Amounts of substance in equations; calculating moles from mass; using moles to balance equations and find masses; the mole calculation triangle.

What this dot point is asking

AQA wants you to convert confidently between mass, moles and number of particles, use the moles equation, and use the mole ratio from a balanced equation to relate amounts of reactants and products. The mole is the central idea in 4.3 Quantitative chemistry: once a measured mass is turned into an amount in moles, every other calculation (reacting masses, concentration, percentage yield) follows from the same simple ratios. The mole-based balancing and calculations are Higher tier (foundation candidates use ratios of formula mass instead).

Why the mole exists

Atoms are far too small and too numerous to count individually, so chemists measure amount in moles. One mole is defined as the number of particles equal to the Avogadro constant, $6.02 \times 10^{23}$. This number is chosen so that one mole of a substance has a mass in grams numerically equal to its relative formula mass. The mole therefore acts as a bridge between the laboratory scale (grams you can weigh) and the particle scale (atoms, molecules and ions that actually react in fixed whole-number ratios).

The moles equation

For an element, use the relative atomic mass $A_r$; for a compound, use the relative formula mass $M_r$ (the sum of the $A_r$ values for every atom in the formula). For example, the moles in $24$ g of carbon ($A_r = 12$) is $24 / 12 = 2$ mol, and the moles in $80$ g of sodium hydroxide ($M_r = 40$) is $80 / 40 = 2$ mol. The "mole triangle" with mass on top and moles times $M_r$ underneath is a memory aid, but understanding the single equation and rearranging it is more reliable in the exam.

Number of particles

The number of particles is found from the moles and the Avogadro constant:

So $2$ mol of carbon contains $2 \times 6.02 \times 10^{23} = 1.204 \times 10^{24}$ atoms. Be careful with what the "particle" is: for an element such as carbon it is atoms, for a molecular compound such as water it is molecules, and for an ionic compound you may be asked for formula units. If a question asks for the number of atoms in a molecular substance, find the number of molecules first, then multiply by the number of atoms in each molecule.

Using mole ratios (Higher)

For example, in $2H_2 + O_2 \rightarrow 2H_2O$, the ratio of $H_2$ to $H_2O$ is $2:2$, so $4$ mol of hydrogen makes $4$ mol of water, while the ratio of $H_2$ to $O_2$ is $2:1$, so the same $4$ mol of hydrogen needs only $2$ mol of oxygen. The mole ratio is the heart of every reacting-mass and titration calculation: masses and volumes are converted to moles precisely so that this ratio can be applied, then converted back.

Method for reacting-amount problems

1. Convert the known mass to moles ($\text{moles} = \text{mass} / M_r$).
2. Use the mole ratio from the balanced equation.
3. Convert the answer back to mass if needed ($\text{mass} = \text{moles} \times M_r$).

Try this

Q1. Calculate the number of moles in $36$ g of water ($M_r = 18$). [1 mark]

• Cue. $36 / 18 = 2$ mol.

Q2. In $N_2 + 3H_2 \rightarrow 2NH_3$, how many moles of ammonia form from $3$ mol of hydrogen? [1 mark]

• Cue. The ratio of $H_2$ to $NH_3$ is $3:2$, so $3$ mol of hydrogen gives $2$ mol of ammonia.

Q3. Calculate the mass of $1.5$ mol of sodium chloride, $NaCl$ ($A_r$: Na = 23, Cl = 35.5). [2 marks]

• Cue. $M_r = 23 + 35.5 = 58.5$; mass $= 1.5 \times 58.5 = 87.75$ g.