In Kjeldahl's estimation of nitrogen, CuSO₄ act as

  $N{H}_4^{+}$  does not have vacant orbitals, therefore it cannot behave as electrophile.

Percentage Br =   $\frac{80*0.094}{188*0.16}*100=25\mathrm{\%}$

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 × no. of moles of H₂O = $\left(\frac{270}{18}\times 2\right)$ moles

 Mass of C = $\frac{330}{44}\times 12$ gm = 90 gm

 Mass of H = $\frac{270}{18}\times 2\times 1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}\times 100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}\times 100\mathrm{\%}=25\mathrm{\%}\end{array}$