Percentages When Solving for Molecular Formula

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Perla Cervantes_1G
Posts: 22
Joined: Fri Apr 06, 2018 11:02 am

Percentages When Solving for Molecular Formula

Postby Perla Cervantes_1G » Sun May 06, 2018 5:10 pm

The following question is #22 from module 1:

Xylitol, a sugar substitute, has a mass composition of C 39.43% O 52.58% H 7.88% and a molar mass of 152.15 g.mol-1 What is its molecular formula?

My question is, how do we know when we should divide the percentages by 100? From what I've read online we can only assume that there are 100g of the sample when we are not given a specific molar mass and thus, use the numbers 39.43g C, 52.58g O, and 7.88g H. In this example however, we are still required to use 39.43g C, 52.58g O, and 7.88g H instead of the numbers we would get if we divide by 100 (.3943g C, .5258gO, & .0788gH) even though we are given the molar mass. Why are we assuming that the sample is out of 100g when we are explicitly told otherwise?????

Someone plz help sos

MeghetyManoyan1A
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am

Re: Percentages When Solving for Molecular Formula

Postby MeghetyManoyan1A » Sun May 06, 2018 6:34 pm

Because you are given the percentages, and they add up to a 100%, you can just use the numbers as they are. However, if you had the numbers expressed in grams, then you would have to find the percent mass composition of each element, and then, assume there is 100g of the sample. This just means that the percentage you obtained is what you use as the number of grams of the element. Then, once you convert from grams to moles and get the empirical formula, you can use the molar mass to find the molecular formula.

Lenaschelzig1C
Posts: 21
Joined: Fri Apr 06, 2018 11:05 am

Re: Percentages When Solving for Molecular Formula

Postby Lenaschelzig1C » Sun May 06, 2018 10:27 pm

Also, assuming that the percentages equal 100 grams will help get the empirical formula!


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