# Question #cfef7

Feb 1, 2015

At STP, 1 mole of any ideal gas occupies exactly $\text{22.4 L}$ - this is known as the molar volume of a gas at STP. You were given $\text{15.0 L}$, which automatically means that you have less than 1 mole.

$\text{15.0 L" * ("1 mole")/("22.4 L") = "0.670 moles oxygen}$

The link between moles and grams (and vice versa) is made through molar mass; a compound's (or an element's) molar mass expresses the weight in grams of 1 mole of that particular compound.

Since oxygen gas exists as a diatomic molecule - ${\text{O}}_{2}$, its molar mass will be twice that of one oxygen atom, which has a molar mass of $\text{16.0 g/mol}$.

Therefore, the mass of oxygen in grams will be

$\text{0.670 moles O"_2 * ("32.0 "g)/("1 mole O"_2) = "21.4 g}$

Feb 2, 2015

The mass of oxygen gas is 21.43g

At STP 1 mole of gas has a volume of 22.4 L

Oxygen has the molecular formula ${O}_{2}$.

The ${A}_{r}$ of oxygen = 16

So the ${M}_{r}$ of ${O}_{2} = \left(2 \times 16\right) = 32$

So 1 mole of ${O}_{2}$ weighs 32g

This means that 22.4 L weighs 32g

So 1L weighs $\frac{32}{22.4} g$

So 15 L weighs $\frac{32}{22.4} \times 15 = 21.43 g$