Stoichiometry

Stoichiometry comes from the Greek words stoicheion (element) and metron (measure). It is the branch of chemistry in which the relationship between the amounts of reactants and products in a balanced chemical equation is studied. A balanced equation has the same number of atoms of each element on both sides, so it has definite ratios of reactants and products — and those ratios are used to calculate the mass or mole of other substances.

Stoichiometric calculations obey two laws: the law of conservation of mass — "matter (mass) can neither be created nor destroyed", so in a balanced equation the total mass of reactants equals the total mass of products — and the law of definite proportions — a pure compound always contains the same elements combined in the same ratio by mass.

The mole is the amount of a substance which contains as many elementary entities as there are atoms in 0.012 kg (12 g) of carbon-12. The elementary entities may be atoms, molecules, ions, electrons or other particles. It is represented by n. The number of entities present in one mole of a substance is a constant, named Avogadro's Number, i.e. 6.02 × 10²³, represented by N_A (after the Italian scientist Amedo Avogadro, 1776–1856).

📌 Did you know? Avogadro's number is a physical constant representing the molar quantity. The exact value of it is 6.02214179 × 10²³ mol⁻¹; in calculations we use the rounded-off value 6.02 × 10²³.

Examples: 1 mole of ¹²C contains 6.02 × 10²³ atoms of ¹²C; 1 mole of H₂O contains 6.02 × 10²³ molecules of H₂O; 1 mole of NaCl contains 6.02 × 10²³ formula units of NaCl; 1 mole of Na⁺ contains 6.02 × 10²³ ions of Na⁺. Chemists use the mole as the SI unit to weigh and count atoms, molecules, formula units or ions.

Molar mass (M) is the mass of one mole of a substance (element, compound or ionic species) equal to its atomic mass, molecular mass, formula mass or ionic mass expressed in grams. The mass of one mole of a substance expressed in grams is called molar mass, in g/mol. The molar mass is the sum of the masses of the component atoms. Example: molar mass of CCl₄ = molar mass of one C + molar mass of Cl × 4 = 12.0 × 1 + 35.5 × 4 = 12.0 + 142.0 = 154.0 g.

Other examples: 1 mole of carbon atoms is 12.0 g; 1 mole of CO₂ molecules is 44.0 g; 1 mole of CaO formula units is 56.1 g; 1 mole of CO₃²⁻ ions is 60.0 g. The number of moles of a substance = given mass ÷ molar mass, i.e. n = m/M.

Worked examples (Sample Problems 4.1–4.3). (4.1) Moles in 20 g of NaOH: n = 20/40 = 0.5 mol. (4.2) Mass of 0.5 moles of HCl: mass = 0.5 × 36.5 = 18.3 g. (4.3) Mass of 10⁻³ mol of MgSO₄: molar mass = 24 + 96 = 120 g mol⁻¹; mass = 10⁻³ × 120 = 0.12 g.

A sample of 12.0 grams of natural carbon contains the same number of atoms as 4.0 grams of natural helium. Both samples contain 1 mole of atoms, i.e. 6.02 × 10²³. It is interesting to know that different masses of elements have the same number of atoms: 1.0 g of hydrogen = 1 mol of hydrogen = 6.02 × 10²³ atoms of H; 23.0 g of sodium = 1 mol of Na = 6.02 × 10²³ atoms of Na; 238.0 g of uranium = 1 mol of U = 6.02 × 10²³ atoms of U. An atom of sodium is 23 times heavier than an atom of hydrogen, so to have an equal number of atoms, sodium should be taken 23 times greater in mass than hydrogen.

Hence, one mole of different compounds has different masses but the same number of molecules: 18.0 g of H₂O = 1 mol of water = 6.02 × 10²³ molecules of water; 180.0 g of glucose = 1 mol of glucose = 6.02 × 10²³ molecules of glucose. Similarly, the number of ions in one mole of different ionic species is always the same, i.e. Avogadro's number: 96.1 g of SO₄²⁻ = 1 mole of SO₄²⁻ = 6.02 × 10²³ ions; 62.0 g of NO₃¹⁻ = 1 mole of NO₃¹⁻ = 6.02 × 10²³ ions.

