Chemical Formulae Equations And Calculations

sodium + water → sodium hydroxide + hydrogen; correct substances on correct sides

methane + oxygen → carbon dioxide + water; all four substances correctly named

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g); correct coefficients (1 mark); correct state symbols (1 mark); all formulae unchanged (1 mark)

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g); correct formulae (1 mark); balanced (1 mark); state symbols (1 mark)

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O; correct coefficients for carbon and hydrogen (1 mark); correct coefficient for oxygen (1 mark)

$${M}_{r}=24+(2\times 14)+(6\times 16)$$ $${M}_{r}=24+28+96=148$$

$${M}_{r}=(2\times 12)+(6\times 1)+16$$ $${M}_{r}=24+6+16=46$$

mole / mol

the Aᵣ of magnesium is 24; the mass of one mole of atoms of an element in grams equals its Aᵣ

$$n=\frac{5.6}{56} \text{(M1)}$$ $$n=0.10\text{ mol} \text{(A1)}$$

$${M}_{r}=40+12+(3\times 16)=100 \text{(M1)}$$ $$m=2.5\times 100 \text{(M1)}$$ $$m=250\text{ g} \text{(A1)}$$

$${M}_{r}({\text{Fe}}_{2}{\text{O}}_{3})=(2\times 56)+(3\times 16)=160 \text{(M1)}$$ $$n({\text{Fe}}_{2}{\text{O}}_{3})=\frac{32.0}{160}=0.20\text{ mol} \text{(M1)}$$ mole ratio 1 : 2 so n(Fe) = 0.40 mol (M1) $$m=0.40\times 56=22.4\text{ g} \text{(A1)}$$

$${M}_{r}({\text{CaCO}}_{3})=100$$ $$n=\frac{10.0}{100}=0.10\text{ mol} \text{(M1)}$$ ratio 1 : 1 so n(CO₂) = 0.10 mol (M1) $$m=0.10\times 44=4.4\text{ g} \text{(A1)}$$

$$\text{percentage yield}=\frac{18.5}{25.0}\times 100 \text{(M1)}$$ $$\text{percentage yield}=74.0\% \text{(A1)}$$

some product is lost during transfer/filtering/evaporation; the reaction may be incomplete / reversible / side reactions may occur

$$n(\text{C})=\frac{40.0}{12}=3.33, n(\text{H})=\frac{6.7}{1}=6.7, n(\text{O})=\frac{53.3}{16}=3.33 \text{(M1)}$$ divide by smallest (3.33) gives C = 1, H = 2, O = 1 (M1) empirical formula = CH₂O (A1)

$${M}_{r}({\text{CH}}_{2})=14$$ $$n=\frac{56}{14}=4 \text{(M1)}$$ molecular formula = C₄H₈ (A1)

heat the hydrated salt and record the mass before and after (1 mark); mass of water = mass before − mass after (1 mark); calculate moles of anhydrous salt and moles of water, then find the simplest ratio (1 mark)

$${m}_{\text{water}}=4.44-2.28=2.16\text{ g}$$ $$n({\text{FeSO}}_{4})=\frac{2.28}{152}=0.015\text{ mol}, n({\text{H}}_{2}\text{O})=\frac{2.16}{18}=0.12\text{ mol} \text{(M1)}$$ $$\frac{0.12}{0.015}=8 \text{(M1)}$$ x = 8, formula is FeSO₄·8H₂O (A1)

$$V=\frac{250}{1000}=0.250{\text{ dm}}^{3} \text{(M1)}$$ $$c=\frac{0.30}{0.250}=1.2{\text{ mol dm}}^{-3} \text{(A1)}$$