Question

(a) Calculate the percentage of boron (B) in borax (Na₂B₄O₇·10H₂O). (H = 1, B = 11, O = 16, Na = 23).
(b) (i) The compound A has the following percentage composition by mass:
Carbon = 26.7%, oxygen = 71.1%, hydrogen = 2.2%. Determine the empirical formula of A (H = 1, C = 12, O = 16)
(ii) If the relative molecular mass of A is 90, what is the molecular formula of A?

Answer 1

(a) Molecular mass of borax = [23(2) + 11(4) + 16(7) + 2(10) + 16(10)] g = 382 g
Mass of boron in one mole of borax = [11(4)] g = 44 g
$\mathrm{Percentage}\mathrm{of}\mathrm{B}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{B}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{Borax}}\times 100=\frac{44}{382}\times 100=11.52\%$

(b) (i)

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio
C	12	26.7	$\frac{26.7}{12}=2.2$	$\frac{2.2}{2.2}=1$
O	16	71.1	$\frac{71.1}{16}=4.4$	$\frac{4.4}{2.2}=2$
H	1	2.2	$\frac{2.2}{1}=2.2$	$\frac{2.2}{2.2}=1$

Hence, empirical formula of compound is CO₂H.

(ii) Empirical formula mass of the compound = [12 + 16(2) + 1] g = 45 g
Relative molecular mass of compound = 90 g
Molecular formula = (Empirical formula)_n
n = $\frac{\mathrm{Molecular}\mathrm{formula}\mathrm{mass}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}=\frac{90}{45}=2$
So, the molecular formula of compound is C₂O₄H₂.