Formulae and the Mole

3.1 Formulae

A formula uses chemical symbols and numbers to show the atoms in a substance. The small number after a symbol is the number of atoms of that element, for example $\ce{H2SO4}$ has two hydrogen, one sulfur and four oxygen atoms.

Glucose has the molecular formula $\ce{C6H12O6}$ and the empirical formula $\ce{CH2O}$. Ethene is $\ce{C2H4}$ and its empirical formula is $\ce{CH2}$. For ionic compounds and giant covalent substances such as $\ce{NaCl}$ and $\ce{SiO2}$, the formula is always the empirical formula.

Formulae from a diagram. To deduce a formula from a model or diagram, count the atoms of each element in one unit. For example, a diagram of a molecule with one carbon atom bonded to four hydrogen atoms gives $\ce{CH4}$.

The formula of an ionic compound

An ionic compound is neutral overall, so the total positive charge equals the total negative charge. Use the charges on the ions.

(a) $\ce{Ca^2+}$ needs two $\ce{Cl-}$ to balance: $\ce{CaCl2}$. (b) $\ce{Al^3+}$ and $\ce{O^2-}$: the lowest common multiple of $3$ and $2$ is $6$, so two $\ce{Al^3+}$ ($+6$) and three $\ce{O^2-}$ ($-6$): $\ce{Al2O3}$. (c) $\ce{NH4+}$ and $\ce{SO4^2-}$: two $\ce{NH4+}$ are needed, and brackets are used around a group of atoms: $\ce{(NH4)2SO4}$.

A Roman numeral in a name gives the charge on a metal ion: iron(III) oxide contains $\ce{Fe^3+}$ and $\ce{O^2-}$, so its formula is $\ce{Fe2O3}$.

Equations

A word equation uses names. A symbol equation uses formulae and must be balanced: the same number of each type of atom on both sides. Never change a formula to balance an equation; change the numbers in front.

State symbols: $(\text{s})$ solid, $(\text{l})$ liquid, $(\text{g})$ gas, $(\text{aq})$ dissolved in water.

$\text{magnesium} + \text{oxygen} \to \text{magnesium oxide}$

$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$

$\ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)}$

$\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}$

Ionic equations

An ionic equation shows only the ions (and substances) that actually take part in the reaction. The ions that are unchanged, called spectator ions, are left out. Both the atoms and the charges must balance.

For silver nitrate and sodium chloride solutions, a white precipitate of silver chloride forms:

$\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}$

The $\ce{Na+}$ and $\ce{NO3-}$ ions are spectators, so the ionic equation is:

$\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$

To write an ionic equation: write the balanced symbol equation with state symbols, split everything that is $(\text{aq})$ and ionic into ions, cancel the ions that are the same on both sides, and write what remains.

3.2 Relative masses of atoms and molecules

$A_\text{r}$ and $M_\text{r}$ have no units.

(a) $(2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98$. (b) $40 + 2 \times (16 + 1) = 40 + 34 = 74$. (c) $\ce{CuSO4}$ is $64 + 32 + 64 = 160$ and $5\ \ce{H2O}$ is $5 \times 18 = 90$, so $160 + 90 = 250$.

3.3 The mole and the Avogadro constant

The molar mass is the mass of one mole, in $\text{g/mol}$. It has the same number as the $A_\text{r}$ or $M_\text{r}$. So one mole of carbon is $12\ \text{g}$ and one mole of $\ce{CO2}$ is $44\ \text{g}$.

$\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \qquad n = \frac{m}{M}$

$\text{number of particles} = n \times 6.02 \times 10^{23}$

(a) $M_\text{r} = 40 + 12 + 48 = 100$, so $m = 0.25 \times 100 = 25\ \text{g}$. (b) $M_\text{r} = 23 + 16 + 1 = 40$, so $n = \dfrac{4.0}{40} = 0.10\ \text{mol}$. (c) $0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ molecules.

Gases

$\text{volume of gas (dm}^3\text{)} = n \times 24 \qquad \text{or} \qquad n = \frac{\text{volume in dm}^3}{24} = \frac{\text{volume in cm}^3}{24\,000}$

$1\ \text{dm}^3 = 1000\ \text{cm}^3$, so divide by $1000$ to change $\text{cm}^3$ into $\text{dm}^3$.

Concentration

The concentration of a solution is the amount of solute in $1\ \text{dm}^3$ of solution, in either $\text{g/dm}^3$ or $\text{mol/dm}^3$.

