Module Empirical Question

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Princess Jereza 3C
Posts: 46
Joined: Sat Jul 20, 2019 12:16 am

Module Empirical Question

Postby Princess Jereza 3C » Mon Sep 30, 2019 12:34 am

I apologize that this may be an inappropriate time to post a question from the post-assessment module, but I am having trouble with this one question. So say Xylitol, which has 39.43% Carbon, 52.58% Oxygen, and 7.88% Hydrogen, and the molar mass is 152.15 g/mol, and I have to find the molecular formula. I followed the steps from Professor Lavelle's lecture which is to convert the mass % composition to mass, and then find the moles of each element, divide each amount by the smallest number and then we would have our ratios for the empirical formula. The empirical formula that I got was C3H3O6, and when I divided Xylitol's molar mass by my empirical formula's (in order to get the number to multiply the empirical formula to get the molecular formula), I got 1, which does not make any sense. None of the answers match with my formula. I know this is a lot, but I am stuck with this problem, and I am not sure how to solve this problem. Thanks.

Mashkinadze_1D
Posts: 87
Joined: Sat Aug 24, 2019 12:15 am

Re: Module Empirical Question

Postby Mashkinadze_1D » Mon Sep 30, 2019 1:39 am

To solve this you would want to proceed as follows:
So assume that we have 100g of Xylitol to make this easier and the continue to

39.43 g Carbon x 1mol C/12.10 g C = 3.259 mol C

52.58 g Oxygen x 1mol O /16.00 g O = 3.286 mol O

7.88 g Hydrogen x 1mol H/1.008 g H = 7.82 mole H

After this continue to find the ratios

C:O = 3.259/3.286 = .992 or for our purposes we can round this to 1. So for every one carbon atom there is one oxygen atom

H:C = 7.82 / 3.259 = 2.40

So we would have C1 O1 H2.4 to get this to the nearest whole number in 2.4 we multiply by a factor of 5 to get 12. Therefore the formula now consits of C5 O5 H12.

After this we continue to add all these values up by doing 5(C) + 5(O) + 12(H) = 5(12.01) + 5(16.00) + 12(1.008) = 152.146g

If we divide the molecular mass by this value, we would get a number extremely close to 1 showing that the empirical formula (with whole digits) is the same as the molecular formula. Hope this helps!

005388369
Posts: 73
Joined: Sat Sep 28, 2019 12:16 am

Re: Module Empirical Question

Postby 005388369 » Tue Oct 01, 2019 3:54 pm

The empirical formula and the molecular formula are the same sometimes which is why you got 1.

Tiffany Chao 2H
Posts: 117
Joined: Fri Aug 09, 2019 12:17 am

Re: Module Empirical Question

Postby Tiffany Chao 2H » Tue Oct 01, 2019 4:16 pm

There's always a possbility that the empircial formula IS the molecular formula. You can tell when you find the molar mass of the empircal formula. Then, divide the mass the problem gives you by the empirical mass. If the factor is 1, then molecular and empirical formula is the same. Say if it's 2 or something for example, you would multiply the empirical formula by 2 and etc.


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