C4.2–3 Equations and Calculations and From Masses to Balanced Equations

A free GCSE Combined Science: Trilogy lesson for AQA (8464) from OBA School: conservation of mass, uncertainty in measurements, reacting masses, balancing equations from masses and limiting reactants. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• Use the law of conservation of mass to explain observed changes in mass in open and closed systems
• Explain what the balancing numbers in a symbol equation tell you about molecules and moles, and estimate the uncertainty of a mean result
• (Higher tier) calculate the mass of a reactant or product from a balanced equation
• (Higher tier) use reacting masses to work out the balancing numbers in an equation
• (Higher tier) identify the limiting reactant and use it to calculate the mass of product

Key words

• Conservation of mass: No atoms are lost or made in a chemical reaction, so the total mass of the products equals the total mass of the reactants.
• Closed system: A container that nothing can enter or leave, such as a sealed flask.
• Reacting masses: The masses of substances that react together and are made, calculated from the balanced equation.
• Limiting reactant: The reactant that is completely used up. It limits the amount of product that can form. (Higher tier)
• In excess: A reactant that is left over when the reaction stops, because there was more than enough of it.
• Thermal decomposition: A reaction in which heat breaks a substance down into simpler substances, e.g. a metal carbonate → metal oxide + carbon dioxide.
• Uncertainty: How far either side of the mean the true value could lie. Estimated as half the range of the repeat results.

Conservation of mass

In a chemical reaction the atoms are rearranged, but none are created or destroyed. So the total mass of the products equals the total mass of the reactants. This is why symbol equations must be balanced.

Sometimes a reaction seems to change mass. This always involves a gas:

• Mass goes up when a reactant is a gas from the air. When magnesium burns, oxygen atoms join on: 2Mg(s) + O₂(g) → 2MgO(s). The solid gets heavier.
• Mass goes down when a product is a gas that escapes. When copper carbonate is heated in an open tube, carbon dioxide leaves: CuCO₃(s) → CuO(s) + CO₂(g).

Key fact: If the mass seems to change, ask: “Did a gas get in, or did a gas get out?” Mass is still conserved if you count the gas.

Uncertainty in measurements

Whenever you measure something there is some uncertainty: repeat readings are spread out because of small random errors (for example, a balance that reads to 0.01 g, or a little solid lost from a crucible). The range of the repeats shows how spread out they are. The mean is the best estimate of the true value, and the uncertainty is about half the range.

mean = sum of readings ÷ number of readings

uncertainty ≈ ± range ÷ 2 (range = highest − lowest)

Worked example: mass of carbon dioxide lost when zinc carbonate is heated

Five repeats: 1.32 g, 1.36 g, 1.30 g, 1.34 g, 1.33 g
mean = 6.65 ÷ 5 = 1.33 g
range = 1.36 − 1.30 = 0.06 g, so uncertainty = ± 0.03 g

Answer: mass of CO₂ = 1.33 ± 0.03 g

Exam tip: Leave out any anomalous result before you work out the mean and the range. The smaller the uncertainty, the more precise the results.

What a balanced equation tells you

The big numbers in front of the formulae are called balancing numbers (or multipliers). They tell you the ratio of particles that react. Because a mole is a fixed number of particles, they also tell you the ratio of moles. So 2H₂O means two molecules of water, or two moles of water.

You may see a fraction in an equation, especially for burning fuels. You cannot have half a molecule, but you can have half a mole. To get whole numbers, multiply every balancing number by 2:

Worked example: removing a fraction

C₄H₁₀ + 6½O₂ → 4CO₂ + 5H₂O (13 O atoms on each side)
× 2: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Answer: still balanced: 8 C, 20 H and 26 O on each side

Calculating reacting masses (Higher tier)

The big numbers in a balanced equation give the ratio of moles. To find a mass from another mass, follow three steps:

1. Moles of the substance you know: moles = mass ÷ Mr

2. Use the equation’s ratio to find moles of the substance you want.

3. Mass of the substance you want = moles × Mr

Worked example: mass of magnesium oxide made from 3.6 g of magnesium

2Mg + O₂ → 2MgO
1. moles Mg = 3.6 ÷ 24 = 0.15 mol
2. ratio Mg : MgO = 2 : 2 = 1 : 1, so moles MgO = 0.15 mol
3. mass MgO = 0.15 × 40 = 6.0 g

