M19

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AronCainBayot2K
Posts: 101
Joined: Fri Aug 30, 2019 12:17 am

M19

Postby AronCainBayot2K » Thu Oct 03, 2019 6:40 pm

M19) A stimulant in coffee and tea is caffeine, a substance of molar mass 194 g*mol^-1. 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.

For this problem, I'm having difficulty writing the equation for its combustion. Would I just need to add CO2 and H20 to the product side?

Tracy Tolentino_2E
Posts: 140
Joined: Sat Sep 07, 2019 12:17 am

Re: M19

Postby Tracy Tolentino_2E » Thu Oct 03, 2019 6:48 pm

You would put CO2, H2O, and N2 to the product side because it stated all three were formed. The balanced equation is
2C8H10N4O2(s) + 19O2(g) --> 16CO2(g) + 10 H2O(l) + 4N2(g)

MMckinney_4H
Posts: 61
Joined: Sat Sep 28, 2019 12:16 am

Re: M19

Postby MMckinney_4H » Thu Oct 03, 2019 7:20 pm

I did it by recognizing first that Nitrogen, Carbon, and Hydrogen could have only come from the formula for caffeine because for combustion, all you need aside from the thing that is being burned is oxygen. With that I set off to find out how much of each. Nitrogen was the easiest as the problem said that nitrogen, the element, is produced meaning that all of the .11 grams were straight from the Caffeine. Next, I figured out how many grams of the Carbon Dioxide was from the Carbon by using the molar mass of Carbon and the molar mass of Carbon Dioxide to set up a ratio then set this equal to a ratio of grams of Carbon over the grams of Carbon Dioxide. That told me that .186 grams of Carbon were on the product side that could only have come from the Caffeine. I did the same for Water to figure out how much Hydrogen was in the equation, all of which had to have come from the caffeine. I added all of the grams from these individual amounts which came out to approx. 0.3153. The rest must be Oxygen (.0607 grams of oxygen in caffeine). I converted all of the amounts of each element in caffeine into mols so that from here they would be on equal playing fields (all in # of atoms as is the empirical formula). I used these amounts to figure out the ratio of each to figure out the empirical formula by dividing by the lowest amount, the oxygen. I got a ratio of 4 C: 2 N: 5 H: 1 O. If you put this in the standard order of Carbon first, Hydrogen second, then in alphabetical order, you get that the empirical formula for caffeine is C4H5N2O.

Jordan Ziegler 2J
Posts: 59
Joined: Sat Aug 17, 2019 12:15 am

Re: M19

Postby Jordan Ziegler 2J » Sun Oct 06, 2019 10:38 pm

I broke down this kind of problem into three major steps:

1. Using the given masses, determine the moles of *ATOMS* of each element. Note that in this problem, oxygen is also on the reactant side.
2. From here, divide the number of moles of each element by the smallest value. If you end up with fractions, multiply by a value that will make each number whole, as well as mutually prime. This is the process for percent composition problems.
3. After you find the empirical formula, check the given molar mass, then multiply as needed to get your molecular formula!


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