Question

Compound X contains only carbon,nitrogen,hydrogen,oxygen and sulfur. Analysis experiments produced the following results: 1. 2.52g of X burned in excess O2 produces 4.25g of carbon dioxide and 1.02g of water at 100% yield. 2. 4.14g of X reacts to produce 1.785g of sulfur trioxide and 85% yield. 3. 5.66g of X reacts to produce 1.680g of nitric acid at 75% yield. What is the empirical formula of compound X?

Answer 1

Compound X

contains = C, H, N, O, S

moles of C = 4.25/44 = 0.0966 mol

mass of C = 0.0966 x 12 = 1.16 g

moles of H = 1.02 x 2/18 = 0.1133 mol

mass of H = 0.1133 x 1 = 0.11 g

total mass of C + H in 2.52 g sample = 1.27 g

remaining (N + O + S) = 1.25 g

moles of SO3 = 2.1/80 = 0.026 g

[2.1 g when 100% reaction gives SO3]

mass of S in sample = 0.026 x 32 = 0.84 g in 4.14 g

So in 2.52 g = 0.51 g S

moles of S in 2.52 g = 0.51/32 = 0.016 mol

moles of N = 2.24/63 = 0.0355 mol

mass of N in 5.66 g = 0.0355 x 14 = 0.50 g

mass of N in 2.52 g sample = 0.22 g

moles of N in 2.52 g = 0.016 mol

mass of O = 1.25-0.73 = 0.52 g

moles of O = 0.52/16 = 0.0325 mol

divide moles of C, H, N, O and S with smallest number

C = 0.0966/0.016 = 6

H = 0.1133/0.016 = 7

S = 0.016/0.016 = 1

N = 0.016/0.016 = 1

O = 0.0325/0.016 = 2

so the empirical formula of X = C6H7SNO2