Question

When 1.65 grams of hydroquinone is subjected to combustion analysis, 3.96 grams of CO2 and 0.81 grams of H2O are produced. What is the empirical formula?

Answer 1

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

= 3.96/44

= 0.09

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

= 0.81/18

= 0.045

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.09

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.045 = 0.09

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 1.65 - 0.09*12 - 0.09*1

= 0.48

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

= 0.48/16.0

= 0.03

Divide by smallest to get simplest whole number ratio:

C: 0.09/0.03 = 3

H: 0.09/0.03 = 3

O: 0.03/0.03 = 1

So empirical formula is:C₃H₃O

Answer: So empirical formula is:C₃H₃O