Determine the molecular formula of a hydrocarbon containing 88.2% carbon and 11.8% hydrogen

Determine the molecular formula of a hydrocarbon containing 88.2% carbon and 11.8% hydrogen, if its vapor density for helium is 17. Indicate the types of hybridization of carbon atoms in a molecule of one of the isomers.

Given:
mass fraction of carbon = 88, 2 percent
mass fraction of hydrogen = 11.8 percent
helium vapor density = 17
find: molecular formula of hydrocarbon

Decision :
M = 4 * 17 = 68 grams / mol
m (С) = 68 grams / mol * 88, 2 percent = 68 * 0, 882 grams = 59, 9 76 = 60 grams
V (C) = 60/12 mol = 5 mol
m (H) = 68 grams / mol * 11.8 percent = 68 * 0, 118 grams = 8.024 = 8 grams
V (H) = 8/1 mol = 8 mol
ratio V (C): V (H) = 5: 8
then we get the formula C5H8
answer: C5H8