Moles, concentrations and chemical equations

What you’ll learn

• Convert between mass, amount and particle number.
• Calculate solution concentration with consistent units.
• Balance full and ionic equations including charge and states.

Understand the idea

A mole counts a specified type of entity: atoms, molecules, ions or formula units. Divide mass by molar mass to obtain amount, then multiply by the Avogadro constant to count entities. Relative atomic and formula masses are ratios and have no units; molar mass has units such as $\mathrm{g\,mol^{-1}}$.

Amount concentration is moles of solute per cubic decimetre of solution. Convert $\mathrm{cm^3}$ into $\mathrm{dm^3}$ by dividing by 1000 before using $n=cV$. Mass concentration in $\mathrm{g\,dm^{-3}}$ equals amount concentration multiplied by molar mass. Parts per million means one part per million on the stated basis; for gas mixtures, volume fraction and mole fraction agree at the same conditions.

Balance an equation by changing coefficients, never the subscripts that identify a substance. Check each element and the overall charge. An ionic equation removes spectator ions that remain unchanged in aqueous solution. Keep solids, liquids and gases intact rather than splitting them into free ions.

Key formulae

$$n=\frac mM,\qquad N=nN_A$$

$$n=cV\quad(V\text{ in }\mathrm{dm^3})$$

$$c_{\mathrm{mass}}=cM$$

$$\mathrm{Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)}$$

Worked example

A solution contains $5.85$ g of sodium chloride in $500\ \mathrm{cm^3}$. Using $M(\mathrm{NaCl})=58.5\ \mathrm{g\,mol^{-1}}$, find its amount concentration.

1. The amount is $n=5.85/58.5=0.100$ mol.
2. Convert the final solution volume: $500\ \mathrm{cm^3}=0.500\ \mathrm{dm^3}$.
3. Divide amount by volume: $c=0.100/0.500=0.200\ \mathrm{mol\,dm^{-3}}$.

Answer: $0.200\ \mathrm{mol\,dm^{-3}}$.