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Topic 7

The mole and chemical calculations

IB MYP Chemistry · Types of chemical reaction · MYP Years 4–5

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Atoms are far too small to count one by one, so chemists count them in huge packs called moles. The mole links the mass you can weigh on a balance to the number of particles in a balanced equation — which is how industry knows exactly how much of each raw material to use.

🎯What you need to be able to do

  • Calculate relative formula mass \( M_r \) from relative atomic masses.
  • Define the mole and use \( n = m/M \) and the Avogadro constant.
  • Use a balanced equation to calculate reacting masses.
  • Calculate percentage yield and percentage by mass of an element.
  • Use the molar gas volume (24 dm3 at room temperature and pressure).
  • Calculate concentrations in mol/dm3 and g/dm3.

⚖️Relative formula mass

The relative formula mass \( M_r \) of a compound is the sum of the relative atomic masses of all the atoms in its formula. For water, H2O: \( M_r = 2(1) + 16 = 18 \). For calcium carbonate, CaCO3: \( 40 + 12 + 3(16) = 100 \). For Ca(OH)2, the bracket doubles everything inside it: \( 40 + 2(16 + 1) = 74 \).

The percentage by mass of an element in a compound is \( \dfrac{A_r \times \text{number of atoms}}{M_r} \times 100\% \). Iron in Fe2O3 (\( M_r = 160 \)): \( \dfrac{2 \times 56}{160} \times 100 = 70\% \).

🧮The mole

One mole is the amount of substance containing \( 6.02 \times 10^{23} \) particles — the Avogadro constant. This number is chosen so that one mole of any substance has a mass in grams equal to its \( A_r \) or \( M_r \): the molar mass \( M \). One mole of carbon has a mass of 12 g; one mole of water, 18 g.

Moles from mass \[ n = \frac{m}{M} \] \( n \) = amount (mol), \( m \) = mass (g), \( M \) = molar mass (g/mol). Number of particles = \( n \times 6.02 \times 10^{23} \).

🔧Reacting masses

The numbers in a balanced equation give the mole ratio of reactants and products. The method is always the same: mass → moles → use the ratio → moles → mass.

✏️Worked example: making lime

Limestone is heated to make quicklime: \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \). What mass of CaO can be made from 25.0 g of CaCO3? If a student actually obtains 11.2 g, what is the percentage yield? (Ar: Ca = 40, C = 12, O = 16)

1. Moles of CaCO3. \( M_r = 100 \), so \( n = 25.0 \div 100 = 0.250 \) mol.

2. Ratio. 1 CaCO3 : 1 CaO, so 0.250 mol of CaO.

3. Mass of CaO. \( M_r = 56 \): \( m = 0.250 \times 56 = 14.0 \) g (the theoretical yield).

4. Percentage yield. \[ \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{11.2}{14.0} \times 100 = 80\% \]

Check with conservation of mass: 25.0 g of CaCO3 gives 14.0 g CaO + 11.0 g CO2 (0.250 × 44) = 25.0 g. ✓
The trap: skipping the moles and assuming 25 g of reactant gives 25 g of product. Masses are not in the equation’s ratio; moles are.

Yields are rarely 100%: the reaction may be reversible or incomplete, some product is lost when it is separated or transferred, and side reactions may make other products.

🎈Gas volumes

At room temperature and pressure (r.t.p., about 20 °C and 1 atmosphere), one mole of any gas occupies 24 dm3 (24 000 cm3). So volume = moles × 24 dm3. 0.250 mol of CO2 in the example above would fill 6.0 dm3.

🧪Concentration

Concentration of a solution \[ c = \frac{n}{V} \] \( c \) in mol/dm3, \( n \) in mol, \( V \) in dm3 (1 dm3 = 1000 cm3). Multiply by \( M_r \) to get g/dm3.

Dissolving 4.0 g of NaOH (\( M_r = 40 \), so 0.10 mol) in water to make 250 cm3 (0.250 dm3) of solution gives \( c = 0.10 \div 0.250 = 0.40 \) mol/dm3, or 16 g/dm3. Titration calculations use the same equation (Topic 8).

🌎Science in context: atom economy and green chemistry

A reaction can have a 100% yield and still waste most of its atoms in by-products. Atom economy = (Mr of the desired product ÷ total Mr of all products) × 100%. Making lime from limestone has an atom economy of only 56% — the rest leaves as CO2, which is why cement making is a major source of emissions. Green chemistry designs reactions with high atom economy, less waste and safer solvents.

🧠Quick check

1. Calculate the Mr of sulfuric acid, H2SO4 (H 1, S 32, O 16).

2(1) + 32 + 4(16) = 98.

2. How many moles are in 9.0 g of water?

\( n = 9.0 \div 18 = 0.50 \) mol.

3. How many molecules are in 2.0 mol of CO2?

\( 2.0 \times 6.02 \times 10^{23} = 1.2 \times 10^{24} \) molecules.

4. 2Mg + O2 → 2MgO. What mass of MgO forms from 6.0 g of Mg? (Mg 24, O 16)

6.0 ÷ 24 = 0.25 mol Mg → 0.25 mol MgO (1:1) → 0.25 × 40 = 10 g.

5. What volume does 0.10 mol of oxygen occupy at r.t.p.?

0.10 × 24 = 2.4 dm3 (2400 cm3).

6. Find the concentration of 0.050 mol of HCl in 500 cm3 of solution.

\( c = 0.050 \div 0.500 = 0.10 \) mol/dm3.

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