Answer: 6.0 g of MgO

Worked example: heating 15 g of calcium carbonate

CaCO₃ → CaO + CO₂
moles CaCO₃ = 15 ÷ 100 = 0.15 mol, so moles CaO = 0.15 mol
mass CaO = 0.15 × 56 = 8.4 g
By conservation of mass, mass CO₂ = 15 − 8.4 = 6.6 g (check: 0.15 × 44 = 6.6 g)

Answer: 8.4 g CaO and 6.6 g CO₂

Exam tip: Give your answer to the same number of significant figures as the data in the question (you do not need to count the significant figures of the Ar values). For example, 3.6 g of magnesium (2 s.f.) gives 6.0 g of MgO, but 3.60 g (3 s.f.) gives 6.00 g.

Balancing equations from masses (Higher tier)

If you know the masses that react and form, you can work out the balancing numbers. Turn every mass into moles, then divide by the smallest number of moles to get a whole-number ratio.

Worked example: potassium burning in oxygen

7.8 g of potassium reacts with 1.6 g of oxygen, O₂, to form 9.4 g of potassium oxide, K₂O. (Ar: K = 39, O = 16)
moles: K = 7.8 ÷ 39 = 0.20; O₂ = 1.6 ÷ 32 = 0.050; K₂O = 9.4 ÷ 94 = 0.10
divide by 0.050: K : O₂ : K₂O = 4 : 1 : 2

Answer: 4K + O₂ → 2K₂O

Limiting reactants (Higher tier)

Reactions are often done with one reactant in excess to make sure all of the other one (often the more expensive one) reacts. The reactant that is used up is the limiting reactant. The amount of product depends only on the limiting reactant: double it and you double the product.

Worked example: 8.4 g of iron is heated with 4.0 g of sulfur

Fe + S → FeS
moles Fe = 8.4 ÷ 56 = 0.15; moles S = 4.0 ÷ 32 = 0.125
The ratio is 1 : 1, so sulfur runs out first. Sulfur is the limiting reactant and iron is in excess.
moles FeS = 0.125, so mass FeS = 0.125 × 88 = 11 g

Answer: 11 g of iron sulfide

Exam tip: To find the limiting reactant, divide the moles of each reactant by its balancing number. The smallest answer is the limiting reactant.

Common mistakes

• Saying “mass is lost” when a gas escapes. The mass is still there – it is in the gas that left the container.
• Using the mass ratio instead of the mole ratio. The balancing numbers compare moles, not grams.
• Multiplying Mr by the balancing number and using the mole ratio – this counts the ratio twice.
• Picking the reactant with the smaller mass as limiting. Compare moles (divided by the balancing numbers), not masses.

Revision questions

Try these, then check your answers below each one.

Question 1. The diagram shows marble chips reacting with dilute acid in an open flask on a balance. Why does the reading go down? (Choose from: Carbon dioxide gas is made and escapes from the flask, Atoms are destroyed in the reaction, The acid evaporates because the reaction is cold, The marble chips get lighter because they get smaller atoms.)

Answer: Carbon dioxide gas is made and escapes from the flask

Working: Calcium carbonate + acid makes carbon dioxide gas, which escapes from the open flask. Mass is conserved – the missing 0.88 g is the mass of the gas that left.

Question 2. (Higher tier) Calcium carbonate decomposes when heated: CaCO₃ → CaO + CO₂. (Aᵣ: C = 12, O = 16, Ca = 40)<br/>Calculate the mass of calcium oxide made from 40 g of calcium carbonate.

Answer: 22.4 g

Working: moles CaCO₃ = 40 ÷ 100 = 0.40 mol, so 0.40 mol CaO. Mᵣ CaO = 56; mass = 0.40 × 56 = 22.4 g.

Question 3. A reactant is left over when a reaction stops. What word describes this reactant?

Answer: excess

Working: It is in excess: there was more of it than needed to react with all of the limiting reactant.

Related lessons

• Previous lesson: C4.1 Relative Masses and Moles
• Next lesson: C4.4 Expressing Concentrations

Continue in the full course

This lesson is section C4.2–3 (AGCSC2102) of our free GCSE Combined Science course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Combined Science course.