f.4 chemistry

Question

on complete combustion, a 3.41 g of compound (CxHyOz) produced 8.6 g of carbon dioxide and 1.51 g of water.WHAT IS The empirical formula of the compound .

HOW TO DO??????

Answer 1

Molar mass of C = 12 g mol-1
Molar mass of H = 1 g mol-1
Molar mass of O = 16 g mol-1
Molar mass of CO2 = 12 + 16x2 = 44 g mol-1
Molar mass of H2O = 1x2 + 16 = 18 g mol-1

Fraction by mass of C in CO2 = 12/44
Fraction by mass of H in H2O = 2/18

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Unbalance equation :
CxHyOz + O2 → CO2 + H2O

In 3.41 g of the compound :
Mass of C
= Mass of C in CO2 formed
= 8.6 x (12/44)
= 2.345 g

Mass of H
= Mass of H in H2O formed
= 1.51 x (2/18)
= 0.168 g

Mass of O
= 3.41 - (2.345 + 0.168)
= 0.897 g

=====
In the compound, mole ratio C : H : O
= (2.345/12) : (0.168/1) : (0.897/16)
= 0.195 : 0.168 : 0.056
= 3.5 : 3 : 1
= 7 : 6 : 2

Hence, empirical formula = C7H6O2

Answer 2

The mass of C = 8.6 x 12/(12+16x2) = 2.35g
The mass of H = 1.51 x 2/(16+2) = 0.168g
The mass of O = 3.41 - 2.35 - 0.168 = 0.892g

The number of mol of C = 2.35/12 = 0.196
The number of mol of H = 0.168/1 = 0.168
The number of mol of O = 0.892/16 = 0.0556


The ratio of mol C:H:O = 0.196/0.0556 : 0.168/0.0556 : 0.0556/0.0556
C:H:O
= 3.53 : 3.02 : 1
= 7:6:2

so the empirical formula of the compound is C7H6O2