# Question #e7283

Nov 30, 2017

$\text{0.32 g}$ methane.

#### Explanation:

Take the ideal gas law

$P V = n R T$

and the equation for molar mass,

$M M = \frac{m}{\text{mol}}$

When you rearrange that equation, you get

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

When you plug in the rearranged molar mass equation into ideal gas law, you get

$P V = \frac{m R T}{M M}$

I wasn't sure what you use for room temp and pressure, so I went with $\text{298 K}$ and $\text{1 atm}$. Typically this would be calculated at STP (standard temperature and pressure, $\text{273 K}$ and $\text{1 atm}$) because room temperature is subjective.

The molar mass of methane is $\text{16.04 g/mol}$; the ideal gas constant in atmospheres $\left(R\right)$, is $0.08206 \left(\text{atm" * "L")/("mol" * "K}\right)$

Plug in all of these values into the rearranged ideal gas law:

$\left(\text{1 atm")("0.48 L") = (m(0.08206 ("atm" * "L")/("mol" * "K"))("298 K")) / ("16.04 g/mol}\right)$

Plug this into a calculator and for $m$ (mass), you get $\text{0.32 g}$.