## Chemistry 12th Edition

We know that: $Molar\space mass\space of \space SnO_2=118.7+2(16.00)=150.7g$ Now: $\%Sn=\frac{118.7}{150.7}\times 100\%=78.77\%$ and $\%$ composition of oxygen is calculated as follows $\%O=\frac{2(16.00)}{150.7}\times 100\%=21.23\%$