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

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

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

Relative atomic mass (A_r) is the average mass of an atom of an element, taking into account the mass numbers and relative abundances of all its isotopes, compared with one twelfth of a carbon-12 atom. It has no units.

Relative formula mass (M_r) is found by adding together the A_r values of every atom shown in the formula. Brackets multiply everything inside them.

For magnesium nitrate the bracket contains one N and three O atoms, and there are two brackets, so there are 2 N and 6 O atoms in total.

Chemists count particles in moles. One mole of any substance contains the same number of the stated particles (atoms, molecules or ions) as one mole of any other substance. That number is the Avogadro constant, 6.02 × 10²³ per mole.

In the laboratory we cannot count particles or measure moles directly. We use a digital balance and measure grams, so we must convert between moles and grams using the molar mass (the M_r or A_r in g/mol):

moles = mass (g) ÷ molar mass (g/mol), which can be rearranged to mass = moles × molar mass and molar mass = mass ÷ moles. A formula triangle with mass at the top and moles and molar mass at the bottom helps: cover the quantity you want.

Example2.64 g of sucrose (molar mass 342.3 g/mol) is 2.64 ÷ 342.3 = 7.71 × 10^-3 mol.

A balanced symbol equation tells us the ratio in which moles of substances react and form. The method is the same each time:

Always show your working. It is easier to check, and you can earn method marks even if the final answer is wrong.

Large industrial quantities work the same way: 1 tonne = 10⁶ g, so 51 tonnes is 51 000 000 g.

The masses of reactants and products also reveal the equation itself. Convert each mass to moles, then divide all the mole values by the smallest to get the simplest whole-number ratio. These ratios become the balancing coefficients.

Remember that mass is conservedthe total mass of reactants equals the total mass of products, which is a useful check on data in a question.

Yield is the amount of product obtained from a reaction. The theoretical yield is the maximum mass predicted from the balanced equation; the actual yield is the mass really obtained.

In practice you never get 100% yield. Typical reasons are that some reactants stay behind in the apparatus, the reaction may be reversible and not go to completion, side reactions form other products, and some product is lost during separation, filtering or transfer.

percentage yield = (actual yield ÷ theoretical yield) × 100

The result must be between 0% and 100% if the data are sensible.

The formulae of simple compounds can be found by careful experiments and accurate mass measurements. The principle is to measure masses before and after a reaction, then convert the masses into moles and find the simplest ratio.

Practical tipavoid overheating, because the salt could decompose and give a larger mass change than the water alone.

The mass of water lost is the difference between the first and last masses; the anhydrous salt mass is the remainder. Divide each by its molar mass, then divide by the smaller value to find the ratio.

To find the empirical formula of magnesium oxide by burning magnesium:

Mass of magnesium = (crucible + magnesium) − (empty crucible). Mass of oxygen = (crucible + magnesium oxide) − (crucible + magnesium). Divide each mass by its A_r to get moles, then divide by the smaller to give the ratio.

The formula of a metal oxide such as copper(II) oxide can be found by reduction, because the mass lost is the mass of oxygen removed.

Mass of oxygen = loss in mass; mass of metal = final mass − mass of empty tube. Convert both to moles and find the ratio.

Chemical manufacturers aim for a high percentage yield because wasted reactants cost money and unwanted by-products must be disposed of.

The molecular formula shows the actual number and type of each atom in one molecule. For example, ethanoic acid is C₂H₄O₂.

The empirical formula is the simplest whole-number ratio of the atoms of each element in one molecule or formula unit of the compound. Ethanoic acid has the empirical formula CH₂O.

Ionic compounds and giant structures are always written as empirical formulae, because they do not contain separate molecules.

The molar ratio must always be whole numbers. A ratio of 1 : 1.5 becomes 2 : 3 on doubling.

Worked case

• a sample contains 10 g of hydrogen and 80 g of oxygen.
• H = 10 ÷ 1 = 10 mol and O = 80 ÷ 16 = 5 mol.
• Dividing by 5 gives H : O = 2 : 1, so the empirical formula is H₂O.

The molecular formula is a whole-number multiple of the empirical formula and gives the actual numbers of atoms of each element in the compound. To find it you also need the relative formula mass (M_r) of the compound.

Example

• compound X has empirical formula C₄H₁₀S and a relative formula mass of 180.
• The empirical M_r = (12 × 4) + (1 × 10) + (32 × 1) = 90.
• Then 180 ÷ 90 = 2, so the molecular formula is C₈H₂₀S₂.
• A compound whose M_r equals the empirical M_r already has the molecular formula equal to the empirical formula.

Exam questions about hydrated salts follow a set pattern.

A table with one column per substance makes errors easy to spot. Always show every step, because credit is given for method.

A solute is a substance that dissolves in a liquid solvent to make a solution. The amount of solute can be given in grams (g) or in moles (mol). The concentration is the amount of solute in a given volume of solution.

Volumes are often measured in cm³, but dm³ must be used when calculating concentration. To convert cm³ to dm³, divide by 1000. To convert dm³ to cm³, multiply by 1000.

A formula triangle with moles at the top and concentration and volume below helps you rearrange.

Avogadro's law states that at the same temperature and pressure, equal amounts (moles) of gases occupy the same volume. At room temperature and pressure (RTP: 20 °C and 1 atmosphere) one mole of any gas occupies 24 dm³ (24 000 cm³). This is the molar gas volume.

Use moles = volume ÷ molar volume, where volume and molar volume are both in dm³ (24) or both in cm³ (24 000). Cover the quantity you want in the formula triangle.

Because equal moles of gases occupy equal volumes, the coefficients in a balanced equation also give the ratio of gas volumes (when measured under the same conditions). You do not need to convert to moles.

For the complete combustion of propane, C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g), 150 cm³ of propane needs 5 × 150 = 750 cm³ of oxygen. The total gaseous products are 3 + 4 = 7 volumes, so 7 × 150 = 1050 cm³.