Question

1. Combustion analysis of 0.150 g of an unknown compound containing carbon, hydrogen, and oxygen produces 0.343 g CO2 and 8.76×10−2 g H2O. What is the empirical formula of the compound?

2. If the molar mass is 770.8 g/mol what is the molecular formula?

Answer 1

moles of CO2 = 0.343 / 44 = 7.79 x10^-3

moles of C= 7.79 x10^-3

mass of Carbon = 12 x 7.79 x10^-3= 0.09354 g

moles of H2O = 8.76×10−2 / 18 = 4.87x10^-3

moles of H = 2 x 4.87x10^-3 = 9.73 x10^-3

mass of hydroegen = 9.73 x10^-3 g

Oxygen mass = 0.150 - (9.73 x10^-3 +9.73 x10^-3) = 0.046734 g

moles of O = 0.046734 /16 = 2.92 x 10^-3

C              O              H

7.79 x10^-3       2.92 x 10^-3        9.73 x10^-3

3                         1                                     3

C3H3O -----------------------> empirical formula

empirical formula mass = 36 + 3 + 16 = 55

n = molar mass / empirical formula mass

    = 770.8 / 55

     = 14

molecular formula = n x empirical formula

                             = C42 H42 O14