What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP?

Jun 18, 2018

The molar mass of the gas is $\text{58.9 g/mol}$.

Explanation:

You will need to use the equation for the ideal gas law:

$P V = n R T$,

where:

$P$ is pressure, $V$ is volume, $n$ is moles, $R$ is gas constant, and $T$ is temperature.

$\text{STP}$ is ${0}^{\circ} \text{C}$ or $\text{273.15 K}$ (required for gas laws), and ${10}^{5}$ $\text{Pa}$ or $\text{100 kPa}$.

Use the equation for the ideal gas law to calculate moles of gas. Then calculate the molar mass by dividing the given mass by the calculated moles.

Known

$P = \text{100 kPa}$

$V = \text{1.35 L}$

$R = \text{8.31446 L kPa K"^(-1) "mol"^(-1)}$

$T = \text{273.15 K}$

Unknown

$n$

Solution

Rearrange the ideal gas law equation to isolate $n$. Plug in the known values and solve.

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

n=(100color(red)(cancel(color(black)("kPa")))xx1.35color(red)(cancel(color(black)("L"))))/(8.31446 color(red)(cancel(color(black)("L"))) color(red)(cancel(color(black)("kPa"))) color(red)(cancel(color(black)("K")))^(-1) "mol"^(-1)xx273.15color(red)(cancel(color(black)("K"))))="0.0594 mol" (rounded to three significant figures)

To calculate the molar mass of the gas, divide its given mass in grams by the calculated number of moles.

$\text{molar mass"=("3.50 g")/("0.0594 mol")="58.9 g/mol}$