# A 5.0 g sample of a pure compound containing H, C, and N contains .19 g of H and 2.22 g of C. Another smaller sample of this pure compound is analyzed. How many grams of N should be found in the new 2.0 g pure sample?

Feb 25, 2016

$\text{1.0 g N}$

#### Explanation:

The idea here is that you need to use the amount of hydrogen and the amount of carbon found in the $\text{5.0-g}$ sample of this unknown compound to determine the percent composition of nitrogen.

Once you know the percent composition of nitrogen, you can determine the amount of nitrogen present in the $\text{2.0-g}$ sample.

So, if your compound contains only hydrogen, $\text{H}$, carbon, $\text{C}$, and nitrogen, $\text{N}$, then you can write the mass of the compound as the sum of the masses of these elements

${m}_{\text{compound}} = {m}_{H} + {m}_{C} + {m}_{N}$

This means that the $\text{5.0-g}$ sample will contain

${m}_{N} = \text{5.0 g" - "0.19 g" - "2.22 g" = "2.59 g N}$

If you divide the mass of nitrogen by the total mass of the compound and multiply the result by $100$, you will get the compound's percent composition of nitrogen

(2.59 color(red)(cancel(color(black)("g"))))/(5.0color(red)(cancel(color(black)("g")))) xx 100 = "51.8% N"

So, what does this tell you?

For every $\text{100 g}$ of this unknown compound, $\text{51.8 g}$ will be nitrogen.

This means that the $\text{2.0-g}$ sample will contain

2.0 color(red)(cancel(color(black)("g compound"))) * overbrace("51.8 g N"/(100color(red)(cancel(color(black)("g compound")))))^(color(purple)("% composition of N")) = "1.036 g N"

Rounded to two sig figs, the answer will be

${m}_{N} = \textcolor{g r e e n}{\text{1.0 g}}$