Question

A 2.52g sample of a compound containing only carbon, hydrogen, nitrogen, oxygen, and sulfur was burned in excess O to yield 3.87g of CO2 and 0.792g of H2O as the only carbon and hydrogen containing products respectively.

Another sample of the same compound, of mass 4.14g , yielded 2.31g of SO3 as the only sulfur containing product.

A third sample, of mass 5.66g , was burned under different conditions to yield 2.49g of HNO3 as the only nitrogen containing product.

Calculate the empirical formula of the compound.
Express your answer as a chemical formula. Enter the elements in the order: C, H, S, N, O.

Answer 1

moles CO2 = 3.87 / 44.009 =0.088
mass C = 0.088 x 12.011 =1.055 g

moles H = 2 x 0.792 / 18.02 =0.088
mass H = 0.088 x 1.008=0.088 g

mass of SO3 yielded by 2.52 g :
2.52 x 2.31/ 4.14 =1.41 g
moles SO3 = 1.41 / 80.063 g/mol= 0.0176
mass S = 0.0176 x 32.066=0.563 g

mass HNO3 yielded by 2.52 g :
2.49 x 2.52 / 5.66 = 1.11 g
moles HNO3 = 1.11 / 63.0117 g/mol=0.0176
mass N = 0.0176 x 14.0067=0.246 g

mass O = 2.52 - (0.246 + 0.563 +0.088 +1.055)=0.568 g
moles O 0.568/ 15.999=0.0355

C0.088 H 0.088 O 0.0355 N 0.0176 S 0.0176
divide by the smallest to get the empirical formula
C5H5NSO