Answer:
SnCl₂·2 H₂O.
Explanation:
Relative atomic mass data from a modern periodic table:
How many moles of SnClâ‚‚ formula units in this sample?
The first mass 4.90 grams contain both the SnClâ‚‚ formula units and a number of water molecules. Luckily, the mass of the dehydrated salt 4.10 grams contains only SnClâ‚‚.
Formula mass of tin (II) chloride SnClâ‚‚:
[tex]M(\rm SnCl_2) = 118.710 + 2\times 35.45 = 189.610\; g\cdot mol^{-1}[/tex].
Number of moles of tin (II) chloride SnClâ‚‚ formula units in this sample:
[tex]\displaystyle n(\mathrm{SnCl_2}) = \frac{m}{M} = \rm \frac{4.10\; g}{189.610\; g\cdot mol^{-1}} = 0.0216233\; mol[/tex].
How many moles of water molecules Hâ‚‚O in this sample?
Water of crystallization exist as Hâ‚‚O molecules in typical hydrated salts. The molar mass of these molecules will be:
[tex]M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g\cdot mol^{-1}[/tex].
The mass of water in the hydrated salt is the same as the mass that is lost when the water molecules are removed and the salt is dehydrated.
In other words,
[tex]\begin{aligned}m(\text{Water of Hydration})&=m(\text{Hydrated Sample}) - m(\text{Anhydrous Sample}) \\ & = \rm 4.90\; g - 4.10\; g \\ &= \rm 0.80\; g\end{aligned}[/tex].
[tex]\displaystyle n(\mathrm{H_2O}) = \frac{m}{M} = \rm \frac{0.80\; g}{18.015\; g\cdot mol^{-1}} = 0.0444074\; mol[/tex].
What's the coefficient in front of water in the formula of this hydrated salt? In other words, how many water molecules are there in the compound for each SnClâ‚‚ formula unit?
[tex]\displaystyle \frac{n(\mathrm{H_2O})}{n(\mathrm{SnCl_2})} = 2.05 \approx 2[/tex].
There are approximately two water molecules for each SnClâ‚‚ formula unit. The formula of this compound shall thus be [tex]\rm SnCl_2 \cdot 2H_2 O[/tex].
