Chemical Calculations: Ratios, Empirical Formulae & Moles

Mass must always be in grams (g) before using this equation. If an exam question provides mass in other units, you must convert it first: multiply kilograms by $1000$ to get grams, or divide milligrams by $1000$ to get grams. For extremely large or small mole values, Edexcel examiners reward the correct use of standard form.

Calculate the number of moles in $4.5 \text{ kg}$ of water ($H_2O$). (Relative atomic masses: $H = 1$, $O = 16$)

1. Convert the mass into grams: $4.5 \text{ kg} \times 1000 = 4500 \text{ g}$
2. Calculate the relative formula mass ($M_r$): $M_r \text{ of } H_2O = (2 \times 1) + 16 = 18$
3. Substitute and calculate: $\text{moles} = \frac{4500 \text{ g}}{18} = 250 \text{ mol}$

Empirical Formulas

You can describe a bicycle by listing every single part it has, or by simply stating the basic ratio of two wheels to one frame. In chemistry, the molecular formula tells you the exact number and type of atoms in a single molecule. In contrast, the empirical formula represents the simplest whole-number ratio of atoms of each element present in a compound.

For ionic compounds, the standard chemical formula provided (such as $NaCl$) is always its empirical formula. If a question provides element percentages instead of masses, assume you have a $100 \text{ g}$ sample; the percentage value then directly equals the mass in grams.

A compound contains $2.4 \text{ g}$ of carbon and $0.8 \text{ g}$ of hydrogen. Determine its empirical formula. (Relative atomic masses: $C = 12$, $H = 1$)

Step	Carbon ($C$)	Hydrogen ($H$)
1. Mass (g)	$2.4$	$0.8$
2. $A_r$	$12$	$1$
3. Moles (Mass $\div A_r$)	$2.4 \div 12 = 0.2$	$0.8 \div 1 = 0.8$
4. Ratio (Divide by smallest)	$0.2 \div 0.2 = 1$	$0.8 \div 0.2 = 4$
Empirical Formula	$CH_4$

Deducing Stoichiometry from Masses (Higher Tier Only)

How do industrial chemists know exactly how much of each reactant to mix without wasting any? They rely on stoichiometry, which is the relationship between the relative quantities of substances in a reaction. This is shown by the stoichiometric coefficients (the large numbers in front of formulas).

You can determine these coefficients by calculating the molar ratio of all substances involved. You must never alter the small subscript numbers in a chemical formula to balance an equation.

In a reaction, $46 \text{ g}$ of sodium ($Na$) reacts with $71 \text{ g}$ of chlorine gas ($Cl_2$) to produce $117 \text{ g}$ of sodium chloride ($NaCl$). Write the balanced equation with state symbols. (Relative atomic masses: $Na = 23$, $Cl = 35.5$)

1. Calculate $M_r$ for each: $Na = 23$, $Cl_2 = 71$, $NaCl = 58.5$
2. Calculate moles:
   • $Na: 46 \div 23 = 2 \text{ mol}$
   • $Cl_2: 71 \div 71 = 1 \text{ mol}$
   • $NaCl: 117 \div 58.5 = 2 \text{ mol}$
3. Determine molar ratio: $2 : 1 : 2$
4. Write equation: $2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)$

Yield and Reacting Masses

The theoretical yield is the maximum mass of product possible, assuming $100\%$ conversion. The actual yield obtained in practice is always less due to side reactions, reversible processes, or product lost during transfer. The maximum amount of product is restricted by the limiting reactant, which is completely used up first.

$\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

A student reacts $1.4 \text{ kg}$ of lithium ($Li$) with excess oxygen to produce lithium oxide ($Li_2O$). The actual yield is $2.5 \text{ kg}$. Calculate the percentage yield. Equation: $4Li + O_2 \rightarrow 2Li_2O$ ($A_r: Li = 7, O = 16$)

1. Reactant Moles: $1400 \text{ g} \div 7 = 200 \text{ mol } Li$
2. Molar Ratio: $4:2$ ratio (from equation) means $100 \text{ mol } Li_2O$ should form.
3. Theoretical Mass: $M_r \text{ of } Li_2O = 30$. Mass $= 100 \times 30 = 3000 \text{ g}$
4. Percentage Yield: $(2500 \text{ g} \div 3000 \text{ g}) \times 100 = 83.33\%$