00:01
So to solve for this, let us first calculate the mass of nitrogen in the compound, assuming that we have 100 grams of the sample.
00:12
So that means the mass of nitrogen will be equivalent to 30 .46 grams.
00:18
And then the mass of oxygen will be equivalent to 69 .54 grams.
00:24
Now we find the number of moles of n and nitrogen that's going to be 30.
00:30
46 grams nitrogen multiply this by this rather by the atomic mass of nitrogen which is 14 .01 grams for every mole of nitrogen so that means 30 .46 divided by 14 .01 that's going to give us 2 point um it's right here we have uh 2 .174 moles of um nitrogen and and then for oxygen, we have 69 .54 grams of oxygen, divide it by the atomic mass of oxygen, which is 16 grams.
01:11
For every mole of oxygen, that means we have 4 .346 moles of oxygen.
01:18
Now we find the mole ratio.
01:21
So, the molecule of the amount ratio of n, that's going to be the number of moles of n, which is 2 .174...