# Mass in a chemical reaction: using moles

I can interpret a balanced symbol equation to predict the mass of either a reactant or a product.

# Mass in a chemical reaction: using moles

I can interpret a balanced symbol equation to predict the mass of either a reactant or a product.

## Lesson details

### Key learning points

- In a chemical reaction, the atoms in reactants are rearranged and are the same atoms that are in the products.
- Chemical equations are most correctly interpreted in terms of molar ratios (i.e. 1 mole of A reacts with 2 moles of B).
- The molar ratios of the substances in a chemical reaction is known as stoichiometry, which means ‘measuring elements’.
- The stoichiometry of a reaction is shown by coefficients in a balanced equation. These molar relationships don't change.
- The moles of substance in a reaction can be determined using: mass(g) = Mr × moles (& the stoichiometry of reaction).

### Common misconception

Pupils struggle to recall the order of steps required to mathematically process the available information in order to answer the question.

Colour-coding the steps can help pupils remember how many steps are involved. Much practice and perseverance is needed to create and maintain the memory pathway of the mathematical processing.

### Keywords

Stoichiometry - Stoichiometry refers to the molar ratio of the reactants and products in a chemical reaction.

Relative formula mass - The relative formula mass of a substance is the sum of the relative atomic masses of all the atoms in a formula.

Mole - A mole of something is 6.02 × 10²³ of it. The mass of a mole of a substance is its relative mass expressed in grams.

Coefficient - A coefficient is the number placed in front of a chemical formula to balance an equation; it multiplies all the atoms in the formula and shows the ratio of substances in a reaction.

Balanced symbol equation - A balanced symbol equation describes a reaction using a symbol equation with coefficients, which ensure there are equal numbers of atoms of each element on both sides of the symbol equation.

### Licence

This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

## Video

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