One can calculate the number of moles by dividing the number of particles by Avogadro's number: Number of moles = No. of particles of substance ÷ Avogadro's Number. In short: Mass in grams ÷ Molar mass ⇌ Moles ⇌ × Avogadro's number → No. of Particles (and back the other way).

Worked examples. (4.4) A glucose sample contains 3.76 × 10²⁴ molecules: moles = 3.76 × 10²⁴ ÷ 6.02 × 10²³ = 6.25 moles. (4.5) Atoms in 2.3 g of sodium metal: moles = 2.3 ÷ 23.0 = 0.1 mol; atoms = 0.1 × 6.02 × 10²³ = 0.602 × 10²³ atoms. (4.6) Juglone, C₁₀H₆O₃, a dye from black walnut husks: (a) molar mass = (10 × 12.0) + (6 × 1.0) + (3 × 16.0) = 120 + 6 + 48 = 174 g/mol; (b) moles in 0.87 g = 0.87 ÷ 174 = 0.005 mol.

📌 Interesting Information! Juglone is a natural herbicide (weed killer). It kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. And isopentyl acetate (C₇H₁₄O₂) is the compound responsible for the scent of bananas; bees release about 1 µg (1 × 10⁻⁶ g) of it when they sting, and the scent attracts other bees to join the attack.

Molar volume (4.3). The volume of one mole of an ideal gas at STP (Standard Temperature and Pressure) is called molar volume. Its value is equal to 22.414 dm³. The value of molar volume is commonly rounded to 22.4 dm³. It is denoted by V_m. By using the molar volume relationship, mass or mole of a gas at STP can be converted into volume, and vice versa.

According to Avogadro's law, "Equal volumes of all ideal gases at the same temperature and pressure contain equal numbers of molecules". This statement is indirectly the same when we say that one mole of an ideal gas at 273.16 K and one atm pressure has a volume of 22.414 dm³. Since one mole of a gas has Avogadro's number of particles, so 22.414 dm³ of various ideal gases at STP will have Avogadro's number of molecules, i.e. 6.02 × 10²³.

So: 22.4 dm³ of a gas at STP = molar mass of a gas = 6.02 × 10²³ particles of a gas = 1 mole of a gas. Examples: 22.4 dm³ of CO₂ at STP = 44.0 g of CO₂ = 6.02 × 10²³ molecules of CO₂ = 1 mole of CO₂; 22.4 dm³ of H₂ gas at STP = 2 g = 6.02 × 10²³ molecules = 1 mole; 22.4 dm³ of NH₃ gas at STP = 17 g = 6.02 × 10²³ molecules = 1 mole.

If the number of moles of a gas is known, one can calculate its volume by multiplying number of moles with molar volume: Volume of a gas = Number of moles × Molar volume, V = n × V_m.

Worked examples. (4.7) Volume of 2.5 moles of chlorine molecules at STP: V = 22.4 × 2.5 = 56.0 dm³. (4.8) Volume of 4.75 mol of methane (CH₄) gas at STP: V = 4.75 × 22.4 = 106.4 dm³.

Molar mass and density of gases (4.4). Density is defined as mass per unit volume: Density = Mass ÷ Volume, d = m/V. As the molar mass of all the gases occupies the same volume at STP, the density of a gas depends on its molar mass. A gas having higher molar mass will have higher density and vice versa. If the density of the gas at STP is known, its molar mass can be calculated: molar mass = density × molar volume.

Worked example (4.9). Molar mass of a gas with density 1.97 g/dm³ at STP: mass of gas at STP = d × V = 1.97 × 22.4 = 44.1 g mol⁻¹ (i.e. CO₂). Similarly, a gas of density 1.34 g/dm³ at STP has molar mass 1.34 × 22.4 = 30.0 g mol⁻¹.

Molar concentration of solutions is given as mol/dm³, which is the number of moles of a substance (reactant or product) dissolved per volume of a solution in dm³. The relationship between number of moles and molar concentration is given by n = C × V, and therefore C = n/V, where C is the molar concentration and V is the volume of the solution.

For example, concentration of 0.2 mol of a substance per dm³ is: Molar Concentration = Number of moles ÷ Volume in dm³ = 0.2 mol ÷ 1 dm³ = 0.2 mol dm⁻³.

