Question

(a) from the equation"
$C+2H_{2}S{O}_4\to C{O}_2+2H_{2}O+2S{O}_2$
Calculate:
(i) hte mass of carbon oxidized by 49 g of sulphuric acid.
(ii) the volume of sulphur dioxide measured at STP, linerated at the same time.
(b) (i) A compound has the following percentage composition by mass; carbon 14.4 % hydrogen 1.2% and chlorine 84.5 % . Determine the empirical formula of this compound. Work correct to 1 decimal place. (H - 1; C = 12; Cl = 35.5)
(ii) The relativ molecular mass of this compound is 168, so what is its molecular formuka?

Answer 1

(a)
(i) From equation:
2 x 98 g of sulphuric acid oxidises 12 g of carbon
therefore,
49g of sulphuric acid will oxidise =3g of C.

ii) From equation:
2 x 98 of sulphuric liberates 2 x 22.4 l of SO2
therefore,

49 g of sulphuric acid will liberate = $\frac{2\times 22.4\times 49}{2\times 98}$
=11.2 L of SO2

(b) (i)

Element	%	At. mass	Relative no of atoms	Simplest ratio
C	14.4	12	14.4/12=1.2	1.2/1.2=1
H	1.2	1	1.2/1=1.2	1.2/1.2=1
Cl	84.5	35.5	84.5/35.5=2.4	2.4/1.2=2

The Empirical Formula = $CHC{l}_2$

ii) Relative molecular mass = 168
Empirical formula mass = 12 +1+71
= 84
so
n=relative molecular mass/empirical formula mass
n= 168/84=2
molecular formula = (empirical formula)x n
($CHC{l}_2$) x2
=$C_2{H}_{2}C{l}_4$