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## Problem 2 Solution

```In 80.042 g (1 mole) of ammonium nitrate (bomb city!), there are 28.014 grams
of nitrogen, 47.997 g of oxygen and the rest is hydrogen.
a. What percent of ammonium nitrate is nitrogen?
b. What percent of ammonium nitrate is oxygen?
c. What percent of ammonium nitrate is hydrogen?
```
```(a) 34.999%      (b) 59.965%     (c) 5.036%
```

### Solution Steps for Part (a):

What percent of ammonium nitrate is nitrogen?

We'll fill in the amounts that we know, and use the clues from the statement:
```What% of 80.042  is  28.014
x    * 80.042   =  28.014
```

dividing by 80.042 gives x = .34998688. Since these are obviously measurements, we will round to 5 digits: 34.999%

### Solution Steps for Part (b):

What percent of ammonium nitrate is oxygen?

We'll use the clues from the statement:
```What% of 80.042  is  47.997
x    * 80.042   =  47.997
```

dividing by 80.042 gives x = .59964769. Since these are obviously measurements, we will round to 5 digits: 59.965%. Notice that the units on 80.042 and 47.997 cancel out when we divide. Percents really are dimensionless!

### Solution Steps for Part (c):

What percent of ammonium nitrate is hydrogen?

Let's take the easy way out:

100% total - 34.999% nitrogen - 59.965% oxygen = 5.036% hydrogen

Next Try It Out Problem.

Developed by
Shodor
in cooperation with the Department of Chemistry,
The University of North Carolina at Chapel Hill