Question

A sample of hydrated tin (II) chloride (SnCl2) has a mass of 4.90 g. When it is dehydrated, it has a mass of 4.10 g. Which is the correct chemical formula for the hydrate?

SnCl2•2H2O     

SnCl2•4H2O     

SnCl2•6H2O

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Answer 1

Answer:

SnCl₂·2 H₂O.

Explanation:

Relative atomic mass data from a modern periodic table:

• Sn: 118.710;
• Cl: 35.45;
• H: 1.008;
• O: 15.999.

How many moles of SnCl₂ formula units in this sample?

The first mass 4.90 grams contain both the SnCl₂ formula units and a number of water molecules. Luckily, the mass of the dehydrated salt 4.10 grams contains only SnCl₂.

Formula mass of tin (II) chloride SnCl₂:

$M(\rm SnCl_2) = 118.710 + 2\times 35.45 = 189.610\; g\cdot mol^{-1}$.

Number of moles of tin (II) chloride SnCl₂ formula units in this sample:

$\displaystyle n(\mathrm{SnCl_2}) = \frac{m}{M} = \rm \frac{4.10\; g}{189.610\; g\cdot mol^{-1}} = 0.0216233\; mol$.

How many moles of water molecules H₂O in this sample?

Water of crystallization exist as H₂O molecules in typical hydrated salts. The molar mass of these molecules will be:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g\cdot mol^{-1}$.

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,

$\begin{aligned}m(\text{Water of Hydration})&=m(\text{Hydrated Sample}) - m(\text{Anhydrous Sample}) \\ & = \rm 4.90\; g - 4.10\; g \\ &= \rm 0.80\; g\end{aligned}$.

$\displaystyle n(\mathrm{H_2O}) = \frac{m}{M} = \rm \frac{0.80\; g}{18.015\; g\cdot mol^{-1}} = 0.0444074\; mol$.

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 SnCl₂ formula unit?

$\displaystyle \frac{n(\mathrm{H_2O})}{n(\mathrm{SnCl_2})} = 2.05 \approx 2$.

There are approximately two water molecules for each SnCl₂ formula unit. The formula of this compound shall thus be $\rm SnCl_2 \cdot 2H_2 O$.