A 1.037 g sample of a substance containing only C, H, and O is burned in excess O₂ and yields 1.90 g CO₂ and 0.521 g H₂O. What is the empirical formula of the substance?

Question

Quizplus · Accepted Answer

1) Calculate the amount of C in the sample:
C = (mass of CO2 produced)/(molar mass of CO2) = 1.90 g/(44.01 g/mol) = 0.0431 mol
2) Calculate the amount of H in the sample:
H = (mass of H2O produced)/(molar mass of H2O) = 0.521 g/(18.02 g/mol) = 0.0289 mol
3) Calculate the amount of O in the sample:
O = (mass of sample) - (mass of C) - (mass of H) = 1.037 g - (0.0431 mol x 12.01 g/mol) - (0.0289 mol x 1.01 g/mol) = 0.965 g
4) Divide each mole amount by the smallest mole amount to obtain the mole ratio:
C: 0.0431 mol/0.0289 mol = 1.49
H: 0.0289 mol /0.0289 mol = 1
O: 0.965 g/(16.00 g/mol) / 0.0289 mol = 2.17
5) Round each ratio to the nearest whole number in order to obtain the empirical formula:
C1.49H1O2.17 ≈ CH1.49O2.17 ≈ C2H3O4
The empirical formula is C2H3O4, which is choice B.

A 1037 G Sample of a Substance Containing Only C, H