3.1 Chemical measurements, conservation of mass and the quantitative interpretation of chemical equations

3.1.1 Conservation of mass and balanced chemical equations

- The law of conservation of mass states that no atoms are lost or made during a chemical reaction, therefore the mass of the product equals the mass of the reactants.
- This means that chemical reactions can be represented by symbol equations which are balanced in terms of the numbers of atoms of each element involved on both sides of the equation.

Simple example of balancing equations

Example 1: Balance the equation for the reaction between hydrogen and oxygen to form water.
Unbalanced: H2 + O2 → H2O

Step 1: Count the number of atoms on each side.
- Left side: 2 H, 2 O
- Right side: 2 H, 1 O

Step 2: Balance oxygen by putting a 2 in front of H2O:
H2 + O2 → 2H2O

Step 3: Now hydrogen is unbalanced (4 on right, 2 on left). Add a 2 in front of H2:
2H2 + O2 → 2H2O

✔ Balanced equation: 2H2 + O2 → 2H2O

3.1.2 Relative formula mass

- The relative formula mass or Mr of a compound is the sum of the relative atomic masses (Ars) of the atoms in the numbers shown in the formula.
- For example, the relative formula mass of water (H2O) is 2 * H (1) + 1 * O (16) = 2 + 16 = 18.
- In a balanced chemical equation, the sum of the relative formula masses of the reactants in the quantities shown equals the sum of the relative formula masses of the products in the quantities shown.

Demonstration of relative formula mass with balanced equations

For example (formula for photosynthesis):
6CO2 + 6H2O → C6H12O6 + 6O2
- Mr of carbon dioxide = 1 * C (12) + 2 * O (16) = 12 + 32 = 44.
- Account for the six carbon dioxide molecules: 44 * 6 = 264.
- Mr of water = 2 * H (1) + 1 * O (16) = 2 + 16 = 18.
- Account for the six water molecules: 18 * 6 = 108.
- Total Mr of reactants = 264 + 108 = 372.
- Mr of glucose = 6 * C (12) + 12 * H (1) + 6 * O (16) = 72 + 12 + 96 = 180.
- Mr of oxygen = 2 * O (16) = 32.
- Accounting for the six oxygen molcules: 32 * 6 = 192.
- Total Mr of products = 180 + 192 = 372.
- As this is a balanced equation, Mr of reactants = Mr of products.

Demonstration: Percentage by mass in a compound

To find the percentage by mass of an element in a compound:
1) Calculate the compound’s relative formula mass (Mr).
2) Find the total mass contributed by the element of interest.
3) Use: % by mass = (mass of that element in the formula ÷ Mr) × 100.

Example 1: H2SO4 (sulfuric acid)

Using Ar values: H = 1, S = 32, O = 16
- Mr(H2SO4) = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98.
- Mass from H = 2 × 1 = 2 → %H = (2 ÷ 98) × 100 ≈ 2.0%.
- Mass from S = 32 → %S = (32 ÷ 98) × 100 ≈ 32.7%.
- Mass from O = 4 × 16 = 64 → %O = (64 ÷ 98) × 100 ≈ 65.3%.
- Check: 2.0 + 32.7 + 65.3 ≈ 100% (allowing for rounding).

Example 2: CaCO3 (calcium carbonate)

Using Ar: Ca = 40, C = 12, O = 16
- Mr(CaCO3) = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100.
- %Ca = (40 ÷ 100) × 100 = 40%.
- %C = (12 ÷ 100) × 100 = 12%.
- %O = (48 ÷ 100) × 100 = 48%.


3.1.3 Mass changes when a reactant or product is a gas

- Some reactions may appear to involve a change in mass but this can usually be explained because a reactant or product is a gas and its mass has not been taken into account.
- For example, when a metal reacts with oxygen the mass of the oxide produced is greater than the mass of the metal. This is because the oxygen has come from the air, and the metal is the same mass as it was before, and oxide mass = metal mass + oxygen mass.
- In the thermal decompositions of metal carbonates carbon dioxide is produced and escapes into the atmosphere leaving the metal oxide as the only solid product. The mass of the carbon dioxide is removed from the metal carbonate, and metal carbonate mass = metal mass - carbon dioxide mass.

Explaining observed changes in mass

- You may be given a question in the exam which asks you to explain why mass increased or decreased in a reaction (with the question of course wanting you to cite the above as the reason).
- The question will contain a balanced chemical equation, for example 2Cu + O₂ → 2CuO.
- Here, mass will be gained from the oxygen and become part of the copper oxide.


3.1.4 Chemical measurements

- Whenever a measurement is made there is always some uncertainty about the result obtained; nothing is completely accurate.
- An error refers to the difference between a measured or estimated value and the true value of a quantity.
- There are two types of errors: systematic error and random error.

Random errors

- In a random error, there is an equal chance the value is too high or too low.
- Random errors at GCSE are most likely to be caused by:

Systematic errors

- In a systematic error, the value will always be too high or too low no matter how many times the experiment is repeated.
- This is because they are errors in the experimental procedure itself.
- Systematic errors at GCSE are most likely to be caused by:

Measuring uncertainty

- Uncertainty tells us how confident we can be in a measurement. It is not the same as an error, but instead describes the range of values within which the true result is likely to lie.
- At GCSE, there are three main ways to measure uncertainty:

- To reduce uncertainty, you can take more measurements and calculate a mean, use more precise instruments, or improve the experimental method.