In: Chemistry
an unkown hydrocarbon contain carbon hydrogen and oxygen combustion of 1.6 g of the hydrocarbon produces 1.0286 g h2o and 3.7681 g co2. what is the empiral formula of this compound?
CO2 moles = mass of CO2 / molar mass of CO2
= 3.7681 / 44
= 0.0856
moles of C = 0.0856
mass of C = 0.0856 x 12
= 1.0276 g
moles of H2O = 1.0286 / 18
= 0.0571
moles of H = 2 x 0.0571
= 0.1143
mass of H = 0.1143 x 1 = 0.1143 g
mass of O = sample mass - (mass of C + mass of H)
= 1.6 - (1.0276 + 0.1143)
= 0.4581 g
moles of O = 0.4581 / 16 = 0.0286
C H O
0.0856 0.1143 0.0286
3 4 1 -----------------> by dividing smallest number
C3H4O
Empirical formula = C3 H4 O