Assessment Question #8

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Joslyn_Santana_2B
Posts: 51
Joined: Wed Sep 21, 2016 2:58 pm

Assessment Question #8

Postby Joslyn_Santana_2B » Sat Sep 24, 2016 6:59 pm

In the Pre-Assessment the question is #8, and states:

"Xylitol, a sugar subsitute, 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? "

In the answers all the Hydrogens were not multiples of 5, which was the number I got to be the multiple of the empirical formula. However, Carbon and Oxygen were multiplied by 5. I am confused on how the values for Hydrogen in the molecular formula were acquired.
Attachments
8 eform work2.jpg

Mirian_Garcia_2G
Posts: 51
Joined: Wed Sep 21, 2016 3:00 pm

Re: Assessment Question #8

Postby Mirian_Garcia_2G » Sun Sep 25, 2016 10:39 am

Dear Joslyn,

From the picture you uploaded, I believe the mistake was made due to the fact that you mistaken the percentages given for the grams of each element. For example, Xylitol is 39.43% Carbon. Since the total molar mass of Xylitol is 152.15 g/mol, then you would multiply the given percentage by the total molar mass to find the molar mass of Carbon:

(152.15 g/mol)(0.3943)= 59.99 g/mol

Therefore, the molar mass of Carbon would be 59.99g, not 39.43 grams. This would be repeated for Oxygen and Hydrogen as the first step in order to acquire the molecular formula of Xylitol. Hope wasn't too confusing and helped.

Sangita_Sub_3H
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: Assessment Question #8

Postby Sangita_Sub_3H » Sun Sep 25, 2016 12:35 pm

Dear Joslyn,

Another issue that I found with your calculation was with the hydrogen. After you divided the 7.88g of hydrogen given by 1.008 (the molar mass of the hydrogen element), you got the number 7.30 mol, whereas I re-did the calculation and got 7.82 mol. Then, after dividing by the lowest moles of the three elements, which is carbon with 3.28 mol, I got the C:H:O ratio of 1 to 1 to 2.38. If you multiply 2.38 by five, you get 11.9, which is close enough to 12 (you need a whole number). So, you then multiply carbon and oxygen by five respectively and end up with the empirical formula of C5H12O5. Once you find the molar mass of C5H12O5, you find that it is the same value of the molar mass given, meaning that the empirical formula and molecular formula are the same.

Hope that helps,
Sangita


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