Chemical formulae, equations and calculations — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

Chemical formulae, equations and calculations Edexcel O Level Linear Chemistry: what to do next

These Chemical formulae, equations and calculations Edexcel O Level Linear Chemistry notes covered the key ideas, worked examples and exam technique for this topic. Check your Chemical formulae, equations and calculations Edexcel O Level Linear Chemistry understanding with the practice questions below, then reread any Chemical formulae, equations and calculations Edexcel O Level Linear Chemistry point you were unsure about before moving on.

The symbol for relative atomic mass is A_r, and for relative formula mass M_r.

Chemical amounts are measured in moles (mol).

The molar mass is the mass of 1 mole in g/mol.

In the laboratory we measure grams, not moles, so we convert using: moles = mass / molar mass. Rearranged: mass = moles x molar mass; molar mass = mass / moles.

Example Amass of 0.250 mol of zinc = 0.250 x 65.4 = 16.3 g.

Always show working, since marks are awarded for method even if the final answer is wrong.

Equal masses of different substances contain different numbers of particles.

Compare amounts

• 18 g of water (1 mol) and 44 g of carbon dioxide (1 mol) hold the same number of molecules even though the masses differ.
• This is the reason why chemists count in moles, not grams, when comparing quantities in a reaction.

To find the mass of a product from a given reactant: convert mass to moles, use the molar ratio in the balanced equation, then convert moles to mass.

Check each stage

• after finding the moles of the given substance, write the ratio line and the moles of the wanted substance before converting to mass.
• Check the final mass is sensible relative to the starting mass and the equation.
• Use the same method for gases, solutions and solids.

The yield is the amount of product obtained. The theoretical yield is the amount that would be obtained if the reaction went to completion with no losses, calculated from the balanced equation. The actual yield is the amount really collected, found by weighing. In practice 100% is never obtained because:

Percentage yield = (actual yield / theoretical yield) x 100. Only the theoretical yield can be calculated from the equation; the actual and percentage yields come from experiment.

Full worked example

• 4.0 g of copper(II) oxide (M_r 80) reacts with excess sulfuric acid to form CuSO₄ (M_r 160).
• Moles of CuO = 0.050;
  • ratio 1:1;
  • theoretical mass of CuSO₄ = 0.050 x 160 = 8.0 g.
• If 6.4 g is collected the percentage yield = 6.4/8.0 x 100 = 80%.

If you get more than 100%, you have inverted the fraction or the product is wet or impure.

For economic reasons, chemical companies want the highest possible percentage yield, to raise profit and reduce cost and waste.

A calculation using yield in reverseif you need 8.0 g of a product and the process has a percentage yield of 80%, the theoretical yield must be 8.0 / 0.80 = 10.0 g, so the reactants required are those that would give 10.0 g.

The molecular formula shows the actual number and type of each atom in a molecule: ethanoic acid is C₂H₄O₂. The empirical formula is the simplest whole-number ratio of atoms of each element in the compound: for ethanoic acid it is CH₂O. Ionic compounds are always written as empirical formulae.

Method

• write the element symbols;
  • write the mass (or percentage) of each;
  • divide by A_r to get moles;
  • divide by the smallest number;
  • if the results are not whole numbers (for example 1.5) multiply all to make whole numbers;
  • write the formula.

Example 1. A compound contains 10 g of hydrogen and 80 g of oxygen. Moles H = 10/1 = 10; moles O = 80/16 = 5. Ratio 10 : 5 = 2 : 1. Empirical formula H₂O.

Example 2. A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Moles: C 40.0/12 = 3.33; H 6.7/1 = 6.7; O 53.3/16 = 3.33. Divide by 3.33: 1 : 2 : 1. Empirical formula CH₂O.

Example 3. The ratio 1 : 1.5 becomes 2 : 3 on doubling; an empirical formula of C₂H₄O₃ shows the ratio must be whole numbers.

The molecular formula is a whole-number multiple of the empirical formula.

Check your answer by calculating the relative formula mass of the molecular formula and comparing with the given M_r: C₂H₄O₂ gives 24 + 4 + 32 = 60, which matches.

Reduce a weighed sample of copper(II) oxide in a horizontal tube with a steady stream of methane, burning off the excess gas.

Another way to find a metal oxide formula is to burn magnesium in a crucible: weigh the crucible, add a coil of magnesium ribbon and weigh again, heat strongly with the lid slightly open to let air in but keep smoke from escaping, cool and reweigh.

The concentration-moles relationship is concentration (mol/dm³) = moles / volume (dm³).

Reaction example

• 20.0 cm³ of 0.500 mol/dm³ sulfuric acid is neutralised by sodium hydroxide, H₂SO₄ + 2NaOH to Na₂SO₄ + 2H₂O.
• Moles acid = 0.0100;
  • moles NaOH = 0.0200;
  • if the NaOH is 1.00 mol/dm³ the volume is 0.0200 dm³ = 20.0 cm³.

At room temperature and pressure (20°C, 1 atm) one mole of any gas occupies 24 dm³ (24 000 cm³), as stated by Avogadro's law: equal volumes of gases under the same conditions contain equal numbers of molecules.

In questions about hydrated salts, find the masses of salt and water, convert to moles of each, and divide by the smaller to get x. Example: a hydrate contains 7.1 g of anhydrous sodium sulfate (M_r 142) and 3.6 g of water; moles are 0.050 and 0.20, ratio 1 : 4, so the formula is Na₂SO₄.4H₂O For a mixed problem such as "what volume of CO₂ forms when 2.0 g of CaCO₃ reacts with 25 cm³ of 2.0 mol/dm³ HCl", find the limiting reactant first: CaCO₃ is 0.020 mol and needs 0.040 mol HCl; the acid supplied is 0.050 mol, so CaCO₃ limits the gas, giving 0.020 mol CO₂ = 0.48 dm³.

Common mistakes to avoid in these problems: using the volume of the solid rather than the solution, forgetting that the same number of moles is present before and after dilution, and not noticing when a reactant is in excess. Always write the units of every number during the calculation so the final unit emerges naturally, and compare your answer to the data to confirm that it is reasonable.