Question

Complete combustion of 3.10 g of a hydrocarbon produced 9.90 g of CO2 and 3.55 g of H2O. What is the empirical formula for the hydrocarbon

Answer 1

Answer: C9H16

Step-by-step explanation:

molar mass C = 12.011, H = 1.008, O = 15.999, CO2 = 44.01, H2O = 18.00

9.9g CO2 = 0.225 mol, 3.55 g H2O = 0.200 mol

long way

molecules CO2 = 1.355x10^23 —> atoms of C

molecules H2O = 1.204x10^23 —> molecules H2

C:H2 = 1.355/1.204 = 1.125. —> C9H16

short way

molar massesCO2:H2O = 0.225:0.2 = 9:8

–> C9H16

verify

molar mass C9H16 =124

3.1g = 0.025 mol

oxidise C9H16 + 13O2 —> 9CO2 + 8H2O

0.025 mol + 13*0.025 mol (0.35mol) —> 9*0.025 + 8*0.025 —> 0.0.225 + 0.2