Maths Skills in Chemistry

At least 20% of the total marks in OCR Gateway Chemistry require mathematical skills — a substantial share of every paper. These are some of the most reliable marks available because each calculation has a fixed method and a definite answer. Yet students lose them constantly by not showing working, forgetting units, mixing up cm³ and dm³, or ignoring the mole ratio in an equation. This lesson works through every type of maths the specification expects, each with a worked example.

By the end of this lesson you should be able to handle moles and relative formula mass, concentration, percentage yield and composition, atom economy, reacting masses and ratios, gas volumes, standard form, rearranging equations, and significant figures.

Exam Tip: A scientific calculator is permitted on both papers, and a periodic table is printed in the paper — read your relative atomic masses off it. Practise with your calculator before the exam, especially the standard-form button (often labelled EXP, EE or ×10ˣ).

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Relative Formula Mass ($M_r$Mr​)

Almost every chemistry calculation starts with a relative formula mass: add up the relative atomic masses ($A_r$Ar​) of all the atoms in the formula.

Worked example: Calculate the $M_r$Mr​ of calcium carbonate, $\text{CaCO}_3$CaCO3​. ($A_r$Ar​: Ca = 40, C = 12, O = 16.)

$M_r = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100$Mr​=40+12+(3×16)=40+12+48=100

Worked example (with brackets): Calculate the $M_r$Mr​ of calcium hydroxide, $\text{Ca(OH)}_2$Ca(OH)2​. ($A_r$Ar​: Ca = 40, O = 16, H = 1.)

$M_r = 40 + 2 \times (16 + 1) = 40 + 2 \times 17 = 40 + 34 = 74$Mr​=40+2×(16+1)=40+2×17=40+34=74

Exam Tip: Watch the brackets and subscripts. In $\text{Ca(OH)}_2$Ca(OH)2​ the subscript 2 multiplies everything inside the bracket — both the O and the H — so there are two oxygens and two hydrogens. Miscounting atoms here is one of the most common $M_r$Mr​ errors.

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Moles and the Mole Triangle

The mole connects mass to amount of substance:

$\text{moles} = \frac{\text{mass}}{M_r}$moles=Mr​mass​

Rearranged: $\text{mass} = \text{moles} \times M_r$mass=moles×Mr​ and $M_r = \dfrac{\text{mass}}{\text{moles}}$Mr​=molesmass​.

Worked example: How many moles are there in $20 \text{ g}$20 g of calcium carbonate ($M_r = 100$Mr​=100)?

$\text{moles} = \frac{20}{100} = 0.20 \text{ mol}$moles=10020​=0.20 mol

Worked example (finding mass): What is the mass of $0.50 \text{ mol}$0.50 mol of water ($M_r = 18$Mr​=18)?

$\text{mass} = 0.50 \times 18 = 9.0 \text{ g}$mass=0.50×18=9.0 g

Exam Tip: A formula triangle with mass on top, and moles and $M_r$Mr​ in the bottom corners, tells you whether to multiply or divide: cover the quantity you want and read off the arrangement of the other two. Always state the unit (mol for moles, g for mass).

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Reacting Masses and Mole Ratios

To find how much product a reaction makes, you use the balanced equation to get the mole ratio, then convert back to mass.

Worked example: What mass of magnesium oxide is formed when $4.8 \text{ g}$4.8 g of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$2Mg+O2​→2MgO. ($A_r$Ar​: Mg = 24, O = 16.)

Step 1 — moles of magnesium. $M_r$Mr​ of Mg $= 24$=24.

$\text{moles Mg} = \frac{4.8}{24} = 0.20 \text{ mol}$moles Mg=244.8​=0.20 mol

Step 2 — use the equation ratio. The ratio Mg : MgO is 2 : 2 = 1 : 1, so moles of MgO $= 0.20 \text{ mol}$=0.20 mol.

Step 3 — convert to mass. $M_r$Mr​ of MgO $= 24 + 16 = 40$=24+16=40.

$\text{mass MgO} = 0.20 \times 40 = 8.0 \text{ g}$mass MgO=0.20×40=8.0 g

Answer: $8.0 \text{ g}$8.0 g of magnesium oxide.

Exam Tip: The step students most often skip is the mole ratio from the balanced equation. Going straight from grams of reactant to grams of product without the ratio gives the wrong answer whenever the ratio is not 1 : 1. The reliable route is always: mass → moles → (×ratio) → moles → mass.

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Concentration of Solutions

Concentration can be given in $\text{g/dm}^3$g/dm3 or $\text{mol/dm}^3$mol/dm3.

$\text{concentration (g/dm}^3) = \frac{\text{mass (g)}}{\text{volume (dm}^3)} \qquad \text{concentration (mol/dm}^3) = \frac{\text{moles}}{\text{volume (dm}^3)}$concentration (g/dm3)=volume (dm3)mass (g)​concentration (mol/dm3)=volume (dm3)moles​

Worked example: $5.0 \text{ g}$5.0 g of sodium chloride is dissolved in water to make $250 \text{ cm}^3$250 cm3 of solution. Calculate the concentration in $\text{g/dm}^3$g/dm3.

Step 1 — convert the volume. $250 \text{ cm}^3 = 0.250 \text{ dm}^3$250 cm3=0.250 dm3 (divide by 1000).

