A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?

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A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?

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Answer 1 · 2015-11-14T22:00:02

Assuming ideality, the molecular formula is $C_{2} H_{2} C l_{2}$ $C_2H_2Cl_2$

Explanation:

From the ideal gas law, $M M = \frac{m R T}{P V}$ $MM = (mRT)/(PV)$ ; where $M M$ $MM$ $=$ $=$ molecular mass; $m$ $m$ $=$ $=$ mass; $P$ $P$ $=$ $=$ pressure in atmospheres; $V =$ $V =$ volume in litres; $R$ $R$ $=$ $=$ gas constant with appropriate units.

So, $(0.800*gxx0.0821*cancelL*cancel(atm)*cancelK^(-1)*mol^(-1)xx373*cancelK)/(0.256*cancelLxx0.987*cancel(atm))$ $=$ $97.0$ $g*mol^(-1)$ .

$nxx(12.01+1.01+2xx35.45)*g*mol^(-1)$ $=$ $97.0*g*mol^(-1)$ .

Clearly, $n$ $=$ $1$ . And molecular formula $=$ $C_2H_2Cl_2$ .

I seem to recall (but can't be bothered to look up) that vinylidene chloride, $H_2C=C(Cl)_2$ is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.

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A compound has the empirical formula CHCl. A 256 mL flask, at 373 K and 750 torr, contains 0.800 g of the compound, what is the molecular formula?

1 Answer

Explanation:

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