Answer to Question #184455 in Organic Chemistry for Daniel Malachy Pius

Question

A 3.1g sample of unknown organic gas molecules composed of carbon, hydrogen and oxygen undergoes complete combustion to produce a 4.4g of carbon dioxide and 2.7g of water. Calculate the empirical and molecular formula of the compound if the molar mass is 62g/mol. Show working.

Answer 1

mass of CO₂ = 4.4 g , molar mass of CO₂ = 44g , mass of H₂O =2.7 g, molar mass of H₂O = 18 g

mass of unknown organic gas = 3.1 g

moles of CO₂ = $\frac{mass}{molar mass}= \frac{4.4}{44}= 0.1$

moles of H₂O = $\frac{mass}{molar mass}= \frac{2.7}{18}=0.15$

mass of oxygen = 4.4 + 2.7 -3.1 =4.0 g

mole of O₂ = $\frac{mass}{molar mass}= \frac{4.0}{32}= 0.125$

so, ratio of O₂: CO₂: H₂O= 0.125:0.1:0.15

O₂: CO₂: H₂O= 5:4:6

so we can write the balanced equation as

$nC_xH_yO_z + 5 O_2 \longrightarrow 4CO_2 + 6H_2O$

from above balanced equation we can conclude that ratio in which C,H and O are present

C:H:O = 4:12:4

C:H:O = 1:3:1

so, empirical formula is CH₃O , and empirical mass = 31

since we are gettting empirical mass as 31 g/mol whereas molar mass = 62 g/mol

so, n= $\frac{62}{31}=2$

Molecular formula of compound = C₂H₆O₂