The book's mole-centre diagram links all four routes to moles: Mass ÷ Molar mass; Volume ÷ Molar Volume; No. of particles ÷ Avogadro's number; and Concentration × Volume. Any one of these gives n, and from n you can reach any other quantity.

Worked example (4.10). Molar concentration of a solution containing 27.64 g of K₂CO₃ dissolved in 1 dm³ of solution: mass = 27.64 g; molar mass of K₂CO₃ = 138.2 g mol⁻¹; moles n = m/M = 27.64 ÷ 138.2 = 0.2 mol; volume = 1 dm³; molar concentration C = 0.2 mol ÷ 1 dm³ = 0.2 mol dm⁻³.

Similarly, a solution of 7.9 g of KMnO₄ (molar mass 158 g mol⁻¹) in 1 dm³ has n = 7.9 ÷ 158 = 0.05 mol, so C = 0.05 mol dm⁻³.

The following types of relationship can be studied with the help of a balanced chemical equation involving quantities of reactant(s) and product(s): (i) Mole-Mole Relationship, (ii) Mass-Mass Relationship, (iii) Volume-Volume Relationship, (iv) Mole-Mass Relationship, (v) Mole-Volume Relationship, (vi) Mass-Volume Relationship.

To understand these relationships, we need to interpret information hidden in a balanced chemical equation, which is used to make stoichiometric calculations. For example: N_2(g) + 3H_2(g) → 2NH_3(g). This equation can be described in different ways: (i) 1 mole of N₂ reacts with 3 moles of H₂ to form 2 moles of NH₃. (ii) 1 molecule of N₂ reacts with 3 molecules of H₂ to form 2 molecules of NH₃. (iii) 22.4 dm³ of N₂ reacts with 67.2 dm³ of H₂ to form 44.8 dm³ of NH₃. (iv) 28.0 g of N₂ react with 6 g of H₂ to form 34.0 g of NH₃.

📌 Keep in Mind! The following assumptions must be made while performing stoichiometric calculations: (1) All the reactants are completely converted into the products. (2) Law of conservation of mass and law of definite proportions are obeyed. (3) No side reaction occurs.

Approach to do stoichiometric calculations (5 steps). (1) Start from mass of known solid, volume of a known gas, or molar concentration of known solution. (2) Calculate number of moles from the mass of known solid, volume of a known gas, or molar concentration of known solution using the relevant formula. (3) Find the ratio of the known and the unknown reactant or product from the balanced chemical equation. (4) Calculate the number of moles of the unknown reactant or product using the relevant formula. (5) Convert the number of moles of the unknown to mass, volume, or concentration of the substance.

Worked examples. (4.11, Mole-Mole) When 3.3 mol of nitrogen reacts with hydrogen to form ammonia, how many moles of hydrogen are consumed? N₂ + 3H₂ → 2NH₃. 1 mole of N₂ needs 3 mol of H₂, so 3.3 mol of N₂ needs 3 × 3.3 = 9.9 mol of H₂. (4.12, Mass-Mass) Mass of Al needed to react completely with 32.0 g of iron(III) oxide: 2Al_(s) + Fe₂O_3(s) → Al₂O_3(s) + 2Fe_(s). Molar mass of Fe₂O₃ = 159.6 g mol⁻¹; moles of Fe₂O₃ = 32.0 ÷ 159.6 = 0.02 mol. From the equation, 1 mol of Fe₂O₃ reacts with 2 moles of Al, so 0.02 mol reacts with 0.04 mol of Al. Mass of Al = 0.04 × 27 = 1.08 g.

Volume-Volume (4.13). Volume of ammonia produced by 100 dm³ of hydrogen with excess nitrogen at STP: N_2(g) + 3H_2(g) → 2NH_3(g). 67.2 dm³ (3 mol) of H₂ produce 44.8 dm³ of ammonia, so 1 dm³ of H₂ produces 44.8 ÷ 67.2 = 2/3 dm³, and 100 dm³ of H₂ produces (2/3) × 100 = 66.7 dm³ of NH₃.

