What are the empirical and molecular formulae of an organic species, whose percentage composition is: C:38.7%;H:9.7%;O,51.6%? A 0.93*g mass of this material occupies 0.336*L at "STP".

1 Answer
May 18, 2017

We get, eventually, $\text{a molecular formula of}$ ${C}_{2} {H}_{6} {O}_{2}$...........

Explanation:

As with all these problems, we assume a mass of $100 \cdot g$, and from this mass we calculate the molar quantity of each element.

$\text{Moles of carbon} = \frac{38.7 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 3.22 \cdot m o l$.

$\text{Moles of hydrogen} = \frac{9.7 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 9.62 \cdot m o l$.

$\text{Moles of oxygen} = \frac{51.6 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 3.22 \cdot m o l$.

(Note that an analyst normally does not give you %O, you are given %C,H,N, and if these percentages do not sum to 100%, the balance is ASSUMED to be %O.)

And now we divide thru by the smallest atomic quantity to get an empirical formula of $C {H}_{3} O$.

And now we use the gas data to estimate a molecular mass. The molar volume an Ideal Gas at $S T P$ is $22.4 \cdot L$.

And thus the molar mass of our vapour is $\frac{0.93 \cdot g}{\frac{0.336 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}} = 62.0 \cdot g \cdot m o {l}^{-} 1$.

Now it is FACT, the molecular formula is ALWAYS a whole number multiple of the empirical formula, i.e.

$\text{Molecular formula"=nxx"Empirical formula}$

And thus................................

$n \times \left(12.011 + 3 \times 1.00794 + 16.00\right) \cdot g \cdot m o {l}^{-} 1 = 62.0 \cdot g \cdot m o {l}^{-} 1$

We solves for $n$, to get $n = 2$, and thus $\text{molecular formula}$ $=$ ${C}_{2} {H}_{6} {O}_{2}$.

$\text{Phew!}$