Q58E - Chemistry (OpenStax)

Question

Heating a sample of \(N{a_2}C{O_3} \cdot x{H_2}O{\rm{ }}\) weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous\(N{a_2}C{O_3}\). What is the formula of the hydrated compound?

Step-by-Step Solution

Answer

The final structure is \({\rm{ N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \cdot 10{{\rm{H}}_2}{\rm{O}}{\rm{.}}\)

 

Step 1: Formula of hydrated compound

In order to determine the formula of the hydrate, the number of moles of water per mole of anhydrous solid (x) will be calculated by dividing the number of moles of water by the number of moles of the anhydrous solid.

Step 2: Finding the number of moles of sodium carbonate

The reaction is\({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \cdot x{{\rm{H}}_2}{\rm{O}} \to {\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} + x{{\rm{H}}_2}{\rm{O,}}\) where \(x\) is the number of moles in the formula.

 

• The molar mass ofNa is 23 g/mol.
• The molar mass of C is 12 g/mol.
• The molar mass of O is 16 g/mol.
• The molar mass of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\)is 106 g/mol.

 

 

The number of moles of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) is evaluated as:

\(n\left( {N{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}} \right) = \frac{{{\rm{mass of N}}{a_2}{\rm{C}}{{\rm{O}}_3}}}{{{\rm{Molar mass of N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}\)

 

The known values are replaced:

 

\(\begin{aligned}{\underline{\phantom{xx}}}n{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} &= \frac{{1.72\;{\rm{g}}}}{{106\frac{{\rm{g}}}{{{\rm{mol}}}}}}\\ &= 106\;{\rm{g/mol}}\\ &= 0.016\;\;{\rm{mol}}{\rm{.}}\end{aligned}\)

Step 3: Finding the number of moles of water

Next, we must calculate the number of moles of \({{\rm{H}}_2}{\rm{O}}\):

• The molar mass ofH is 1 g/mol.
• The molar mass of O is 16 g/mol.
• The molar mass of \({{\rm{H}}_2}{\rm{O}}\) is 18 g/mol.

 

\(\begin{aligned}{\underline{\phantom{xx}}}{\rm{Mass of }}{{\rm{H}}_{\rm{2}}}{\rm{O =  Mass of N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{.x}}{{\rm{H}}_{\rm{2}}}{\rm{O -  Mass of N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\\{\rm{ = 4}}{\rm{.64\;g - 1}}{\rm{.72\;g}}\\{\rm{ = 2}}{\rm{.92 g}}{\rm{.}}\end{aligned}\)

 

\(\begin{aligned}{\underline{\phantom{xx}}}{\rm{Number of moles of }}{{\rm{H}}_2}{\rm{O}} &= \frac{{{\rm{ Mass of }}{{\rm{H}}_2}{\rm{O}}}}{{{\rm{ Molar mass of }}{{\rm{H}}_2}{\rm{O}}}}\\ &= \frac{{2.92\;g}}{{18\frac{g}{{mol}}}}\\ &= 0.16\;\;{\rm{mol}}{\rm{.}}\end{aligned}\)

Step 4: Calculation of final proportion

Now we can calculate the final proportion of the hydrated compound based on the molecular formula of  \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \cdot x{{\rm{H}}_2}{\rm{O}}\).

\({\rm{1 mol N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{:}}\;{\rm{xmol, }}{{\rm{H}}_{\rm{2}}}{\rm{O  =  0}}{\rm{.016\;mol : 0}}{\rm{.16\;mol,}}\) the proportion is 1:10.

 

So, the final structure is \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} \cdot 10{{\rm{H}}_2}{\rm{O}}{\rm{.}}\)