Question

An unknown compound has the formula CxHyOz. When 0,1523 g of the compound is burned in oxygen, 0,3718 g of CO2 and 0,1522 g of H2O is produced. 

(a) What is the empirical formula of the unknown compound? 

(b) If the molar mass of the unknown compound is 72,1 g/mol, what is its molecular formula?  

Answer 1

$Moles (O_2) =\frac{0.1523}{16}=9.5×10^{-3}$

$Moles (Co_2) =\frac{0.3718}{44}=8.5×10^{-3}$

$Moles (O_2) =\frac{0.1522}{18}=8.5×10^{-3}$

Hence

$\frac{9.5×10^{-3}}{8.5×10^{-3}}=1.1$

$\frac{8.5×10^{-3}}{8.5×10^{-3}}=1$

$\frac{8.5×10^{-3}}{8.5×10^{-3}}=1$

Emperical formula = $CHO_2$

$Moles =\frac{ 0.1523}{72.1}=2.1×10^{-3}$

Molecular formula $= C_2H_2O_4$