00:01
Okay, so today we have a compound in its molecular mass is 150 .3 amu.
00:13
And we know its percent composition of carbon, hydrogen, and oxygen.
00:17
And we want to know it's empirical and molecular formulas.
00:20
So the way that we saw for a compound's empirical formula, there's a set of steps that we can follow anytime you're given a problem like this.
00:28
So first we're going to write down the percentage of each of our different atoms.
00:33
So carbon is 40 .002.
00:45
Hydrogen is 6 .7135.
00:49
And oxygen is 53 .284.
00:53
And so what you do is you take each of these percentages and just put them in grams.
00:58
So it becomes 40 grams, 6 grams, 53 grams of each of those elements.
01:06
And the reason why we can do that is because they're a percentage.
01:09
And so we're assuming that we have 100 grams of our sample.
01:13
And so that means in this 100 gram sample, 40 grams of it would be carbon, because 40 % of it is carbon.
01:19
Same thing for oxygen and hydrogen.
01:21
So this will make it a little bit easier when we go to find the ratio of each of these elements to each other in order to get the empirical formula.
01:29
So an empirical formula is a ratio of each of the molecules in a particular compound.
01:35
And so what we can do next is get each of these into moles.
01:40
Order to get that ratio.
01:43
So converting carbon into moles, we can just use the molecular weight.
01:49
So 12 .01 grams of carbon is one mole.
01:56
For hydrogen, 1 .01 grams is equal to 1 mole.
02:04
And for oxygen, 16 grams is equal to 1 mole.
02:11
So now if we go ahead and divide all these, we can get everything into moles...