Fundamentals M.19

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Jasmine Wu 1L
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Fundamentals M.19

Postby Jasmine Wu 1L » Wed Oct 04, 2017 6:21 pm

Hello! I'm having some trouble with fundamentals M.19...

The question is:

A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g/mol. When 0.376 g of caffeine was burned, 0.682 g of carbon dioxide, 0.174 g of water, and 0.110 g of nitrogen were formed. Determine the empirical and molecular formulas of caffeine, and write the equation for its combustion.

(I'm not too sure how to approach this problem. I only know that I have to start out by converting everything in grams to moles, but then I get confused about where to go next...)

Tylor McGrew 1J
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Joined: Fri Sep 29, 2017 7:03 am

Re: Fundamentals M.19

Postby Tylor McGrew 1J » Wed Oct 04, 2017 6:32 pm

First, you need to find the amount of moles in each element. Because the question gives you only the grams of molecules you need to break the molecule down. With the case of CO2, you would break it down to find C. (You would break down Oxygen last because oxygen is also found in H2O.) To break CO2 down you would divide its mass (0.682g) by its molar mass (44g) and then multiply it by (1 mol C)/(1 mol CO2) to get 00155 mol C. Then, to find the atomic mass of C, you would multiply your answer (0.0155 mol C) by its molar mass (12.01 g/mol) to get 0.186g C.You would repeat this procedure with each molecule. After done repeating, you would find the mass of Oxygen by subtracting the mass of C, H, and N by 0.376g (given) to get 0.060g ( which is also 0.0038 mol O).
Now that you have all the moles, divide each mole by the lowest mole (Oxygen at 0.0038 mol). Thus, you would get C4H5N2O. This is the empirical formula and has a molar mass of 97.1g. Since the molecular formula's mass (given) is double that, we would double the amounts of moles in the empirical formula to get your answer of C8H10N4O2.

Michelle Steinberg2J
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Joined: Fri Sep 29, 2017 7:04 am

Re: Fundamentals M.19

Postby Michelle Steinberg2J » Wed Oct 04, 2017 11:08 pm


We did this question in office hours today and it took up the whole white board. Hopefully this explanation helps!

Step 1.
Convert grams of CO2, H2O, and N2 to moles. (You do this by dividing the number of grams by the molar mass) You should get .0155 mol CO2, .00965 mol H2O, and .003926 mol N2.
Step 2.
Convert mol of CO2, H2O, and N2 to mol of JUST C, H, and N. To do this you must multiply by the number of INITIAL moles. For example, there is 1 mol of H2O produced, but there is initially 2 mol of H. So, you would multiply .00965 by 2. You should then get .0155 mol C, .01931 mol H, and .007852 mol N.
Step 3.
Turn the mol of C,H, and N to grams. Then, subtract the total of the grams from the total burned - this will give you the grams of oxygen. To get grams, you will multiply the number of moles by its molar mass. You will get .186171 g C, .019466 g H, and .109983 g N. When added, you get .31562 total grams. Then, subtract that from .376. You will get .06038 g O.
Step 4.
Convert the grams of Oxygen to moles. You should divide the number of grams by its molar mass. You will get .003774 mol O.
Step 5.
To find the empirical formula, you will need to divide by the smallest number of MOLES. Don't use grams!! In this case, it is the moles of oxygen (.003774 mol O). When you divide and round, you will get 0=1, C=4, H=5, and N=2. Therefore, the empirical is C4H5N2O.
Step 6.
To find the molecular formula, you will need to find the total number of grams and compare it to the molar mass that they give you. If it is the same, then the empirical formula IS the molecular formula. That is unfortunately not the case here. The molar mass of the empirical, C4H5N2O, is 97.0834 g. Divide the molar mass that they give you by that number and you will get 1.99 AKA 2. Multiply the empirical formula by 2: C8H10N4O2 - that is the molecular formula.

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