It is often possible to change a hydrate into an anhydrous compound by heating it to drive off the water (dehydration). A 43.19 gram sample of a hydrate of MgBr2 was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, 27.21 grams of the anhydrous compound remained. What is the formula of the hydrate?

Let [tex]x[/tex] denote the number of [tex]\rm H_{2}O[/tex] formula units for every [tex]\rm MgCl_{2}[/tex] formula unit. The formula of the hydrate would be [tex]{\rm MgBr_2} \cdot x\, {\rm H_2O}[/tex].

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

[tex]\rm Mg[/tex]: [tex]24.305[/tex].
[tex]\rm Br[/tex]: [tex]79.904[/tex].
[tex]\rm H[/tex]: [tex]1.008[/tex].
[tex]\rm O[/tex]: [tex]15.999[/tex].

Formula mass:

[tex]\begin{aligned}M({\rm MgBr_{2}}) &= 24.305 + 2 \times 79.904 \\ &= 184.113\; \rm g\cdot mol^{-1}\end{aligned}[/tex].

[tex]\begin{aligned}M({\rm H_{2}O}) &= 2 \times 1.008 + 15.999 \\ &= 18.015\; \rm g\cdot mol^{-1}\end{aligned}[/tex].

The anhydrous compound would be [tex]\rm MgBr_{2}[/tex]. There was [tex]27.21\; \rm g[/tex] of this compound. Calculate the number of moles of formula units in that much of this compound:

[tex]\begin{aligned}n({\rm MgBr_{2}}) &= \frac{m({\rm MgBr_{2}})}{M({\rm MgBr_{2}})} \\ &= \frac{27.21\; \rm g}{184.113\; \rm g \cdot mol^{-1}} \\ &\approx 0.1478\; \rm mol\end{aligned}[/tex].

During heating, [tex]43.19\; \rm g - 27.21\; \rm g = 15.98\; \rm g[/tex] of water [tex]\rm H_{2}O[/tex] escaped from the hydrate. Calculate the number of formula units in that [tex]15.98\; \rm g[/tex] of [tex]\rm H_{2}O\![/tex]:

[tex]\begin{aligned}n({\rm H_{2}O}) &= \frac{m({\rm H_{2}O})}{M({\rm H_{2}O})} \\ &=\frac{43.19\; \rm g - 27.21\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \\&= \frac{15.98\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \\ &\approx 0.887\; \rm mol\end{aligned}[/tex].

In other words, approximately [tex]0.1478\; \rm mol[/tex] of [tex]\rm MgBr_{2}[/tex] formula units (and hence the same number of moles of [tex]{\rm MgBr_2} \cdot x\, {\rm H_2O}[/tex] formula units) contain approximately [tex]0.887\; \rm mol[/tex] of [tex]\rm H_{2}O[/tex] formula units. Calculate the ratio between the two:

[tex]\begin{aligned}x &= \frac{n({\rm H_{2}O})}{n({\rm MgBr_2})} \\ &\approx \frac{0.1478\; \rm mol}{0.887\; \rm mol} \approx 6\end{aligned}[/tex].

Hence, the formula of the hydrate would be [tex]\rm MgBr_2 \cdot 6\, H_2O[/tex].