## If we know the composition of a compound in terms of the masses (or mass percentages) of theelements present, we can calculate the empirical formula but not the molecular formula. Toobtain the molecular formula we must know the molar mass.

Composition

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##### Allea C.

University of Maryland - University College

##### Jake R.

University of Toronto

### Video Transcript

Okay, So we have been tasked to find the molecular formula and to the empirical formula for a compound containing boron and hydrogen. Just these two elements. So what we have been given is, um, a mass percent. So the mass percent of on in this compound is 78 0.14 and the mass percent of hydrogen is 21 point tooks. So what that basically means is that in one mole of this compound, whatever, Xing, why, maybe one mole of it, we would have 78% of that compound of the weights of a compound belonging to the boron element and 21% of the weight of the compound wall entitlement. All right, so since this is a percent and out of 100 we can assume that we know that our 100 grams of this compound would have 78% 780.1 grand sport in 21 grams hydrogen. Then we can use these mass persons as our mass. So to determine the amount of elements or the man of Adams going to bone and 100. We can't use the masses, but we need to convert them into malls. So we know that mole is equal to mass over molar mass. So since we already have the mass from the mass percent and you just have to look at her pure and table toe, figure out the the molar mass of each element. Just do this simple division, but no so shopping for most. So just take this manse by by the molar mass of boron from our, um oppose table just 10.81 And this equals we do the division hiking. Either 70 are 7.22 moles. If we do that, the exact same thing with the hydrogen divide this by the molar mass of hydrogen. We will get that. It's, um, most equals 21 0.68 Okay, So, technically, we already have the molar ratio of each element in this compound, and this could be the X. And this could be the y. However, because, um, X and wife must be introduced in whole values. We're going to have the simplify this further to find three empirical formula. So all we have to do is then divide the ratios by usually police and take a larger mall. I remember this one is belong to the hydrogen and we divided by the smaller one to find the ratio of each element molar issue of each other in the compound. Well, just by even looking at it, we can tell that the molar issue was basically 27 7 is approximately three. So for every it's got the over one. So for every home, like more off bar on, we have three moles of hydrogen. You have three moles of hydrogen. Therefore, the empirical formula is be H three. This being the X, this being a why from via boat. So we have the empirical formula on. The only way we can get the molecular formula is why using more maps. So they told us that the molar mass was in between 27 28 grams per mole. Remember the imperial formulas the simplest formula of, like the mole ratios of each element the compound, But the local form letters, the actual you know, I'm out and they're often they can or may not be like the same thing. So here we have it. So basically have to do is find the one thing formula of this compound and compare it to this like you were from here. So if we just quickly do this summation, we get that this compound, Beach three, has a molecular formula of 13.83 All we need to do now is find the ratio between the smaller ass and this small amounts. If they were to have been the same numbers and we know the molecular formula equals perform, but because just by looking at it, it appears to be double we can tell that the motile formula is twice as big as this, um, empirical formula giving us so we would basically have to do is then multiplied the X and Y's both by two to give us for every two molecules off boron. We have six moles of hydrogen, Mr Kuba. So therefore the molecular formula is be to each six.

McMaster University

#### Topics

Composition

##### Allea C.

University of Maryland - University College

##### Jake R.

University of Toronto