IGCSE Chemistry 0620: Formulae, Equations and Relative Mass - Study Guide PDF with Answers

Read the formula before calculating

A subscript belongs to the element immediately before it, unless it follows a bracketed group. In Mg(OH)2, the 2 multiplies both O and H, not Mg. A coefficient in front of a formula multiplies the entire formula: 3H2O represents six hydrogen atoms and three oxygen atoms. In these plain-text expressions, trailing digits represent subscripts; a leading digit is a coefficient. Ionic substances form lattices rather than separate molecules, so use the term formula unit when appropriate.

Balance charge to construct an ionic formula

An ionic formula must have zero total charge. For Mg2+ and Cl-, two chloride ions balance one magnesium ion, giving MgCl2. Use the simplest whole-number ratio, not necessarily one of each ion. A polyatomic ion is kept together: three nitrate ions, each NO3-, combine with Al3+ to give Al(NO3)3. Brackets show that the entire nitrate group is repeated. The charges supplied in a question are the starting point; do not guess a charge from the final formula.

Balance atoms without changing the substances

A balanced equation conserves the number of atoms of each element. Change whole-number coefficients in front of formulae, never the subscripts inside a correct formula. Changing H2O to H2O2 would change water into a different substance. Make a left/right atom-count table, balance one element at a time and recheck every element at the end. Report the smallest whole-number coefficients. State symbols, when supplied, describe physical state: (s), (l), (g) and (aq) for solid, liquid, gas and aqueous solution.

Calculate relative mass and use mass ratios

Relative molecular mass is found by adding the relative atomic masses of all atoms in a molecule. For an ionic solid, use relative formula mass instead. Both are relative quantities with no unit. Molar mass has a similar numerical value but different meaning and units; this guide does not use the mole method. When a balanced reaction supplies a known mass ratio, scale it proportionally, provided other reactants are in excess and the reaction goes to completion. Use the atomic masses stated in the question rather than mixing different rounded tables.

1. Handle a bracketed formula

Find the relative formula mass of Ca(OH)2 using Ca=40, O=16 and H=1.

1. Count the atoms: one Ca, two O and two H.
2. Relative formula mass = 40 + 2(16 + 1) = 40 + 34 = 74.
3. The 2 applies to the whole OH group. The answer is 74, with no unit, not 74 g.

2. Balance combustion on paper

Balance the equation C3H8 + O2 -> CO2 + H2O.

1. Balance carbon first: one C3H8 requires 3CO2.
2. Balance hydrogen next: eight H atoms require 4H2O.
3. The right-hand side now contains 6 + 4 = 10 O atoms, requiring 5O2. Final equation: C3H8 + 5O2 -> 3CO2 + 4H2O.

3. Use a reacting mass ratio

For 2Mg + O2 -> 2MgO, take Mg=24 and O=16. What mass of MgO forms from 6.0 g Mg with oxygen in excess?

1. The balanced equation corresponds to 48 mass units of Mg forming 80 mass units of MgO.
2. Scale factor = 6.0 / 48 = 0.125.
3. Mass of MgO = 80 x 0.125 = 10 g. The product is heavier than the starting magnesium because oxygen is added. This assumes complete reaction.