M19

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MCracchiolo 1C
Posts: 55
Joined: Sat Jul 22, 2017 3:00 am

M19

Postby MCracchiolo 1C » Thu Oct 05, 2017 11:23 pm

Looking for an explanation for the steps in problem M19 as you go from the grams of product to the formula of the reactant. Thank you!

soniatripathy
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: M19

Postby soniatripathy » Fri Oct 06, 2017 2:21 am

When caffeine is burned, the carbon in CO2, the hydrogen in water, and the nitrogen in nitrogen gas come directly from caffeine. This means that the first step is to calculate the moles of C, H, and N from the given masses. You would then need to convert the moles of each element to grams. The sum of the grams of C, H, and N would then be subtracted from the original sample to figure out how many grams of oxygen are in the sample. Once the grams of O are calculated, you can calculate the moles of O. You now have the moles of each element in caffeine. From here you can calculate the empirical formula by dividing by the smallest mole value and getting the ratio.

AdrienneKashay1J
Posts: 2
Joined: Fri Sep 29, 2017 7:07 am

Re: M19

Postby AdrienneKashay1J » Fri Oct 06, 2017 7:06 am

Also remember to convert the empirical formula into the molecular formula! Divide the molar mass by the empirical mass to determine the conversion multiplicity. In this case it is 2.

Sohini Halder 1G
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
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Re: M19

Postby Sohini Halder 1G » Fri Oct 06, 2017 9:44 am

Can someone explain why all of the oxygen in the products seems to come from the original compound and not the surrounding excess oxygen?


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