Question

Calculate the percentage of water of crystallization in a sample of washing soda, ${\mathrm{Na}}_2{\mathrm{CO}}_3.10{\mathrm{H}}_2\mathrm{O}$. $\left[\mathrm{Na}=23,\mathrm{C}=12,\mathrm{O}=16,\mathrm{H}=1\right]$

Answer 1

Step 1: Given information

• The given compound is washing soda, ${\mathrm{Na}}_2{\mathrm{CO}}_3.10{\mathrm{H}}_2\mathrm{O}$.
• Atomic mass of sodium $\left(\mathrm{Na}\right)=23\mathrm{g}$
• Atomic mass of carbon $\left(\mathrm{C}\right)=12\mathrm{g}$
• Atomic mass of hydrogen $\left(\mathrm{H}\right)=1\mathrm{g}$
• Atomic mass of oxygen $\left(\mathrm{O}\right)=16\mathrm{g}$

Step 2: The formula used

$\mathrm{Percentage}\mathrm{of}\mathrm{water}\mathrm{of}\mathrm{crystallization}=\frac{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{washing}\mathrm{soda}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{washing}\mathrm{soda}}\times 100$

Step 3: Determine molecular mass

• Molecular masses are as follows:

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{Na}}_2{\mathrm{CO}}_3.10{\mathrm{H}}_2\mathrm{O}& =& \left(23\mathrm{g}\times 2\right)+12\mathrm{g}+\left(16\mathrm{g}\times 3\right)+10\times \left(1\mathrm{g}\times 2+16\mathrm{g}\right)\\ & =& 286\mathrm{g}\end{array}$

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{mass}\mathrm{of}10{\mathrm{H}}_2\mathrm{O}\mathrm{molecule}& =& 10\times \left(1\mathrm{g}\times 2+16\mathrm{g}\right)\\ & =& 180\mathrm{g}\end{array}$

Step 4: Calculate percentage of the water of crystallization

• The percentage of the water of crystallization in washing soda is as follows:

$\begin{array}{rcl}\mathrm{Percentage}\mathrm{of}\mathrm{water}\mathrm{of}\mathrm{crystallization}& =& \frac{180\mathrm{g}}{286\mathrm{g}}\times 100\\ & =& 62.94\%\end{array}$

Therefore, $62.94\%$ of the water of crystallization in washing soda.