Step 2 — divide mass by volume.

$\text{concentration} = \frac{5.0}{0.250} = 20 \text{ g/dm}^3$concentration=0.2505.0​=20 g/dm3

To convert between the two units: $\text{concentration (g/dm}^3) = \text{concentration (mol/dm}^3) \times M_r$concentration (g/dm3)=concentration (mol/dm3)×Mr​.

Exam Tip: The single most common concentration error is forgetting to convert cm³ to dm³. There are 1000 cm³ in 1 dm³, so always divide the volume in cm³ by 1000 before dividing into it. A litre and a dm³ are the same thing.

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Percentage Yield

$\text{percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical (maximum) mass}} \times 100$percentage yield=theoretical (maximum) massactual mass of product​×100

Worked example: The maximum (theoretical) mass of product is $8.0 \text{ g}$8.0 g, but only $6.0 \text{ g}$6.0 g is obtained. Calculate the percentage yield.

$\text{percentage yield} = \frac{6.0}{8.0} \times 100 = 75\%$percentage yield=8.06.0​×100=75%

Exam Tip: Percentage yield can never exceed 100% — if yours does, you have divided the wrong way round. Reasons for a yield below 100% include product lost in filtering or transfer, an incomplete or reversible reaction, and side reactions.

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Percentage Composition

The percentage by mass of an element in a compound:

$\% \text{ of element} = \frac{A_r \times \text{number of atoms of that element}}{M_r \text{ of the compound}} \times 100$% of element=Mr​ of the compoundAr​×number of atoms of that element​×100

Worked example: Calculate the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$NH4​NO3​. ($A_r$Ar​: N = 14, H = 1, O = 16.)

Step 1 — find the $M_r$Mr​. $M_r = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80$Mr​=(2×14)+(4×1)+(3×16)=28+4+48=80.

Step 2 — find the total mass of nitrogen. There are 2 nitrogen atoms: $2 \times 14 = 28$2×14=28.

Step 3 — divide and multiply by 100.

$\% \text{ N} = \frac{28}{80} \times 100 = 35\%$% N=8028​×100=35%

Answer: ammonium nitrate is $35\%$35% nitrogen by mass.

Exam Tip: Count every atom of the element — here there are two nitrogens (one in $\text{NH}_4$NH4​ and one in $\text{NO}_3$NO3​), so the mass of nitrogen is $2 \times 14 = 28$2×14=28, not 14. Missing an atom in the formula is the usual slip.

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Atom Economy

Atom economy measures how much of the mass of the reactants ends up as the useful product — important for sustainable, efficient chemistry.

$\text{atom economy} = \frac{M_r \text{ of the useful product}}{\text{total } M_r \text{ of all products}} \times 100$atom economy=total Mr​ of all productsMr​ of the useful product​×100

Worked example: Hydrogen is made by reacting carbon with steam: $\text{C} + 2\text{H}_2\text{O} \rightarrow \text{CO}_2 + 2\text{H}_2$C+2H2​O→CO2​+2H2​. Calculate the atom economy for making hydrogen. ($A_r$Ar​: C = 12, O = 16, H = 1.)

Step 1 — $M_r$Mr​ of the useful product. Useful product is $2\text{H}_2$2H2​: $2 \times (2 \times 1) = 4$2×(2×1)=4.

Step 2 — total $M_r$Mr​ of all products. $\text{CO}_2 = 12 + 32 = 44$CO2​=12+32=44; $2\text{H}_2 = 4$2H2​=4; total $= 44 + 4 = 48$=44+4=48.

Step 3 — divide and multiply by 100.

$\text{atom economy} = \frac{4}{48} \times 100 = 8.3\%$atom economy=484​×100=8.3%

Answer: $8.3\%$8.3% (to 2 s.f.) — a low atom economy, because most of the mass becomes waste carbon dioxide.

Exam Tip: A high atom economy means less waste and is more sustainable and often more economical. Use the totals from the balanced equation (including the big numbers in front), and remember atom economy has no units because it is a ratio of masses.

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Ratios and Empirical Formulae

A ratio compares quantities; in chemistry it links reacting amounts.

Worked example (simplifying a ratio): A reaction uses $0.4 \text{ mol}$0.4 mol of hydrogen for every $0.2 \text{ mol}$0.2 mol of oxygen. Express this as a whole-number ratio.

$0.4 : 0.2 = 2 : 1 \quad (\text{dividing both by } 0.2)$0.4:0.2=2:1(dividing both by 0.2)

Worked example (empirical formula): A compound contains $0.6 \text{ mol}$0.6 mol of carbon and $1.2 \text{ mol}$1.2 mol of hydrogen. Find the empirical formula.

Divide both by the smaller (0.6): C $= \frac{0.6}{0.6} = 1$=0.60.6​=1; H $= \frac{1.2}{0.6} = 2$=0.61.2​=2. Ratio C : H $= 1 : 2$=1:2, so the empirical formula is $\text{CH}_2$CH2​.

Exam Tip: To find an empirical formula, divide each amount by the smallest of them to get the simplest whole-number ratio. If you end up with a number like 1.5, multiply all the values by 2 to clear the fraction.

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Gas Volumes