Complete combustion of 7.60 g of a hydrocarbon produced 23.4 g of co2 and 10.8 g of h2o. what is the empirical formula for the hydrocarbon?

Mass of hydrocarbon = 7.60 g

All the C gets converted to 23.4 g of CO2

All the H gets converted to 10.8 g of H2O

Now:

Molecular mass of CO2 = 48 g/mol

Molecular mass of H2O = 18 g/mol

# moles of CO2 = 23.4 g/48 gmol-1 = 0.4875  = moles of C

# moles of H2O = 10.8/18 = 0.6 moles = moles of H

Now:

Mass of C = 0.4875 moles * 12 g/mole = 5.85 g

Mass of H = 0.6 moles * 1 g/mole = 0.6 g

Total mass = C + H = 5.85 + 0.6 = 6.45 g. Ideally, this must be equal to the given  mass of the hydrocarbon, 7.60 g.

Nevertheless, in order to find the empirical formula, divide the # moles of C and H by the smallest value:

C = 0.4875/0.4875 = 1

H = 0.6/0.4875 = 1.23 (approximately = 1)

Empirical formula = CH

the correct answer is:

a. formation of precipitation