# Question f957c

Apr 23, 2015

Your unknown gas is ethane and its molecular formula is ${C}_{2} {H}_{6}$.

So, you know that you're dealing with a hydrocarbon, which is a compound that contains only carbon and hydrogen. This means that you can use the percentage of carbon to determine the percentage of hydrogen in the compound

"%H" = 100% - "%C" = 100 - 80 = 20%

Before determining the compound's molecular formula, you're going to have to determine its empirical formula. To make the calculations easier, assume that you have a 100-g sample of your unknown gas.

Since you know the percent composition of the compound to be 80 g carbon and 20 g hydrogen, you can use the molar masses of the two elements to determine how many moles of each you'd get in that sample.

$\text{For C": (80.0cancel("g"))/(12.0cancel("g")/"mol") = "6.67 moles}$

$\text{For H": (20.0cancel("g"))/(1.00cancel("g")/"mol") = "20.0 moles}$

Divide these two numbers by smallest one to determine the mole ratio that exists between carbon and hydrogen

"For C": (6.67cancel("moles"))/(6.67cancel("moles")) = 1

"For H": (20.0cancel("moles"))/(6.67cancel("moles")) = 3#

This means that your empirical formula will be ${\left(C {H}_{3}\right)}_{n}$.

In order to determine the value of $n$, you need the molar mass of the compound. Since you know what the weight of 1.145 moles is, you can use that to determine the weight of 1 mole of gas

${M}_{M} = \frac{m}{n} = \text{43.5 g"/"1.45 moles" = "30.0 g/mol}$

Now determine the value of $n$ by

$\left(12.0 + 3 \cdot 1.00\right) \cdot n = 30.0 \implies n = \text{30.0"/"15.0} = 2$

Therefore, the molecular formula for your compound is

${\left(C {H}_{3}\right)}_{2} = \textcolor{g r e e n}{{C}_{2} {H}_{6}}$ $\to$ ethane.