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Ch.3 - Molecules and Compounds

Chapter 3, Problem 97a

Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

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Hi everyone today, we have a question telling us based on the elemental mouse percent composition of the following stimulant determine its empirical formula pseudoephedrine. Used as a nasal slash sinus de congestion Has 72.67% carbon 9.17% hydrogen, 8.48% nitrogen and 9.68% oxygen. So our first step is going to get our molds of all of these. And because we have percentages, we're just going to assume we have 100 g so we can use the percentages as grams. So for carbon We have 72 .67 g. And we're going to multiply that by one mole over Its molar mass, which is 12.01 g. And our grams here are going to cancel out leaving us with 6.05 moles of carbon. Next we have hydrogen And we have 9.17 g of it. And we're going to multiply that by one mole Over its molar mass, which is 1.101g. And our grams are going to cancel out Leaving us with 9.08 moles of hydrogen. Then we have nitrogen And we have 8.48 g of it. We're going to multiply that by one mole Over its smaller mass, which is 14. And our g are going to cancel out. And that's going to give us 0. moles of nitrogen. Lastly we have our oxygen We have 9.68 g. We're going to multiply that by one mole over 16 grams. And our g are going to cancel out. And that gives us zero . moles of oxygen. Now we need to divide by our lowest number. So our lowest number here is our 0.605. So we have our carbon Which is 6.05 moles. And we're going to divide that by 0.605. And that gives us 10. And for hydrogen, we have nine point 08 moles. And that gives us 15. And of course our 0.605 moles of nitrogen Is divided by 0.605, is going to give us one. And our 0.605 moles of oxygen, divided by 0.6605, is also going to give us one. So our empirical formula here is C tin H 15 in. Oh, and that is our final answer. Thank you for watching. Bye.