# Identifying a chloride of the unknown element from the masses of chlorine and compound as well as vapor density [closed]

1 g of a chloride of an element contains 0.835 g of chlorine. If vapour density of the chloride is 85, find weight of the element and its valency.

I tried to solve it but couldn't get to the answer then I saw its solution and I also could not understand what they want to explain. Can anyone please help me out. By the way, it's not my homework.

• First find empirical formula of the chloride. Then, use vapor density to find the molecular formula. According to Wikipedia vapor density is the density of a vapor in relation to that of hydrogen. So, you can tell what is the valency of the element. – Mathew Mahindaratne May 27 '19 at 4:18

I agree with the starting point of the Matthew's solution, namely the definition of vapor density, but I fail to understand the follow-up math, so I just post my approach.

$$M(\ce{ECl_x}) = D\cdot M(\ce{H2}) = 85\cdot\pu{2 g mol-1} = \pu{170 g mol-1}\tag{1}$$

By definition, molar mass $$M(\ce{ECl_x})$$ is also

$$M(\ce{ECl_x}) = M(\ce{E}) + x\cdot M(\ce{Cl})$$

$$\implies M(\ce{E}) = M(\ce{ECl_x}) - x\cdot M(\ce{Cl}) = (170 - x\cdot 35.5)~\pu{g mol-1}\label{eqn:2}\tag{2}$$

The unknown variable $$x$$ is defined by the molar ratio between the elements:

$$x = \frac{n(\ce{Cl})}{n(\ce{E})} = \frac{m(\ce{Cl})\cdot M(\ce{E})}{m(\ce{E})\cdot M(\ce{Cl})} = \frac{\pu{0.835 g}\cdot (170 - x\cdot 35.5)~\pu{g mol-1}}{(1 - 0.835)~\pu{g} \cdot \pu{35.5 g mol-1}} = 24.14 - 5.04\cdot x\label{eqn:3}\tag{3}$$

Solving \eqref{eqn:3}:

$$6.04\cdot x= 24.14 \quad\implies\quad x = 4.0 \tag{4}$$

Using $$x$$, we can now determine the element from \eqref{eqn:2}:

$$M(\ce{E}) = (170 - 4.0\cdot 35.5)~\pu{g mol-1} = \pu{28.0 g mol-1}\tag{5}$$

which appears to be silicon, valency 4. The unknown compound is silicon tetrachloride $$\ce{SiCl4}$$, a fuming volatile liquid at NTP.

Suppose the element is $$\ce{Q}$$, its chloride is $$\ce{QCl_x}$$ (since the oxidation number of $$\ce{Cl}$$ is $$-1$$, the oxidation number of $$\ce{Q}$$ would be $$+x$$), and the atomic weight of $$\ce{Q}$$ is $$W$$.

Since $$\pu{1.0 g}$$ of $$\ce{QCl_x}$$ contains $$\pu{0.835 g}$$ of $$\ce{Cl}$$, the mass of $$\ce{Q}$$ is $$\pu{(1.0-0.835) g} = \pu{0.165 g}$$. Also:

$$\frac{x}{1} = \frac{n_\ce{Cl}}{n_\ce{Q}}$$

But, $$n_\ce{Cl} = \frac{\pu{0.835 g}}{\pu{35.5 gmol^{-1}}}= \pu{0.0235 mol}$$ and $$n_\ce{Q} = \frac{\pu{0.165 g}}{\pu{W gmol^{-1}}}= \pu{\frac{0.165}{W} mol}$$. Therefore,

$$\frac{x}{1} = \frac{n_\ce{Cl}}{n_\ce{Q}} = \frac{\pu{0.0235 mol}}{\pu{\frac{0.165}{W} mol}}=0.1424W$$ $$\therefore x=0.1424W$$

Now, it is known that $$\text{vapor density} \approx \frac12 \times \text{molar mass}$$ (Wikipedia). $$\therefore W + x \times 35.5 = 2 \times 85= 170$$ Apply, $$x=0.1424W$$ here, and hence, $$W + 0.1424W \times 35.5 = (1+5.055)W = 6.055W = 170$$ $$\therefore W=\frac{170}{6.055}=28.08$$ $$\therefore x=0.1424 \times 28.08=4.0 \approx 5$$

Thus, the valency of the element is $$+4$$. Since its calculated atomic weight is $$\pu{28.08 gmol^{-1}}$$, it should be silicon ($$\ce{Si}$$). Thus, I conclude that the compound is $$\ce{SiCl4}$$.

• The answer looks logical, but I think you might want to check the math part. For some reason I arrive at $\ce{SiCl4}$; also the molar mass of $\ce{PCl5}$ (208 g/mol) is way off from 170 g/mol in my opinion. – andselisk May 27 '19 at 20:12
• @andselisk: I checked and corrected the answer. Thank you for the careful reading. My problem was not counting the forth decimal point. I have been sleepy that night and been sloppy not to check back the calculation. Thanks again. – Mathew Mahindaratne May 27 '19 at 20:46