Question 103f4

May 29, 2015

The ratio between the number of molecules will be 2:7.

You know that the mixture contains oxygen and nitrogen in a 1:4 mass ratio. You can write this as

${m}_{{O}_{2}} / {m}_{{N}_{2}} = \frac{1}{4}$

The mass of the two gases can be expressed using their respective molar masses and the number of moles of each present in the mixture

$m = \text{number of moles"/"molar mass}$

${m}_{{O}_{2}} = {n}_{{O}_{2}} / \text{32.0 g/mol}$

${m}_{{N}_{2}} = {n}_{{N}_{2}} / \left(\text{28.0 g/mol}\right)$

This means that the mass ratio that exists between the two gases can be written as

${m}_{{O}_{2}} / {m}_{{N}_{2}} = {n}_{{O}_{2}} / \frac{32.0 \cancel{\text{g/mol")) * (28.0cancel("g/mol}}}{n} _ \left({N}_{2}\right) = {n}_{{O}_{2}} / {n}_{{N}_{2}} = \frac{1}{4}$

$\frac{7 \cdot {n}_{{O}_{2}}}{8 \cdot {n}_{{N}_{2}}} = \frac{1}{4} \implies {n}_{{O}_{2}} / {n}_{{N}_{2}} = \frac{8}{28} = \frac{2}{7}$

One mole of any substance contains exactly $6.022 \cdot {10}^{23}$ atoms or molecules of that substance - this is known as Avogadro's number.

In order to have 1 mole of oxygen, you need to have $6.022 \cdot {10}^{23}$ molecules of oxygen. Likewise, 1 mole of nitrogen must contain $6.022 \cdot {10}^{23}$ molecules of nitrogen.

So, if your gaseous mixture contains oxygen and nitrogen in a $2 : 7$ mole ratio, then the ratio between the number of molecules of each gas must be equal to 2:7 as well.

Regardless of how many moles of oxygen you actually have, the 2:7 ratio you have between the two gases applies to the number of molecules too.

For example, if you have 0.5 moles of oxygen and 1.75 moles of nitrogen, you'd get

0.5cancel("moles"O_2) * (6.022 * 10^(23)"molecules")/(1cancel("mole"O_2)) = 3.011 * 10^(23)"molecules" ${O}_{2}$

and

1.75cancel("moles"N_2) * (6.022 * 10^(23)"molecules"N_2)/(1cancel("mole"N_2)) = 10.5385 * 10^(23)"molecules"# ${N}_{2}$

DIvide these two numbers to get

$\frac{3.011 \cdot \cancel{{10}^{23}}}{10.5385 \cdot \cancel{{10}^{23}}} = \frac{2}{7}$ $\to$ the molecule ratio is equal to the mole ratio