Similarly, when 4.5 dm³ of methane is burnt with excess oxygen (CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g)), 1 volume of CH₄ gives 1 volume of CO₂, so 4.5 dm³ of CO₂ is produced.

Mole-Mass (4.14). Solid lithium hydroxide LiOH is used in space vehicles to remove exhaled carbon dioxide, forming solid lithium carbonate and liquid water. Mass of Li₂CO₃ produced by 20.0 mol of LiOH: 2LiOH_(s) + CO_2(g) → Li₂CO_3(s) + H₂O_(l). 2 mol of LiOH produces 1 mol of Li₂CO₃, so 20.0 mol produces (1/2) × 20.0 = 10.0 mol of Li₂CO₃. Mass = 10.0 mol × 73.9 g mol⁻¹ = 739.0 g.

Similarly, 2.25 moles of chlorine with excess sodium hydroxide (2NaOH_(aq) + Cl_2(g) → NaOCl_(aq) + NaCl_(aq) + H₂O_(l)) produces 2.25 mol of sodium hypochlorite NaOCl, mass = 2.25 × 74.5 = 167.6 g.

Mass-Mole (4.15). Baking soda (NaHCO₃) acts as an antacid; it neutralizes excess hydrochloric acid (HCl) in the stomach: NaHCO_3(g) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(aq). Moles of HCl neutralized by 2.1 g of baking soda: molar mass of NaHCO₃ = 84.0 g mol⁻¹; moles = 2.1 ÷ 84.0 = 0.025 mol. The mole ratio of HCl to NaHCO₃ is 1 : 1, hence moles of HCl used = 0.025 mol.

Mass-Volume (4.16). Volume of hydrogen at STP produced when 7.0 g of iron reacts with excess sulphuric acid: Fe_(s) + H₂SO_4(aq) → FeSO_4(aq) + H_2(g). Molar mass of Fe = 55.8 g/mol; moles of iron = 7.0 ÷ 55.8 = 0.125 mol. From the equation, 1 mol of iron produces 1 mole of hydrogen, so moles of H₂ = 0.125 mol. Volume = 22.4 dm³ mol⁻¹ × 0.125 mol = 2.8 dm³.

In many chemical processes, the quantities of the reactants are usually not present in the proportions indicated by the balanced chemical equation. Frequently, a large amount of an inexpensive reactant is supplied because of the following reasons: (a) To ensure that the whole of the mass of the expensive reactant is completely converted to the desired product. (b) To produce the maximum amount of product. (c) To increase the rate of reaction.

We know that a large quantity of oxygen in a chemical reaction makes things burn more rapidly. In this way, excess of oxygen is left behind at the end of the reaction and the other reactant, i.e. fuel, is consumed earlier. This reactant which is consumed earlier is called the limiting reactant. In this way, the amount of product that is formed is limited by the reactant that is completely used. Once this reactant is consumed, the reaction stops and no additional product is formed. The reactant which controls the amount of products formed in a chemical reaction and is consumed earlier is called the limiting reactant or reagent. The maximum amount of the product formed depends upon the amount of limiting reactant in the reaction mixture.

4.7.1 Strategy for the identification of limiting reactant. To identify a limiting reactant, the following three steps are performed: (i) Calculate the number of moles from the given amounts of reactants. (ii) Find out the number of moles of product with the help of a balanced chemical equation. (iii) Identify the reactant which produces the least amount of product as limiting reactant and the other as an excess reactant.

📌 Did You Know? Fire is a combustion reaction in which fuel and oxygen, O₂, combine, usually at high temperatures, to form water and carbon dioxide. Once the fire has started, it is self-supporting. An effective way to quench a fire is smothering, which reduces the amount of available oxygen below the level needed to support combustion. In other words, smothering decreases the amount of the excess reactant. Foams, inert gas, and CO₂ are effective substances for smothering.

