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Problem 81 Hard Difficulty

A compound having an approximate molar mass of $165-170$ g has the following percentage composition by mass: carbon, 42.87$\%$ ; hydrogen, $3.598 \% ;$ oxygen, 28.55$\%$ ; nitrogen, 25.00$\% .$ Determine the empirical and molecular formulas of the compound.

Answer

$\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{ON}\right)_{3}$ or $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{3} \mathrm{N}_{3}$

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Top Chemistry 101 Educators
Stephanie C.

University of Central Florida

Nadia L.

Rice University

Allea C.

University of Maryland - University College

Morgan S.

University of Kentucky

Video Transcript

Okay, so now we have four elements carbon, hydrogen, oxygen and nitrogen. And we have been given mass per cents for each of these moments and beer to find it's empirical formula and molecular formula. So but to do so, first, let me explain how what a mass percent is. So what it is basically is that these are the mass percent, by the way that I'm writing here is that in one mole of this compound, the weight percent of the cart of carbon in that compound and one more than compound it's 42 weight percent of hydrogen is three. So So yes. So, basically, if we were to have 100 grams, if the with one mole of his compound where we under grams 42 grams would be carbon three games would be hydrogen in still wanted so forth. So we can use mass percents as mass on its own because it represents, like the ratio of each element is to two awards like the other compares into the total. Yes, so this is the mass percent. But also the Mets, because of what it represents, remember, like the mass of this is basically 42 grams in the total 100 grams of this element is parting. Does that make sense? So Yeah. So because this is mass and need to find the molar ratios of each element means to convert this mass into moles. And the way we do that is by taking the mass of each own and over the molar mass. So all right, the most along this cold. So we already have mass here, and we can find Mullah mass from our periodic table. So if we do that quick calculation, these air, the moles that we get just Seagram's here. So let me know the units for everything in this column with grams, I mean everything in this column. Ismael's Yeah, 69 So and as we compare the malls of each element, we need to look for, um, patterns because these mole, least like moles over here will give us the ratio. Are giving us the ratio of from, you know, of elements and Makhan from so, like the number of elements in each In this compound number of them get the number of bones in this compound is a ratio of the moles of the amount of most in the compound. So that this is basically saying is that for every mole of carbon and I, like the moles of carbon and hydrogen are equivalent in the mold of author. Gin and hydrogen are equivalent. So if we have three moles of nitrogen, we will have three moles of carbon. And if we divide 3.469 by 1.78 we conceive Adam, it's double. So I'm just going to write to the ratio again in this column, but in a more you know, easier to understand or more in a simplified Inter jer Yes, innocents interject. So if this makes sense so these two are molar equivalents oxygen, nitrogen and these two molar equivalents. But this thes Mueller equivalents are double of this because of, you know, if you do that simple division So here we have found the empirical formula to be C two h two. Oh, and so now the molecular formula would be the exact scene. But I'm going out of X toe. All of the's sub scripts. Let's just see ECs h. So the molecular formula is basically C two x h two eps Oh x in an ETS So now, um, to figure out what X is If you see my pervious previous videos, you know that you just need toe multiply the molecular weight off the most of this compound divided by the molecular weight of this compound to the motel rate of the molecular formula over the molecular rate of the empirical formula. So they told us that the Markkula rate of this compound is in between 1 65 and 1 70 So this is a very, you know, broad range, I guess I was looking to find the ratio in an interviewer form, and we will under Browning, so I'll just stick with 1 70 grams from all and to find the grams, the molar mass of this impure coal formula shares a simple addition. So we know the molecular rates of each individual element. We just add it up, making sure we, you know, take into effect the twos here. So you that calculation we get 56 basically 56 cramps pool. So now we're just looking for an interview We're going to round anyways in this number is basically street. So if it's if we were to use, like 1 65 we would probably get like 2.98 We'll just till around into three. And this year was like 3.0, something which is still going toe. It's wasabi because these must be integers we can t know work with anything else. So basically, now we need to multiply three, you know, into each of these elements. So three times to 60 times 263 times one is just three degree to give us the molecular formula C six h six 03 and three. So this is the molecular formula. This is the empirical formula.

McMaster University
Top Chemistry 101 Educators
Stephanie C.

University of Central Florida

Nadia L.

Rice University

Allea C.

University of Maryland - University College

Morgan S.

University of Kentucky