# Question #1fd4c

Feb 25, 2018

The molecular formula is ${C}_{4} {H}_{10}$

#### Explanation:

$n = \frac{V}{24} = \frac{9.6}{24} = 0.4 m o l$

$M r = \frac{M}{n} = \frac{23.2}{0.4} = 58 g m o {l}^{-} 1$

Since the formula mass of ${C}_{2} {H}_{5}$ is 29, it follows that the molecule contains $\frac{58}{29}$, or $2$ times the number of atoms.

Hence, the molecular formula is ${C}_{4} {H}_{10}$, which is butane.

Feb 25, 2018

The molecular formula is ${\text{C"_4"H}}_{10}$, which is probably butane.

#### Explanation:

We have ${\text{C"_2"H}}_{5}$ as the empirical formula. We must find its molar mass. To do so, we need to know how many moles of it evaporated. Use the Ideal Gas Law:

$P V = n R T$, and here:

$P = 1 \text{atm}$

$V = 9.6 \text{L}$

$R = 0.0821 {\text{L atm K"^-1"mol}}^{-} 1$

$T = 293.15 \text{K}$.

To solve for $n$, moles, we rearrange:

$n = \frac{P V}{R T}$, and inputting:

$n = \frac{1 \cdot 9.6}{293.15 \cdot 0.0821}$

$n = 0.399 \text{mol"~~0.4"mol}$

Another formula for moles is $n = \frac{m}{M}$, where $m$ is the mass and $M$ the molar mass.

Rearranging to solve for $M$, we get:

$M = \frac{m}{n}$

We have $m = 23.2 \text{g}$. Inputting:

$M = \frac{23.2}{0.399}$

$M = 58.16 \text{g/mol}$

We must calculate the molar mass of the empirical formula:

${M}_{e} = \left(12 \cdot 2\right) + \left(1 \cdot 5\right)$

${M}_{e} = 29 \text{g/mol}$

We see a $2 : 1$ ratio.

Therefore, the molecular formula is ${\text{C"_4"H}}_{10}$, which is probably butane.