Question

(i) Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and 46.6% fluorine by mass. (Atomic weight of Be = 9; F = 19; K = 39).
(ii) Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide?
The equation for the reaction is:
3CuO + 2NH₃ $\stackrel{}{\to }$ 3Cu + 3H₂O + N₂
(Volume occupied by 1 mole of gas at STP is 22.4 litres).

Answer 1

(i)

Element	Atomic mass	Percentage	Relative number of atoms	Simplest whole number ratio
K	39	47.9	$\left(\frac{47.9}{39}\right)=1.23$	$\left(\frac{1.23}{0.61}\right)=2.02\approx 2$
Be	9	5.5	$\left(\frac{5.5}{9}\right)=0.61$	$\left(\frac{0.61}{0.61}\right)=1$
F	19	46.6	$\left(\frac{46.6}{19}\right)=2.45$	$\left(\frac{2.45}{0.61}\right)=4.02\approx 4$

Empirical formula of the compound = K₂BeF₄

(ii) The relative molecular mass of CuO is 80.
The given reaction states that,
Moles of CuO reacting with two moles of ammonia = 3
Mass of three moles of copper oxide = (3$\times$80) g = 240 g
Volume of two moles of ammonia = (2 $\times$ 22.4) L = 44.8 L
Volume of ammonia react with 240 g of copper oxide = 44.8 L
Therefore, volume of ammonia needed to react with 120 g of copper oxide = $\left(\frac{44.8\times 120}{240}\right)\mathrm{L}=22.4\mathrm{L}$