Skip to main content
Ch.3 - Mass Relationships in Chemical Reactions

Chapter 3, Problem 93

An unknown liquid is composed of 34.31% C, 5.28% H, and 60.41% I. The molecular weight found by mass spectrometry is 210.06. What is the molecular formula of the compound?

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
583
views
Was this helpful?

Video transcript

Hey everyone. So here I asked the empirical molecular formula for ascorbic acid, Which has a molar mass 176.12 g per mole. That has a mass percent composition of % 40.92 carbon 4.58% hydrogen And 54.51% oxygen. And we know that our empirical formula is this the relative number of the atoms. Then I'm a lack of a formula. Gonna give us our actual number of atoms. So we're gonna send, we have 100 g of the compound. We're gonna convert our percentages into grams. We have 40.92 fans of carbon 4.5, 8 brands of hydrogen and 0.51 fans of oxygen. Now we're going to convert our masses of each into malls using the molar masses in one mall of carbon 12.011g In one mole of hydrogen. You have 1.008g And one mole of oxygen. We have 15 .999 g Bleeding at 3.41 moles of carbon 4.54 moles of hydrogen In 3.41 moles of oxygen. And now we're gonna divide each by the smallest number of moles. To get the smallest whole number ratios for carbon 3.41 by about 3.41, which is one for hydrogen 4.54. About 3.41. It's gonna give us 1.33 and for oxygen 3.4, 1 by about 3.41. I'm Gonna Get one. We're gonna see H. 1.33. Oh. And these sub scripts need to be whole numbers. So we need to multiply each subscript by three to get our smallest one number ratios. So we're gonna get seed three age four 03. So far. Molecular formula We have c. three H. 403. Now we need to find what number. To multiply the sub scripts by to get the whole number ratios for the molecular formula. We're gonna do this by using the formula X. It was a Mueller mass. About about the empirical mass. The empirical mass fc. Three Age 403. It's three. I'm 12.011 last four That was 1.008 plus three. I was 15 .999. It's gonna give us 88.06 grams. They're gonna have X. Because 176.12. What about 88 06? We're gonna get to They're gonna multiply the subscript and the empirical formula by two. We're going to get c. six H. Eight oh six. This is actually the empirical formula. So our molecular formula is C. Six H. Eight oh six. Thanks for watching my video and I hope it was helpful