0
$\begingroup$

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.

$\endgroup$

closed as off-topic by M.A.R., Tyberius, Todd Minehardt, Mithoron, Nilay Ghosh May 28 at 3:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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. $\endgroup$ – Mathew Mahindaratne May 27 at 4:18
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\ce{SiCl4}$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – andselisk May 27 at 20:12
  • 1
    $\begingroup$ @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. $\endgroup$ – Mathew Mahindaratne May 27 at 20:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.