Answer:
Mass ratio:
a. Approximately [tex]51.2\%[/tex].
b. Approximately [tex]35.3\%[/tex].
c. Approximately [tex]43.2\%[/tex].
Explanation:
Look up the relative atomic mass of each element on a modern periodic table:
- [tex]\rm Mg[/tex]: [tex]24.305[/tex].
- [tex]\rm S[/tex]: [tex]32.06[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
- [tex]\rm H[/tex]: [tex]1.008[/tex].
- [tex]\rm Li[/tex]: [tex]6.94[/tex].
- [tex]\rm C[/tex]: [tex]12.011[/tex].
- [tex]\rm Al[/tex]: [tex]26.982[/tex].
- [tex]\rm N[/tex]: [tex]14.007[/tex].
For example, the relative atomic mass of [tex]\rm Al[/tex] is [tex]26.982[/tex]. Hence, the mass of one mole of [tex]\rm Al\![/tex] atoms would be (approximately) [tex]26.982\; \rm g[/tex].
Calculate the formula mass of [tex]\rm H_2O[/tex]:
[tex]\begin{aligned} & M({\rm H_2O}) \\ &= 2 \times 1.008 + 15.999 \\ &= 18.015\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the formula mass for each of the hydrates:
[tex]\begin{aligned}& M({\rm MgSO_4 \cdot {7\, H_2O}}) \\ &= 24.305 + 32.06 + 4 \times 15.999 \\ & \quad\quad + 7 \times (2 \times 1.008 + 15.999) \\ &= 246.466\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
[tex]\begin{aligned}& M({\rm LiC_2H_3O_2 \cdot {2\, H_2O}}) \\ &= 6.94 + 2 \times 12.011 + 3 \times 1.008 + 2 \times 15.999 \\ & \quad\quad + 2 \times (2 \times 1.008 + 15.999) \\ &= 102.014\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
[tex]\begin{aligned}& M({\rm Al(NO_3)_3 \cdot {9\, H_2O}}) \\ &= 26.982 + 3 \times (14.007 + 3 \time 15.999) \\ &\quad \quad + 9 \times (2 \times 1.008 + 15.999) \\ &= 375.129\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
The mass of [tex]1\; \rm mol[/tex] of [tex]{\rm MgSO_4 \cdot {7\, H_2O}}[/tex] formula units is [tex]246.466\; \rm g[/tex].
There are seven mole of [tex]\rm H_2O[/tex] formula units in that many [tex]{\rm MgSO_4 \cdot {7\, H_2O}}[/tex] formula units. The mass of that [tex]7\; \rm mol[/tex] of [tex]\rm H_2O\![/tex] formula units would be [tex]7 \; \rm mol \times 18.015\; \rm g \cdot mol^{-1} = 127.105\; \rm g[/tex].
Hence, the percentage mass of [tex]\rm H_2O[/tex] in [tex]{\rm MgSO_4 \cdot {7\, H_2O}}[/tex] would be:
[tex]\begin{aligned}\frac{127.015\; \rm g}{246.466\; \rm g} \approx 51.2\%\end{aligned}[/tex].
Similarly:
- Mass of [tex]1\;\rm mol[/tex] of [tex]\rm LiC_2H_3O_2 \cdot {2\, H_2O}[/tex] formula units: [tex]102.014\; \rm g[/tex].
- Mass of the [tex]2\; \rm mol[/tex] of [tex]\rm H_2O[/tex] in that [tex]1\;\rm mol[/tex] of [tex]\rm LiC_2H_3O_2 \cdot {2\, H_2O}[/tex] formula units: [tex]2 \; \rm mol \times 18.015\; \rm g \cdot mol^{-1} = 36.030\; \rm g[/tex].
- Percentage mass of [tex]\rm H_2O[/tex] in [tex]\rm LiC_2H_3O_2 \cdot {2\, H_2O}[/tex]: [tex]\begin{aligned}\frac{36.030\; \rm g}{102.014\; \rm g} \approx 35.3\%\end{aligned}[/tex].
- Mass of [tex]1\;\rm mol[/tex] of [tex]\rm Al(NO_3)_3 \cdot {9\, H_2O}[/tex] formula units: [tex]375.129\; \rm g[/tex].
- Mass of the [tex]9\; \rm mol[/tex] of [tex]\rm H_2O[/tex] in that [tex]1\;\rm mol[/tex] of [tex]\rm Al(NO_3)_3 \cdot {9\, H_2O}[/tex] formula units: [tex]9 \; \rm mol \times 18.015\; \rm g \cdot mol^{-1} = 162.135\; \rm g[/tex].
- Percentage mass of [tex]\rm H_2O[/tex] in [tex]\rm Al(NO_3)_3 \cdot {9\, H_2O}[/tex]: [tex]\begin{aligned}\frac{162.135\; \rm g}{375.129\; \rm g} \approx 43.2\%\end{aligned}[/tex].
