The percent by mass of H₂O in the hydrated compound, Na₂CO₃.xH₂O = 64.13%
Further explanation
Given
90 g sample Na₂CO₃.xH₂O
32.28 g Na₂CO₃
Required
The percent by mass of H₂O
Solution
mass of H₂O = mass of sample - mass of the anhydrous salt, Na₂CO₃
![\tt mass~H_2O=90-32.28=57.72~g](https://img.qammunity.org/2021/formulas/chemistry/high-school/lo66vsedz6uu7rq1fgd0hgnicu557bhioh.png)
The percent by mass :
![\tt \%mass~H_2O=(mass~H_2O)/(mass~sample)* 100\%\\\\\%mass~H_2O=(57.72)/(90)* 100\%\\\\\%mass~H_2O=64.13\%](https://img.qammunity.org/2021/formulas/chemistry/high-school/e2tfd35bn33qt15aky3969crhger3ohhn5.png)