Mole ratios are taken from the coefficients of a balanced chemical equation. Convert the known quantity to moles, apply the appropriate coefficient ratio, and then convert the resulting moles into the quantity requested.
This is required at both SL and HL. It applies the mole concept from Structure 1.4 and connects directly to stoichiometric calculations in Reactivity 2.1.
A balanced chemical equation shows the relative amounts, in moles, of all reactants and products. For example:
The coefficients give these mole ratios:
These ratios describe reacting amounts, not necessarily the amounts initially placed in a container. For every two moles of hydrogen consumed, one mole of oxygen is consumed and two moles of water are formed.
The ratio acts as a conversion factor: units for the known substance cancel, leaving moles of the required substance. It is valid only because the equation is balanced and therefore conserves the number of atoms of each element.
Suppose of oxygen reacts with excess hydrogen:
If mass is required, use :
Because hydrogen is in excess, oxygen determines the amount of water formed. The result has three significant figures, matching .
Choose the initial mole conversion from the data supplied:
| Given quantity | Conversion to moles |
|---|---|
| Mass | ; use grams and |
| Solution | ; convert volume to |
| Gas | ; use units consistent with |
Then multiply by the required coefficient ratio and convert into the requested quantity. If two reactant amounts are given, calculate how much product each could form. The reactant producing less product is the limiting reactant; use that result for theoretical yield.
A common misconception is to use subscripts as mole ratios. Subscripts describe one particle's composition; coefficients determine the reacting mole ratio.
For a calculate question, show the mole conversion and coefficient ratio explicitly. Include units throughout and check whether a limiting reactant must be identified first.
