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
Loading...