$\text{concentration (g/dm}^3\text{)} = \frac{\text{mass (g)}}{\text{volume (dm}^3\text{)}} \qquad \text{concentration (mol/dm}^3\text{)} = \frac{n}{\text{volume (dm}^3\text{)}}$

$\text{concentration in mol/dm}^3 = \frac{\text{concentration in g/dm}^3}{M_\text{r}} \qquad n = \text{concentration} \times \text{volume in dm}^3$

Volume $= 250 \div 1000 = 0.250\ \text{dm}^3$. In $\text{g/dm}^3$: $\dfrac{5.0}{0.250} = 20\ \text{g/dm}^3$. In $\text{mol/dm}^3$: $\dfrac{20}{40} = 0.50\ \text{mol/dm}^3$.

Reacting masses and gas volumes

The balanced equation gives the ratio of moles. The method:

1. Write the balanced equation.
2. Work out the moles of the substance you are given ($n = m \div M_\text{r}$, or from volume or concentration).
3. Use the mole ratio from the equation to find the moles of the substance you want.
4. Convert to a mass, a volume of gas, or a concentration.

$n(\ce{CaCO3}) = \dfrac{10}{100} = 0.10\ \text{mol}$. The ratio is $1:1$, so $n(\ce{CaO}) = 0.10\ \text{mol}$. $M_\text{r}(\ce{CaO}) = 56$, so mass $= 0.10 \times 56 = 5.6\ \text{g}$.

$n(\ce{H2}) = \dfrac{120}{24\,000} = 0.0050\ \text{mol}$. The ratio $\ce{Mg}:\ce{H2}$ is $1:1$, so $n(\ce{Mg}) = 0.0050\ \text{mol}$ and the mass is $0.0050 \times 24 = 0.12\ \text{g}$.

Limiting reactants

The limiting reactant is the one that is used up first, and it decides how much product forms. The other reactant is in excess. To find it, work out the moles of each reactant and compare with the equation.

$n(\ce{Mg}) = \dfrac{2.4}{24} = 0.10\ \text{mol}$ and $n(\ce{HCl}) = 1.0 \times 0.100 = 0.10\ \text{mol}$. The equation needs $2$ mol of $\ce{HCl}$ for every $1$ mol of $\ce{Mg}$, so $0.10\ \text{mol}$ of $\ce{Mg}$ needs $0.20\ \text{mol}$ of $\ce{HCl}$. The acid is limiting. $n(\ce{H2}) = 0.10 \div 2 = 0.050\ \text{mol}$, so the volume is $0.050 \times 24 = 1.2\ \text{dm}^3$.

Titration calculations

In a titration, the volumes of two solutions that react exactly are measured, and the concentration of one is found from the other.

$n(\ce{HCl}) = 0.100 \times \dfrac{20.0}{1000} = 0.00200\ \text{mol}$. The ratio is $1:1$, so $n(\ce{NaOH}) = 0.00200\ \text{mol}$. Concentration $= \dfrac{0.00200}{0.0250} = 0.0800\ \text{mol/dm}^3$.

Empirical and molecular formulae

To find the empirical formula from masses or percentages: divide each by its $A_\text{r}$ to get moles, then divide by the smallest to get the simplest whole-number ratio.

Carbon: $\dfrac{85.7}{12} = 7.14$. Hydrogen: $\dfrac{14.3}{1} = 14.3$. Dividing by $7.14$ gives the ratio $1:2$, so the empirical formula is $\ce{CH2}$ (mass $14$). $\dfrac{56}{14} = 4$, so the molecular formula is $\ce{C4H8}$.

$\ce{Mg}$: $\dfrac{4.8}{24} = 0.20$. $\ce{O}$: $\dfrac{3.2}{16} = 0.20$. The ratio is $1:1$, so the formula is $\ce{MgO}$.

Percentage yield, composition and purity

$\text{percentage yield} = \frac{\text{actual mass obtained}}{\text{maximum (theoretical) mass}} \times 100$

$\text{percentage composition by mass of an element} = \frac{\text{mass of the element in one mole of the compound}}{M_\text{r}} \times 100$

$\text{percentage purity} = \frac{\text{mass of pure substance}}{\text{mass of the impure sample}} \times 100$

(a) $\dfrac{4.2}{5.6} \times 100 = 75\%$. (b) $M_\text{r} = 14 + 4 + 14 + 48 = 80$, and there are $2$ nitrogen atoms ($28$), so $\dfrac{28}{80} \times 100 = 35\%$. (c) $n(\ce{CO2}) = \dfrac{0.36}{24} = 0.015\ \text{mol}$, so $n(\ce{CaCO3}) = 0.015\ \text{mol}$ and its mass is $0.015 \times 100 = 1.5\ \text{g}$. Purity $= \dfrac{1.5}{2.0} \times 100 = 75\%$.