Worked examples. (4.17) Mass of N₂ produced from 18.1 g of NH₃ (molar mass 17.0 g mol⁻¹) and 90.4 g of CuO (molar mass 79.5 g mol⁻¹): 2NH_3(g) + 3CuO_(s) → N_2(g) + 3Cu_(s) + 3H₂O_(g). Moles of NH₃ = 18.1 ÷ 17.0 = 1.06 mol; moles of CuO = 90.4 ÷ 79.5 = 1.14 mol. From CuO: CuO : N₂ = 3 : 1, so N₂ = (1/3) × 1.14 = 0.38 mol. From NH₃: NH₃ : N₂ = 2 : 1, so N₂ = (1/2) × 1.06 = 0.53 mol. CuO gives less product, so CuO is the limiting reactant and N₂ produced = 0.38 mol. The book prints the answer as 9.0 g of N₂, but its solution line is slipped: it multiplies by 17.0 g mol⁻¹ (that is NH₃'s molar mass, not N₂'s), and even 0.38 × 17.0 = 6.5 g, not 9.0 g. The correct working uses M(N₂) = 28.0 g mol⁻¹: 0.38 × 28.0 = 10.6 g. The limiting-reactant identification (CuO) is unaffected.

(4.18) When aqueous solutions of Na₂SO₄ and Pb(NO₃)₂ are mixed, PbSO₄ precipitates down. Mass of PbSO₄ formed when 1.25 dm³ of 0.05 mol dm⁻³ Pb(NO₃)₂ and 2.00 dm³ of 0.025 mol dm⁻³ Na₂SO₄ are mixed: Na₂SO_4(aq) + Pb(NO₃)_2(aq) → PbSO_4(s) + 2NaNO_3(aq). Net ionic equation: Pb²⁺_(aq) + SO₄²⁻_(aq) → PbSO_4(s). Moles of Pb²⁺ = 0.05 × 1.25 = 0.0625 mol; moles of SO₄²⁻ = 0.025 × 2.00 = 0.05 mol. As Pb²⁺ and SO₄²⁻ react in a 1 : 1 ratio, SO₄²⁻ (0.05 mol) will be consumed earlier than Pb²⁺ (0.0625 mol), so SO₄²⁻ is the limiting reactant. Only 0.05 mole of solid PbSO₄ will be formed. Mass = 0.05 × 303.3 = 15.2 g.

4.7.2 Amount of Product and Unreacted Excess Reagent. The reactants which are in larger amounts (according to stoichiometry of reaction) and remain unreacted at the end of the reaction are called "excess reagents" (or excess reactants). Consider 2H_2(g) + O_2(g) → 2H₂O_(l): when we take 2 moles of hydrogen (4 g) and allow it to react with 2 moles of oxygen (64 g), we will get only 2 moles (36 g) of water, because 2 moles (4 g) of hydrogen react with 1 mole (32 g) of oxygen according to the balanced equation. When 1 mole of O₂ and 1 mole of H₂ are mixed, all the H₂ will react completely and O₂ will be left unreacted, because for 1 mole of H₂, 1/2 mole of O₂ is required; the remaining 1/2 mole will be excess.

Worked example (4.19, Excess Reactant). Natural gas is primarily methane (CH₄). Complete combustion of methane gives carbon dioxide and water: CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g). (a) Grams of CO₂ produced when 30 g of CH₄ and 50 g of O₂ are allowed to combine: moles of CH₄ = 30 ÷ 16 = 1.875 mol; moles of O₂ = 50 ÷ 32 = 1.563 mol. From CH₄: 1.875 mol of methane produces 1 × 1.875 = 1.875 mol of CO₂. From O₂: 1.563 mol of oxygen produces 0.5 × 1.563 = 0.7815 mol of CO₂. O₂ produces less product, so O₂ is the limiting reactant. Mass of CO₂ = 0.7815 × 44 = 34.39 g. (b) Grams of excess reagent left unreacted: 2 mol of O₂ reacts with 1 mol of CH₄, so 1.563 mol of O₂ reacts with (1/2) × 1.563 = 0.7815 mol of CH₄. Moles of CH₄ in excess = 1.875 − 0.7815 = 1.0935 mol. Excess mass of CH₄ = 1.0935 × 16.0 = 17.5 g.

The amount of the products obtained in a chemical reaction is called the actual yield of that reaction. The amount of the products calculated from the balanced chemical equation represents the theoretical yield. The theoretical yield is the maximum amount of the product that can be produced by a given amount of a reactant, according to balanced chemical equation.

