Question

A 2.52-g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess oxygen to yield 4.23 g of CO2 and 1.01 g of H2O. Another sample of the same compound, of mass 4.14 g, yielded 2.11 g of SO3. A third sample, of mass 5.66 g, was burned under different conditions to yield 2.27 g of HNO3 as the only nitrogen containing product. Determine the empirical formula of the compound.

Answer 1

Empirical formula of the compund is C₆H₇SNO₂

Step-by-step explanation:

The moles produced of CO₂ are the same than moles of C and the moles of H₂O are half moles of H.

4,23 g CO₂ ×
`(1mol)/(44,01g)` = 0,096 moles of CO₂ ≡ moles of C×12,01 g/mol = 1,15296 g of C/ 2,52*100 = 45,80 wt%

1,01 g H₂O ×
`(1mol)/(18,02g)` = 0,056 moles of H₂O = 0,112 moles of H×1,01 g/mol= 0,1132 g of H/2,52*100 = 4,49 wt%

Moles of SO₃ are the same than moles of S and moles of HNO₃ are the same than moles of N.

2,11 g SO₃ ×
`(1mol)/(80g)` = 0,026 moles of SO₃ = 0,026 moles of S×32,065 g/mol= 0,8457 g of S/4,14*100 = 20,43 wt%

2,27 g HNO₃ ×
`(1mol)/(63,01g)` = 0,036 moles of H₂O = 0,036 moles of N×14 g/mol= 0,504g of H/5,56*100 = 9,07 wt%

wt% of Oxygen is 100-9,07-20,43-4,49-45,80 = 20,21 wt%

With wt% it is possible to obtain relative moles dividing in molar mass of each atom thus:

moles of C: 45,80/12,01 = 3,81

moles of H: 4,49/1,01 = 4,44

moles of S: 20,43/32,05 = 0,64

moles of N: 9,07/14 = 0,65

moles of O: 20,21/16 = 1,26.

Dividing in the least number:

C: 3,81/0,64 = 6

H: 4,44/0,64 = 7

S: 0,64/0,64 = 1

N: 0,65/0,64 = 1

O: 1,26/0,64 = 2

Thus, empirical formula of the compound is:

C₆H₇SNO₂

I hope it helps!