Moles, molar mass and calculations from equations

What you'll learn

This revision guide covers quantitative chemistry for CIE IGCSE Co-ordinated Science (Double Award). You'll master how to calculate relative formula masses, understand the mole concept, and perform reacting mass calculations from balanced chemical equations. These fundamental skills underpin all quantitative chemistry at IGCSE level and beyond.

Key terms and definitions

Relative atomic mass (Ar) — the average mass of atoms of an element compared to 1/12 of the mass of a carbon-12 atom; found on the periodic table for each element.

Relative formula mass (Mr) — the sum of the relative atomic masses of all atoms in a chemical formula; also called relative molecular mass for covalent compounds.

Mole — the unit of amount of substance; one mole contains 6.02 × 10²³ particles (Avogadro's constant) and has a mass in grams numerically equal to the relative formula mass.

Molar mass — the mass of one mole of a substance, measured in grams per mole (g/mol); numerically equal to the relative formula mass.

Stoichiometry — the quantitative relationship between reactants and products in a chemical reaction, shown by the balancing numbers in a chemical equation.

Limiting reactant — the reactant that is completely used up in a chemical reaction, determining the maximum amount of product that can form.

Empirical formula — the simplest whole number ratio of atoms of each element in a compound.

Reacting masses — the masses of reactants and products involved in a chemical reaction, calculated using mole ratios from balanced equations.

Core concepts

Calculating relative formula mass

The relative formula mass (Mr) is found by adding together the relative atomic masses of all atoms in a formula. You must use the periodic table to find Ar values and remember to multiply by subscript numbers.

Step-by-step method:

1. Write out the formula clearly
2. Identify each element and how many atoms of each are present
3. Find the Ar for each element from the periodic table
4. Multiply each Ar by the number of atoms
5. Add all values together

Example calculations:

• Water (H₂O): (2 × 1) + (1 × 16) = 2 + 16 = 18
• Carbon dioxide (CO₂): (1 × 12) + (2 × 16) = 12 + 32 = 44
• Calcium carbonate (CaCO₃): (1 × 40) + (1 × 12) + (3 × 16) = 40 + 12 + 48 = 100
• Magnesium nitrate (Mg(NO₃)₂): (1 × 24) + (2 × 14) + (6 × 16) = 24 + 28 + 96 = 148

Note the last example carefully: the subscript 2 outside the brackets means everything inside is multiplied by 2, giving 2 nitrogen atoms and 6 oxygen atoms.

Understanding the mole

The mole is the chemist's counting unit for particles. Just as a dozen means 12, a mole means 6.02 × 10²³ particles (atoms, molecules, ions or electrons). This number is called Avogadro's constant.

Key relationships:

• 1 mole of any substance contains 6.02 × 10²³ particles
• 1 mole of any substance has a mass in grams equal to its Mr
• For example: 1 mole of H₂O (Mr = 18) has a mass of 18 g

The mole formula:

The fundamental equation connecting moles, mass and molar mass is:

Number of moles = mass (g) ÷ molar mass (g/mol)

Or rearranged:

• Mass = number of moles × molar mass
• Molar mass = mass ÷ number of moles

This triangular relationship is crucial for all mole calculations.

Practical examples:

• How many moles in 44 g of CO₂ (Mr = 44)?

  • Moles = 44 ÷ 44 = 1 mole

• What mass is 0.5 moles of CaCO₃ (Mr = 100)?

  • Mass = 0.5 × 100 = 50 g

• How many moles in 9 g of water (Mr = 18)?

  • Moles = 9 ÷ 18 = 0.5 moles

Reacting masses from equations

Balanced chemical equations show the mole ratios in which substances react. The balancing numbers (coefficients) tell you the ratio of moles of each substance.

Method for reacting mass calculations:

1. Write the balanced equation
2. Identify the substances you're interested in
3. Write the mole ratio from the equation
4. Calculate the Mr for relevant substances
5. Use the formula: moles = mass ÷ Mr
6. Apply the mole ratio from the equation
7. Convert back to mass using: mass = moles × Mr

Important principle: The ratio of moles is always the same as the ratio of coefficients in the balanced equation.

For example, in: 2H₂ + O₂ → 2H₂O

• 2 moles of H₂ react with 1 mole of O₂ to form 2 moles of H₂O
• The ratio is 2:1:2
• If you start with 4 moles of H₂, you need 2 moles of O₂ and will make 4 moles of H₂O

Mass conservation in reactions

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction. The total mass of reactants always equals the total mass of products.

In practice:

• All atoms present at the start must be present at the end
• They may be rearranged into different compounds
• No atoms are lost or gained
• For closed systems, you can weigh before and after to verify this

Exception for open systems:

If a gas escapes or is absorbed during a reaction in an open container, the mass appears to change. However, if you could capture and weigh the gas, conservation would still apply.

Examples:

• Heating copper carbonate releases CO₂ gas; the solid residue weighs less
• Burning magnesium in air gains oxygen; the product weighs more

Percentage composition by mass

You can calculate what percentage of a compound's mass comes from a particular element.

Formula:

Percentage by mass = (Ar × number of atoms of element ÷ Mr of compound) × 100

Example: Percentage of carbon in carbon dioxide (CO₂)

• Mr of CO₂ = 12 + (2 × 16) = 44
• Mass of carbon = 12
• Percentage = (12 ÷ 44) × 100 = 27.3%

This calculation is useful for:

• Determining the purity of samples
• Finding which compound contains more of a desired element
• Working out empirical formulas from experimental data

Empirical formula calculations

The empirical formula shows the simplest whole number ratio of atoms in a compound. You might be given masses or percentages of each element.

