Empirical Formula of a Hydrocarbon?

Question

Combustion of 200mg of a hydrocarbon results in 600mg of CO2 and 327.27mg of water.

What is the empirical formula of the compound?

Answer 1

600 mg CO2 contains 164 mg C (600 mg x 12.01/44.01 = 163.7)

327.27 mg H2O contains 36.6 mg H (327.27 mg x 2.016/18.015 = 36.6)

163.7 mg C / 12.01 = 13.6 millimol C atoms

36.6 mg H / 1.008 = 36.3 millimol H atoms

ratio of C:H = 1 : 2.67 or 3 : 8

empirical formula = C3H8

(extra credit for West Highland White Terrier lovers)

Answer 2

Determine the moles of carbon in 600.00 mg of CO2 and the moles of hydrogen in 327.27 mg of water.  Then divide both by the smaller number of moles.

CxHy + zO2 --> xCO2 + y/2H2O

0.600 g CO2 x (1 mol C / 44.0g CO2) = 0.01364 mol C

0.32727 g H2O x (2 mol H / 18.0 g H2O) = 0.03636 mol H

0.01364 / 0.01364 = 1 mol C

0.03636 / 0.01364 = 2.667 mol H

Multiply each number of moles by 3 to make them into whole numbers

1 x 3 = 3 mol C

2.667 x 3 = 8 mol H

Empirical formla is C3H8

Answer 3

write the equation and balance it in terms of 'x' and 'y'.

CxHy+ (x+1/4y)O2-----> xCO2+ 1/2yH2O

1) find the number of moles of CO2 adn equate it to x

2) find the numbee of moles of H2O and equate it to y

3) apply to CxHy and simpilfy if possible