Question

Calculate the percentage of water for each of the hydrates listed:

a. MgSO4 ● 7H2O




b. LiC2H3O2 ● 2H2O



c. Al(NO3)3 ● 9H2O

Answer 1

Mass ratio:

a. Approximately
`51.2\%`.

b. Approximately
`35.3\%`.

c. Approximately
`43.2\%`.

Step-by-step explanation:

Look up the relative atomic mass of each element on a modern periodic table:

• 
  `\rm Mg`:
  `24.305`.
• 
  `\rm S`:
  `32.06`.
• 
  `\rm O`:
  `15.999`.
• 
  `\rm H`:
  `1.008`.

• 
  `\rm Li`:
  `6.94`.
• 
  `\rm C`:
  `12.011`.

• 
  `\rm Al`:
  `26.982`.
• 
  `\rm N`:
  `14.007`.

For example, the relative atomic mass of
`\rm Al` is
`26.982`. Hence, the mass of one mole of
`\rm Al\!` atoms would be (approximately)
`26.982\; \rm g`.

Calculate the formula mass of
`\rm H_2O`:

`\begin{aligned} & M({\rm H_2O}) \\ &= 2 * 1.008 + 15.999 \\ &= 18.015\; \rm g \cdot mol^(-1)\end{aligned}`.

Calculate the formula mass for each of the hydrates:

`\begin{aligned}& M({\rm MgSO_4 \cdot {7\, H_2O}}) \\ &= 24.305 + 32.06 + 4 * 15.999 \\ & \quad\quad + 7 * (2 * 1.008 + 15.999) \\ &= 246.466\; \rm g \cdot mol^(-1)\end{aligned}`.

`\begin{aligned}& M({\rm LiC_2H_3O_2 \cdot {2\, H_2O}}) \\ &= 6.94 + 2 * 12.011 + 3 * 1.008 + 2 * 15.999 \\ & \quad\quad + 2 * (2 * 1.008 + 15.999) \\ &= 102.014\; \rm g \cdot mol^(-1)\end{aligned}`.

`\begin{aligned}& M({\rm Al(NO_3)_3 \cdot {9\, H_2O}}) \\ &= 26.982 + 3 * (14.007 + 3 \time 15.999) \\ &\quad \quad + 9 * (2 * 1.008 + 15.999) \\ &= 375.129\; \rm g \cdot mol^(-1)\end{aligned}`.

The mass of
`1\; \rm mol` of
`{\rm MgSO_4 \cdot {7\, H_2O}}` formula units is
`246.466\; \rm g`.

There are seven mole of
`\rm H_2O` formula units in that many
`{\rm MgSO_4 \cdot {7\, H_2O}}` formula units. The mass of that
`7\; \rm mol` of
`\rm H_2O\!` formula units would be
`7 \; \rm mol * 18.015\; \rm g \cdot mol^(-1) = 127.105\; \rm g`.

Hence, the percentage mass of
`\rm H_2O` in
`{\rm MgSO_4 \cdot {7\, H_2O}}` would be:

`\begin{aligned}(127.015\; \rm g)/(246.466\; \rm g) \approx 51.2\%\end{aligned}`.

Similarly:

• Mass of
  `1\;\rm mol` of
  `\rm LiC_2H_3O_2 \cdot {2\, H_2O}` formula units:
  `102.014\; \rm g`.
• Mass of the
  `2\; \rm mol` of
  `\rm H_2O` in that
  `1\;\rm mol` of
  `\rm LiC_2H_3O_2 \cdot {2\, H_2O}` formula units:
  `2 \; \rm mol * 18.015\; \rm g \cdot mol^(-1) = 36.030\; \rm g`.
• Percentage mass of
  `\rm H_2O` in
  `\rm LiC_2H_3O_2 \cdot {2\, H_2O}`:
  `\begin{aligned}(36.030\; \rm g)/(102.014\; \rm g) \approx 35.3\%\end{aligned}`.

• Mass of
  `1\;\rm mol` of
  `\rm Al(NO_3)_3 \cdot {9\, H_2O}` formula units:
  `375.129\; \rm g`.
• Mass of the
  `9\; \rm mol` of
  `\rm H_2O` in that
  `1\;\rm mol` of
  `\rm Al(NO_3)_3 \cdot {9\, H_2O}` formula units:
  `9 \; \rm mol * 18.015\; \rm g \cdot mol^(-1) = 162.135\; \rm g`.
• Percentage mass of
  `\rm H_2O` in
  `\rm Al(NO_3)_3 \cdot {9\, H_2O}`:
  `\begin{aligned}(162.135\; \rm g)/(375.129\; \rm g) \approx 43.2\%\end{aligned}`.