Formulae Worked ExamplesIGCSE Chemistry 0620

(a) The molecular formula gives the number and type of different atoms in one molecule.

(b) Count the atoms: 4 carbon atoms in the chain; hydrogen atoms: $3 + 2 + 2 + 3 = 10$.

(c) The empirical formula is the simplest whole-number ratio of atoms. Divide both numbers by their highest common factor, 2: $4 : 10 = 2 : 5$.

Potassium ion $\text{K}^+$, carbonate ion $\text{CO}_3^{\,2-}$. Two $\text{K}^+$ are needed to balance one $2-$ charge: $\text{K}_2\text{CO}_3$.

Copper(II) ion $\text{Cu}^{2+}$, nitrate ion $\text{NO}_3^{\,-}$. Two nitrate ions balance one $2+$ charge; the whole nitrate group is doubled, so it goes in brackets: $\text{Cu(NO}_3)_2$. The answer is C.

A treats copper(II) as $1+$. B treats potassium as $2+$ (or carbonate as $1-$). D swaps the ratios round.

(a) The Roman numeral is the oxidation number of chromium in the compound, and its value is +3. (So the chromium ion is $\text{Cr}^{3+}$.)

(b)(i) $\text{Cr}^{3+}$ and $\text{O}^{2-}$. Lowest common multiple of 3 and 2 is 6: two $\text{Cr}^{3+}$ ($+6$) balance three $\text{O}^{2-}$ ($-6$).

(ii) $\text{Co}^{2+}$ needs two $\text{Cl}^-$ ions.

(iii) $\text{Fe}^{3+}$ and $\text{SO}_4^{\,2-}$: two iron(III) ions ($+6$) balance three sulfate ions ($-6$). Brackets round the sulfate.

(iv) Ammonium $\text{NH}_4^{\,+}$ and carbonate $\text{CO}_3^{\,2-}$: two ammonium ions per carbonate.

(a) Count each type of ion in the section shown: 12 ions of M and 6 ions of N. The ratio is $12 : 6 = 2 : 1$, so the formula is $\text{M}_2\text{N}$.

(b) A Group VI atom has 6 outer electrons and gains 2 to fill its shell, so N forms $\text{N}^{2-}$.

The compound has no overall charge. Two M ions balance one $2-$ charge, so each M ion is $\text{M}^+$. A $1+$ ion is formed by losing one electron: M is in Group I.

R is in Group III (Period 3): it is a metal that loses 3 electrons to form $\text{R}^{3+}$.

T is in Group VI: it is a non-metal that gains 2 electrons to form $\text{T}^{2-}$.

Balance the charges: $2 \times (+3) = +6$ and $3 \times (-2) = -6$. The formula is $\text{R}_2\text{T}_3$: C.

B comes from swapping the group numbers (6 and 3) and simplifying, instead of using the ion charges. D swaps the charges the wrong way round. A assumes the charges are equal.

(a) Fe: 3 on the left, so $3\text{Fe}$. O: the oxide has 4 O atoms; each CO takes one O to become $\text{CO}_2$, so $4\text{CO}$ and $4\text{CO}_2$. Check C: 4 = 4; O: $4 + 4 = 8 = 4 \times 2$.

(b) C: 5, so $5\text{CO}_2$. H: 12, so $6\text{H}_2\text{O}$. O on the right: $5 \times 2 + 6 = 16$ atoms, which is $8\text{O}_2$.

(c) With one ZnS, the right side has $1 + 2 = 3$ O atoms, needing $1\tfrac{1}{2}\,\text{O}_2$. Double everything to remove the fraction.

Check (c): Zn $2 = 2$; S $2 = 2$; O $6 = 2 + 4$. ✓

(a) Formulae: Na, $\text{H}_2\text{O}$, NaOH, $\text{H}_2$. Hydrogen is diatomic, so two NaOH are needed to release one $\text{H}_2$. States: sodium solid, water liquid, the hydroxide dissolved (aq), hydrogen gas.

(b) Zinc sulfate: $\text{Zn}^{2+}$ with $\text{SO}_4^{\,2-}$ gives $\text{ZnSO}_4$. The equation is already balanced 1 : 1 : 1 : 1.

(c) The black solid is copper(II) oxide, CuO; the gas that turns limewater milky is carbon dioxide.

(a) Identify the molecules from the key: CO (one C, one O), $\text{O}_2$ (two O), $\text{CO}_2$ (one C, two O).

Count what actually reacted. Before: 4 CO and 3 $\text{O}_2$. After: 4 $\text{CO}_2$ and 1 $\text{O}_2$ left over. So 4 CO reacted with $3 - 1 = 2$ $\text{O}_2$ to form 4 $\text{CO}_2$. Ratio $4 : 2 : 4 = 2 : 1 : 2$.

(b) Oxygen was in excess: one $\text{O}_2$ molecule is still present after the reaction, while all the CO molecules have reacted.

(a) Copper(II) hydroxide: $\text{Cu}^{2+}$ with two $\text{OH}^-$, so $\text{Cu(OH)}_2$. Two NaOH are needed; the other product is sodium sulfate. The precipitate is (s); everything else is (aq).

(b) Write the aqueous compounds as separate ions and cross out ions that appear unchanged on both sides ($\text{Na}^+$ and $\text{SO}_4^{\,2-}$). Only the ions that form the precipitate remain.

Check the charges: left $+2 + 2(-1) = 0$; right $0$. ✓

(c) The spectator ions are the sodium ions and the sulfate ions.

(a) Full equation: $\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$. The sulfate ion is unchanged, so it is a spectator ion and is removed.

(b) Full equation: $\text{Cl}_2\text{(aq)} + 2\text{KI(aq)} \rightarrow 2\text{KCl(aq)} + \text{I}_2\text{(aq)}$. The potassium ion is the spectator ion. The brown colour is iodine.

Check charges: (a) $+2$ on each side; (b) $-2$ on each side. ✓

(a) Reactants ZnO and NaOH; products $\text{Na}_2\text{ZnO}_2$ and $\text{H}_2\text{O}$. Two Na on the right need 2 NaOH; then H: 2 on each side and O: $1 + 2 = 3$ on the left, $2 + 1 = 3$ on the right.

(b) Each sodium ion is $\text{Na}^+$, so two of them carry $+2$. The rest of the formula must carry $-2$: the zincate ion is $\text{ZnO}_2^{\,2-}$.

(c) Oxygen has oxidation number $-2$. Let zinc be $x$: the total must equal the ion's charge.

(d) In $\text{MnO}_4^{\,-}$: $x + 4(-2) = -1$, so $x = +7$. Manganese has oxidation number +7, matching the (VII) in the name.

(a) The gas is ammonia, which is alkaline: the damp red litmus paper turns blue.

(b) Formulae: ammonium sulfate $(\text{NH}_4)_2\text{SO}_4$ (two $\text{NH}_4^{\,+}$ per $\text{SO}_4^{\,2-}$), sodium sulfate $\text{Na}_2\text{SO}_4$. The two ammonium ions give 2 $\text{NH}_3$ and need 2 NaOH, forming 2 $\text{H}_2\text{O}$.

Check: N 2 = 2; H $8 + 2 = 10 = 6 + 4$; S 1 = 1; O $4 + 2 = 6 = 4 + 2$; Na 2 = 2. ✓

(c) Sodium and sulfate ions are spectators. What reacts is the ammonium ion with the hydroxide ion.