AQA Chemistry 5.3.2.2 · Higher only

Amounts of substances in equations

The big numbers in a balanced equation are a ratio of moles. Turn a mass into moles, cross the ratio, turn the moles back into a mass: three steps, each learned on its own first.

Triple Science · Higher tier

Calculates

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A recipe.

A cake recipe needs 2 eggs per cake. You are making 3 cakes.

If you had to guess, do you need 6 eggs or 3 eggs?

Coefficients are a mole ratio

Scale one bar and the others follow.

The balancing numbers in a symbol equation give the ratio of moles of each substance. Set the moles of magnesium and watch the oxygen and the oxide. To go from one substance to another, multiply by the wanted coefficient and divide by the given one.

N₂ + 3H₂ 2NH₃. From 0.6 mol of H₂, how many moles of NH₃?

Pick one.

Masses in a balanced equation

Read the equation in grams.

Multiply each coefficient by the M_r and the equation reads as masses. Conservation of mass checks it: the two sides add up.

2Mg + O₂ 2MgO

2 × 24 + 1 × 32 2 × 40

48 g + 32 g 80 g

N₂ + 3H₂ 2NH₃. A_r: N = 14, H = 1. Which line is right?

Pick one.

Equation · moles from mass

The same triangle as Moles. It is used twice in a reacting-mass question: forwards for the substance you are given (Step 1), backwards for the one you want (Step 3). The ratio step between them is not a triangle.

Equation

Moles from mass

Side by side: multiply.
One over the other: divide.

Tap the quantity you want to find.

Written out

n = m ÷ Mr

m: mass; g
n: amount of substance; mol
Mr: relative formula mass (Ar for an element); no unit

Step 2 · n(wanted) = n(given) × wanted coefficient ÷ given coefficient

Think again · which way is the ratio?

“3H₂ makes 2NH₃, so 0.6 mol H₂ × 3/2 = 0.9 mol NH₃.”

Your call

Choose

N₂ + 3H₂ → 2NH₃. A pupil gets 0.9 mol NH₃ from 0.6 mol H₂. What went wrong?

Sort

Which step is this line?

Place each line of working.

Tap a card, then tap the box it belongs in. Tap a placed card to take it back.

Not on this page

Percentage yield and atom economy are Chemistry-only and have their own lessons: Percentage yield and Atom economy. Balancing an equation from reacting masses, and limiting reactants, are in Using moles in calculations.

Key fact

Mass to moles (÷ M_r), cross the mole ratio from the coefficients, moles to mass (× M_r). Grams throughout. Use each coefficient once, in the ratio.

Key note · AQA 5.3.2.2 · Higher only

Amounts of substances in equations

Chemical equations can be interpreted in terms of moles: Mg + 2HCl → MgCl₂ + H₂ shows that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of magnesium chloride and one mole of hydrogen gas.
Masses in a balanced equation: coefficient × Mr for each substance; the two sides add up (2Mg + O₂ → 2MgO reads 48 g + 32 g → 80 g).
Reacting masses in three steps. Step 1: n(given) = m ÷ Mr (grams; convert kg or tonnes first). Step 2: n(wanted) = n(given) × wanted coefficient ÷ given coefficient. Step 3: m(wanted) = n × Mr.
Use each coefficient once, in the ratio. Hydrogen and oxygen gases are H₂ (Mr 2) and O₂ (Mr 32).
Worked: 1.2 g Mg with excess acid gives 0.05 mol Mg, 0.05 mol H₂, 0.1 g of hydrogen.

Photograph this card. Tomorrow, cover it and say each line first.

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Flipping the ratio the wrong way?

Relative atomic masses are rounded as given in each question.