00:01
Calculates the percent composition, the percent mass of each element for the following elements.
00:07
So we have chloroform, potassium nitrate, c6 -h6, fenzene, and we have sulfuric acid, and we have ammonia.
00:17
So the first thing we have to do is determine the molar mass for each one of these.
00:24
And all you do is you add up all of the masses for each one of the elements.
00:28
So for chloroform, it weighs 119 .38, and potassium nitrate is 101 .101 .10.
00:40
Benzyme, 78 .11, sulfuric acid, 98 .08, and ammonia, 17 .03.
00:50
So the percent mass for each element is just the mass of that element, divided by the total mass of the compound.
01:00
So for each one of these, the units are grams per mole.
01:05
I'm just going to write down the first one.
01:08
For carbon, for the first one, that's 12 .01 divided by 119 .38, and you will get a certain number.
01:20
And then you have hydrogen, which is 1 .008 divided by 119 .38.
01:28
And chlorine is three times the mass of chlorine, which is 35 .453, all over 119 .38.
01:46
So that mass i'm actually going to put right down here.
01:52
And i'm going to jump cut to filling in all of the rest.
01:57
So here we have all of our values.
01:59
I will go ahead and separate them out a little bit by drawing a line where they are all separate.
02:09
Now i left these as a decimal.
02:12
You can go ahead and leave them as a decimal or you can put them into a percentage.
02:18
And if you are going to put them into a percentage, all you have to do is move the decimal point to places to the right.
02:27
So for example, this 0 .106, all you have to do is move it two places to the right, and it becomes 10 .06%.
02:40
Same thing with this hydrogen.
02:44
It's 8 .4 times 10 to the minus 3.
02:48
So 0 .84, and that's 8 .44%.
02:56
This 89 .09 will become, 89 .1%.
03:03
So you can leave them as a decimal point or you can put them into a percentage.
03:08
It's up to you and what your instructor wants.
03:11
Question two has two questions, but they are virtually identical.
03:17
So let's get started.
03:18
A compound contains carbon, hydrogen, and oxygen, and it was shown to have 53 .3 % carbon, 11 .2%.
03:31
Hydrogen, and then the rest is oxygen.
03:34
So all you do for oxygen is you take 100 minus 53 .3 3 minus 11 .2.
03:43
So let's do that.
03:44
We have 100 minus 53 .3 minus 11 .2.
03:50
That will give you the rest remaining oxygen, which is just 35 .5%.
03:55
Now from here, you're going to want to assume you have 100 grams of your compound.
04:02
So these percentages will actually translate immediately to grams.
04:08
Because if you have 100 grams of something and 35 % of that is oxygen, for example, 35 .5 grams will become oxygen.
04:18
Next, you want to convert these to moles by dividing by the molar mass for each one of these compounds.
04:27
I'm going to round oxygen to 16.
04:29
So we have 50.
04:30
33 .3 divided by 12 .01.
04:34
This becomes 4 .43797.
04:40
And then we have 11 .2 divided by 1 .08.
04:45
This becomes 11 .11 .11.
04:47
And then we have 35 .5 divided by 16.
04:52
And this is 2 .219.
04:55
So now we divide everything by the smallest number of moles, because this is moles, after all.
05:10
So oxygen will become one.
05:12
Then we have 11 .11 divided by our previous answer of 2 .219.
05:17
This comes out to be a 5, and then we have 4 .43 divided by 2 .219, and this comes out to be approximately 2 .2.
05:27
These are approximately two and approximately five.
05:30
We say approximately because there is some inherent rounding in the percentages and in the molar masses.
05:36
So your empirical formula is c2h5o.
05:42
So if the molecular mass is 90 .12 grams per mole, what is the molecular formula? well, let's figure out how much this weighs.
05:52
We have 12 .01 times 2, 1 .008.
05:57
Times 5, and then 16 times 1.
06:01
So let's add all of these up, 12 .01 times 2, plus 16 plus 1 .008 times 5.
06:11
And this comes out to be 45 .06.
06:16
Now this is grams per mole.
06:18
This is approximately half of 90 .12.
06:22
So if we take, i'm going to move here in the middle, if we have 90 .12 divided by 45 .06, we get approximately two.
06:36
Therefore, your empirical formula must be multiplied by two to get to the molecular formula.
06:45
So your molecular formula now becomes c4h1002.
06:53
So this is your molecular.
06:55
And this is your empirical over here.
07:02
So now let's take a look at the second part.
07:10
This time we have carbon, we have hydrogen, we have oxygen, and we have nitrogen.
07:17
Carbon was 40 .4%.
07:21
Oxygen was 36 .0 %.
07:27
Nitrogen was 15 .7%.
07:30
So for our hydrogen, that's 100 minus.
07:34
40 .4 minus 36 minus 15 .7.
07:40
That's gonna be 7 .9%...