# Question fb71c

Mar 15, 2015

Your mistery gas has a molar mass equal to $\text{28 g/mol}$; given the fact that it's diatomic, nitogen gas, ${N}_{2}$, is the closest match, its molar mass being $\text{28 g/mol}$.

So, you know that you must determine the identity of a diatomic gas by using a 6.1-L bulb. The first thing you must calculate is exactly how many moles of your mistery gas you added to the empty bulb once the experiment started.

Since you have all the data you need - temperature, pressure, and volume - the ideal gas law equation will help you determine the number of moles of gas

$P V = n R T \implies n = \frac{P V}{R T}$

n = ("1.00 atm" * "6.1 L")/(0.082("atm" * "L")/("mol" * "K") * (273.15 + 21)"K") = "0.253 moles"#

You know that the difference in the weight of the bulb between when it's empty and when it's filled with gas is 7.1 g. What this means is that the mass of the gas must be 7.1 g, since it caused the weight of the bulb to change (increase) by that many grams.

Since molar mass is defined by the mass of 1 mole, you can determine the gas' molar mass by dividing its mass by the number of moles you added

${M}_{M} = \text{mass"/"number of moles" = "7.1 g"/"0.253 moles" = "28.06 g/mol}$

If you round this to two sig figs, the number of sig figs in 7.1 g and 6.1 L, the answer will be

${M}_{M} = \text{28 g/mol}$

Rounded to two sig figs as well, the molar mass of nitrogen gas, ${N}_{2}$, is

${M}_{M \left({N}_{2}\right)} = \text{28 g/mol}$