Question

Question #f957c

Answer 1

Your unknown gas is ethane and its molecular formula is `C_2H_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.

`"For C": (80.0cancel("g"))/(12.0cancel("g")/"mol") = "6.67 moles"`

`"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 `(CH_3)_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 = m/n = "43.5 g"/"1.45 moles" = "30.0 g/mol"`

Now determine the value of `n` by

`(12.0 + 3 * 1.00) * n = 30.0 => n = "30.0"/"15.0" = 2`

Therefore, the molecular formula for your compound is

`(CH_3)_2 = color(green)(C_2H_6)` `->` ethane.