Question

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?

Answer 1

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 1.9/44

= 0.0432

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.521/18

= 0.0289

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.0432

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.0289 = 0.0579

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 1.037 - 0.0432*12 - 0.0579*1

= 0.4609

number of mol of O = mass of O / molar mass of O

= 0.4609/16.0

= 0.0288

Divide by smallest:

C: 0.0432/0.0288 = 1.5

H: 0.0579/0.0288 = 2

O: 0.0288/0.0288 = 1

multiply by 2 to get simplest whole number ratio:

C : 1.5*2 = 3

H : 2*2 = 4

O : 1*2 = 2

So empirical formula is:C3H4O2