Question

When 5.895 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 18.50 grams of CO₂and 7.574 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 42.08 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Empirical formula?

Molecular formula?

Answer 1

mass of HYdrocarbon = 5.895 grams

mass of CO2 = 18.50 grams

molar mass of CO2 = 44 grams/mole

% by mass of C = 12/44 xmass of CO2/mass of O.C x100

                          = 12/44 x18.50/5.895 x100 = 22200/ 259.38

                           = 85.59%

mass of H2O= 7.574 grams

Molar mass of H20= 18 grams/mole

% by mass of H = 2/18 xmass of H2O/massof OC x100

                        = 2/18 x7.574/5.895 x100=1514.8/106.11

                      = 14.27%

element      % by mass      atomic wt      relative number                simple ratio

C                        85.59         12              85.59/12 = 7.13              7.13/7.13 = 1.0

H                         14.27         1.0            14.27/1.0 =14.27           14.27/7.13 = 2.0

    Emperical formula = CH2

Emperical forumala mass = 12+2 = 14

Molar mass of a compound = 42.08grmas

n = molar mass of cimpound/emperical formula mass

= 42.08/14 = 3.005= 3.0

Molecular formula = nxemperical formula

                           = 3 xCH2

                                 = C3H6

Molecular formula of the compound = C3H6