The complete combustion of 0.492g of an organic compound containing 'C', 'H' and 'O' given 0.793g of CO₂ and 0.442g H₂O. The percentage of oxygen composition in the organic compound is __________. (nearest integer)

Moles of C = moles of CO₂

=  $\frac{0.793}{44}mol$

Mass of C =  $\frac{0.793}{44}*12g$

= 0.261g

Moles of H = 2 *  $\frac{0.442}{18}mol$

Mass of H =  $2*\frac{0.442}{18}*1g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492}*100$  

= 46.14%

the nearest integer = 46