# A 1.50-g sample of nitrous oxide contains 2.05 * 10^22 N_2O molecules. How many nitrogen atoms are in this sample? How many nitrogen atoms are in 44.0 g of nitrous oxide?

Feb 18, 2016

There are $4.1 \cdot {10}^{22}$ nitrogen atoms in 1.50g and $1.20 \cdot {10}^{24}$ in a 44.0g sample.

#### Explanation:

Note: This answer may seem to be quite long, but that is because I have tried to explain each of the steps fully and clearly. If you want, you can just read the first method for finding the second part of the question but the second method will be useful to you in future if you are not already learning it, and I would recommend at least reading through and trying to understand the concepts involved.

The answer for the first part of the question is quite simple. In 1 ${N}_{2} O$ molecule there are 2 nitrogen atoms, indicated by the ${N}_{2}$ in the formula. To find the number of nitrogen atoms in any amount of ${N}_{2} O$, we simply multiply the number of ${N}_{2} O$ molecules by 2:

$\text{Number of N atoms} = 2.05 \cdot {10}^{22} \times 2 = 4.1 \cdot {10}^{22}$

There are two ways of finding the answer to the next part of the question. The easiest way to find this is by using the number given to us in the question. We know that there are $2.05 \cdot {10}^{22} {N}_{2} O$ molecules in 1.50g. We can divide this this number by 1.5 to find the number of ${N}_{2} O$ in 1g, and then multiply that by 44 to find the number in 44g.

$\frac{2.05 \cdot {10}^{22}}{1.5 g} = 1.367 \cdot {10}^{22} {g}^{-} 1$

$1.367 \cdot {10}^{22} {g}^{-} 1 \times 44 g = 6.013 \cdot {10}^{23}$

Now we know the number of ${N}_{2} O$ in 44g , we multiply that by 2 to get the number of nitrogen atoms:

$6.013 \cdot {10}^{23} \times 2 = 1.203 \cdot {10}^{24}$ nitrogen atoms.

The second, more accurate way to find this out is by using the relative atomic masses of the atoms in ${N}_{2} O$ molecules.

If you look on a periodic table , you see that Nitrogen has a relative atomic mass of about 14. This is not completely accurate, but using the correct value of 14.007 makes our calculations harder and does not make a large difference to the final answer we get (Ask your teacher which value you should use and then use that in future). We can also see that the relative atomic mass of oxygen is about 16 (actually it's 15.999).

To find out the amount of substance in a sample we use the equation $n = \frac{m}{M}$ where:
$n$ is the amount of substance in in the sample in moles ($m o l$)
$m$ is the mass of the sample in grams ($g$) and
$M$ is the molar mass of the molecules in the sample in grams per mole ($g m o {l}^{-} 1$)

To find the molar mass ($M$) of ${N}_{2} O$, we have to add the relative atomic masses of the atoms that it contains:

$M \left({N}_{2} O\right) = \left(2 \times 14\right) + \left(1 \times 16\right) = 44 g m o {l}^{-} 1$, since there are 2 nitrogen atoms and 1 oxygen atom in 1 ${N}_{2} O$ molecule.
We can now substitute these values into the equation to get:

$n \left({N}_{2} O\right) = \frac{44 g}{44 g m o {l}^{-} 1} = 1 m o l$

Now we see we have 1 mole of ${N}_{2} O$ in the 44g sample, but we need to know the number of molecules in a mole. This number is called Avogadro's Number and is equal to approximately $6.02 \cdot {10}^{23}$, so there are $6.022 \cdot {10}^{23}$ molecules of ${N}_{2} O$ in our sample. Now we multiply this number by 2, the same as in method 1, to get:

$6.022 \cdot {10}^{23} \times 2 = 1.204 \cdot {10}^{24}$ nitrogen atoms.

Both methods produced a similar answer, and both could be reasonably rounded to $1.20 \cdot {10}^{24}$nitrogen atoms in 44g of ${N}_{2} O$