What is the mass percent of oxygen in ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3.18{\mathrm{H}}_2\mathrm{O}? (\mathrm{MW}=666.43)$

Hydrous aluminum sulphate formula =  ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3.18{\mathrm{H}}_2\mathrm{O}$

Given M.Wt = 666.43 grams

% of element in compound $=\frac{\mathrm{No}.\mathrm{of} \mathrm{atoms}\times \mathrm{A}.\mathrm{Wt}}{\mathrm{M}.\mathrm{Wt} \mathrm{of} \mathrm{compounds} }\times 100$

$=\frac{30\times 16}{666.4}\times 100$