In most chemical reactions the amount of the product obtained is less than the theoretical yield. There are following reasons for that: (1) The processes like filtration, separation by distillation, separation by a separating funnel, washing, drying and crystallization, if not properly carried out, decrease the actual yield. (2) Some of the reactants might take part in a competing side reaction and reduce the amount of the desired product. So, in most of the reactions the actual yield is less than the theoretical yield. (3) A reaction may be reversible. Therefore, the amount of the product will be reduced by the backward reaction.

A chemist is usually interested in the efficiency of a reaction. This is expressed by comparing the actual and theoretical yields in the form of percentage (%) yield: % Yield = (Actual yield ÷ Theoretical yield) × 100. Greater the % age yield, higher will be the efficiency of reaction and vice versa.

Worked example (4.20, % age Yield). Aspirin (C₉H₈O₄) is prepared by heating salicylic acid, C₇H₆O₃ (molar mass 138.12 g mol⁻¹), and acetic anhydride, C₄H₆O₃ (molar mass 163.93 g mol⁻¹): C₇H₆O_3(s) + C₄H₆O_3(l) → C₉H₈O_4(s) + CH₃COOH_(aq). Calculate the theoretical yield of aspirin (molar mass 180.16 g mol⁻¹) when 3.00 g of salicylic acid is heated with 6.00 g of C₄H₆O₃. What is % yield when actual yield is 3.15 g?

Solution: 1 mol of salicylic acid (138.12 g) + 1 mol of acetic anhydride (163.93 g) → 1 mol of aspirin (180.16 g). Moles of salicylic acid = 3.00 ÷ 138.12 = 0.022 mol. Moles of acetic anhydride = 6.00 ÷ 163.93 = 0.037 mol. Here, salicylic acid is limiting reactant while acetic anhydride is an excess reactant. The amount of salicylic acid controls the yield of product, i.e. aspirin: 0.022 mol of salicylic acid produces 0.022 mol of aspirin. Mass of aspirin = 0.022 × 180.16 = 3.96 g (theoretical yield). % age yield = (3.15 ÷ 3.96) × 100 = 79.5%.

📌 Note on the book's printed solution: the book's solution line uses an actual yield of 2.85 g and gives 71.97%, although the question itself states the actual yield as 3.15 g. Using the question's own figure, 3.15 g, the correct % yield is 79.5%, as worked above. (Similarly, when limestone CaCO₃ is roasted: CaCO_3(s) → CaO_(s) + CO_2(g); 4.5 kg of limestone = 45 mol → theoretical CaO = 45 × 56 = 2520 g = 2.52 kg; with actual yield 2.5 kg, % yield = 2.5 ÷ 2.52 × 100 = 99.2%.)

While preparing the required dose of a medicine, the optimum amount of the active ingredient in a medicine is essential to produce desired effects in the patient. Stoichiometry ensures the accuracy of drug synthesis. Any deviation can result in incomplete reaction or contamination with un-reacted reactants or by-products. Stoichiometry allows chemists to precisely control chemical reactions to produce drugs, to ensure its efficiency, effectiveness and safe use.

4.9.1 Significance of Stoichiometry in Medicine. Stoichiometry is very important in the field of medicine and is used: (1) In the preparation of antibiotics, the stoichiometry ensures that each dose matches the active ingredient and target bacteria. (2) To determine the cholesterol level in the blood of patients. Cholesterol is a form of fat that is not all bad. However, cholesterol can have harmful effects. (3) To determine the glucose level in the blood of a diabetic patient. Use of insulin relies on the stoichiometry to precise control of blood sugar levels. (4) To determine the steroid and other stimulants in the urine of athletes. Athletes use steroids and other stimulants to enhance performance and increase strength. (5) To determine the concentration of viral antigens in the preparation of vaccine for effective results. (6) To determine the amount and number of drugs to give a dosage to a patient. The medicine has no effect when given in small amounts and can cause toxic state or death when given in large amounts. Paracetamol is used as a pain killer and to decrease fever. An overdose may result in blood thinning, organ damage and severe liver damage.

This is why the SLO on medicine matters for the paper: stoichiometry is not only about equations — it is the calculation that makes a dose safe, a vaccine effective, and a drug pure.