Standard method:

1. List the mass (or percentage) of each element
2. Divide each by the element's Ar to get moles
3. Divide all mole values by the smallest mole value
4. If necessary, multiply all numbers to get whole number ratios
5. Write the empirical formula

Example: A compound contains 2.4 g carbon and 0.8 g hydrogen

• Moles of C = 2.4 ÷ 12 = 0.2
• Moles of H = 0.8 ÷ 1 = 0.8
• Ratio: C = 0.2 ÷ 0.2 = 1, H = 0.8 ÷ 0.2 = 4
• Empirical formula: CH₄

Worked examples

Example 1: Calculating moles and mass

Question: Calculate the mass of 0.25 moles of calcium hydroxide, Ca(OH)₂. [Ar: Ca = 40, O = 16, H = 1]

Solution:

Step 1: Calculate Mr of Ca(OH)₂

• Mr = 40 + (2 × 16) + (2 × 1) = 40 + 32 + 2 = 74

Step 2: Use the formula mass = moles × molar mass

• Mass = 0.25 × 74 = 18.5 g

Answer: 18.5 g [2 marks: 1 for correct Mr, 1 for correct final answer]

Example 2: Reacting masses calculation

Question: Magnesium reacts with hydrochloric acid according to this equation:

Mg + 2HCl → MgCl₂ + H₂

Calculate the mass of magnesium chloride produced when 12 g of magnesium reacts completely with excess hydrochloric acid. [Ar: Mg = 24, Cl = 35.5]

Solution:

Step 1: Calculate Mr values

• Mr of Mg = 24
• Mr of MgCl₂ = 24 + (2 × 35.5) = 24 + 71 = 95

Step 2: Calculate moles of magnesium

• Moles of Mg = 12 ÷ 24 = 0.5 moles

Step 3: Use mole ratio from equation

• From equation: 1 mole Mg produces 1 mole MgCl₂
• Therefore: 0.5 moles Mg produces 0.5 moles MgCl₂

Step 4: Calculate mass of MgCl₂

• Mass = 0.5 × 95 = 47.5 g

Answer: 47.5 g [4 marks: 1 for Mr of MgCl₂, 1 for moles of Mg, 1 for applying mole ratio, 1 for final mass]

Example 3: Limiting reactant calculation

Question: Iron reacts with sulfur to form iron sulfide:

Fe + S → FeS

If 5.6 g of iron is mixed with 2.4 g of sulfur, which reactant is in excess and what mass of iron sulfide is produced? [Ar: Fe = 56, S = 32]

Solution:

Step 1: Calculate moles of each reactant

• Moles of Fe = 5.6 ÷ 56 = 0.1 moles
• Moles of S = 2.4 ÷ 32 = 0.075 moles

Step 2: Compare with equation ratio

• Equation shows 1:1 ratio needed
• We have 0.1 moles Fe but only 0.075 moles S
• Sulfur is the limiting reactant; iron is in excess

Step 3: Calculate mass of product

• Mr of FeS = 56 + 32 = 88
• 0.075 moles S will produce 0.075 moles FeS (1:1 ratio)
• Mass of FeS = 0.075 × 88 = 6.6 g

Answers: Sulfur is the limiting reactant, iron is in excess, 6.6 g of FeS is produced [5 marks total]

Common mistakes and how to avoid them

• Forgetting to multiply by subscripts in brackets — When calculating Mr of compounds like Ca(NO₃)₂, remember the subscript 2 outside brackets means 2 nitrogen atoms AND 6 oxygen atoms, not 2 oxygen atoms. Always expand brackets first.

• Using Ar instead of Mr in mole calculations — The formula requires molar mass (Mr) not atomic mass (Ar). For elements that exist as molecules (like O₂, Cl₂, H₂), you must use the molecular mass, not the atomic mass.

• Ignoring the mole ratio from the equation — The coefficients in balanced equations are crucial. If the equation shows 2:3:1 ratio, you cannot assume 1:1. Always identify and apply the correct ratio.

• Mixing up mass and moles — Keep track of whether you're working with mass (grams) or amount (moles). Write down units in every step to avoid confusion.

• Rounding too early — Keep full calculator values until the final step. Only round your final answer to 2-3 significant figures or as instructed in the question.

• Not checking if reactants are in excess — Questions may state "excess" of one reactant, meaning it won't limit the reaction. Base your calculations on the non-excess (limiting) reactant.

Exam technique for "Moles, molar mass and calculations from equations"

• Show all working clearly — Even if you make an arithmetic error, you can gain method marks by showing the correct approach. Write out formulas, substitution and units.

• Identify command words — "Calculate" requires numerical work with units. "Determine" and "find" are similar. "Show that" means you must arrive at a given answer through clear steps. "State" needs just the answer with no working.

• Use the mark allocation as a guide — A 4-mark question typically needs 4 distinct steps or pieces of information. If you've only written one line for a 3-mark question, you're missing something.

• Check answer sensibility — If you calculate that 1 g of reactant produces 1000 g of product, something is wrong. Mass can't be created. Use approximate Mr values to check your answer is reasonable before moving on.

Quick revision summary

The mole is the chemist's unit for amount of substance; one mole has a mass in grams equal to the relative formula mass. Calculate Mr by adding Ar values for all atoms in a formula. Use the formula: moles = mass ÷ molar mass for conversions. In reacting mass calculations, use the balanced equation to find mole ratios, convert masses to moles, apply the ratio, then convert back to mass. Always show clear working and include units throughout your calculations.