Question

Question 13 (4 points) In a quantitative analysis a 3.2798 g sample of a hydrocarbon was when combusted with excess oxygen produced 11.085 grams of carbon dioxide and 2.2785 grams of water. What is a possible molecular formula? CH CgH20 CoH18 OCoHo C9H16 Page 13 of 30 Next Page Previous Page

Answer 1

44 g CO2 contain 12 g C

11.085 g CO2 contain 12×11.085/44 g C

= 3.023 g C

18 g H2O contain 2 g H

2.2785 g H2O contain 2×2.2785/18 g H

= 0.2532 g H

Mass % of c = mass of C×100/total mass

= 3.023×100/3.2798

= 92.17%

Mass % of H = 100- 92.17 = 7.83%

Moles of C = mass/molar mass = 92.17/12 = 7.68 moles

Moles of H = 7.83/1 = 7.83 moles

Molar ratio = C : H = 7.68 : 7.83 = 1 : 1

Answer is C ₉H₉