass of the excess reactant left, (d) the percentage yield if 27.2 g of NH₃ is actually obtained.

Step 1: moles.

• N₂: 28 g / 28 g mol⁻¹ = 1 mol.
• H₂: 10 g / 2 g mol⁻¹ = 5 mol.

Step 2: find the limiting reactant. The equation needs 3 mol of H₂ for each mole of N₂. For 1 mol N₂ only 3 mol H₂ are needed, but 5 mol is available. So N₂ is the limiting reactant and H₂ is in excess.

Step 3: product. From the equation, 1 mol N₂ gives 2 mol NH₃, so the theoretical yield is 2 mol × 17 g mol⁻¹ = 34 g of NH₃.

Step 4: excess left. H₂ used = 3 mol; left = 5 - 3 = 2 mol = 4 g.

Step 5: percentage yield. = 27.2/34 \times 100 = \mathbf{80\%}.

Species	Initial (mol)	Used or formed (mol)	Final (mol)
N₂	1	-1	0
H₂	5	-3	2
NH₃	0	+2	2

Check by mass: 28 g + 10 g = 38 g before; after the reaction there are 34 g NH₃ + 4 g H₂ = 38 g. ✓

Why yield is below 100%. The reaction may be reversible (Haber's process reaches equilibrium), side reactions occur, some product is lost during separation, and the reactants may be impure.

Importance. The idea guides industry: Nepal's fertiliser import calculations use it, and a chemical engineer always chooses the cheaper reactant in excess so that the costly one is completely used.

Conclusion. The limiting reactant fixes the maximum product. Converting grams to moles, comparing them with the ratio of the equation, and then calculating the yield from the limiting reactant is the standard method.

Model answer D3: empirical and molecular formula (10 marks)

Introduction. The empirical formula shows the simplest whole-number ratio of atoms in a compound, and the molecular formula shows the actual number of atoms in a molecule. The molecular formula is a whole-number multiple n of the empirical formula:

Procedure (six steps).

1. Write the percentage of each element (take 100 g of the compound, so the percentages become grams).
2. Divide each mass by the atomic mass to get moles.
3. Divide all the mole values by the smallest one.
4. If any ratio is not a whole number, multiply all by a small integer (2, 3...) to make them whole (for example 1.5 x 2 = 3).
5. Write the formula; this is the empirical formula.
6. Find n from the molar mass and multiply to get the molecular formula.

Worked example. A compound has 40.0% C, 6.67% H and 53.33% O, and a molar mass of 180 g mol⁻¹ (C = 12, H = 1, O = 16).

Element	Mass (g in 100 g)	Moles	Divide by smallest (3.33)	Ratio
C	40.0	40.0/12 = 3.33	1.00	1
H	6.67	6.67/1 = 6.67	2.00	2
O	53.33	53.33/16 = 3.33	1.00	1

The empirical formula is CH₂O, with an empirical mass of 12 + 2 + 16 = 30. Then n = 180/30 = 6 and the molecular formula is C₆H₁₂O₆, which is glucose.

Second example (fractions). A sample of an iron oxide has 69.9% Fe and 30.1% O. Moles: Fe 69.9/55.8 = 1.253, O 30.1/16 = 1.881. Dividing by 1.253 gives 1 : 1.50. Multiplying by 2 gives 2 : 3, so the formula is Fe₂O₃ (haematite).

Uses. The method identifies unknown compounds from a combustion analysis (percent C and H by mass of CO₂ and H₂O produced) and is used in pharmaceutical quality control.

Compound	Empirical formula	Molecular formula	n
Glucose	CH₂O	C₆H₁₂O₆	6
Acetic acid	CH₂O	C₂H₄O₂	2
Benzene	CH	C₆H₆	6
Hydrogen peroxide	HO	H₂O₂	2

Conclusion. The empirical formula comes from the percentage composition, and the molecular formula needs the molar mass in addition. Different compounds can share the same empirical formula, such as glucose and acetic acid.