00:01
This question is a combination of using the ideal gas law and determining the empirical formula through mass percent.
00:11
So let's start with determining the empirical formula through mass percent.
00:16
If it's 11 .79 percent carbon, then if we have 100 grams of the material, we would have 11 .79 grams of carbon, which we can convert into moles of carbon by dividing by its smaller mass.
00:29
If it's 69 .57 % chlorine, then 100 grams of the compound would contain 69 .57 grams chlorine, which we can convert to moles by dividing by chlorine's molar mass.
00:45
Then if we sum up the two percentages that were given to us and subtract them from 100, what will be left over will be the percent that is fluorine, which ends up being 18 .64.
00:57
We divide those grams by the molar mass of fluorine in order to get the moles fluorine.
01:05
These then become the moles that are present in 100 grams of the sample.
01:11
If we divide by the smallest, we can then convert them to whole numbers, and we'll get one mole carbon, two moles chlorine, and one mole fluorine.
01:21
So the empirical formula is ccl2f.
01:26
Now we can use the ideal gas level, in order to determine the moles...