2013 Midterm Q2A

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Armo_Derbarsegian_3K
Posts: 35
Joined: Sat Jul 09, 2016 3:00 am

2013 Midterm Q2A

Postby Armo_Derbarsegian_3K » Tue Nov 01, 2016 5:19 pm

For this problem:

I solved for moles of each element by taking the given values and dividing by their atomic masses.

i.e.: Carbon 3.27g x 1molC/12.011gC = 0.272molesC

then I divided by the smallest value, got the ratio, multiplied by a number to get rid of fractions to get the answer, found grams of empirical, divided by molar mass to get the molecular formula.

Notice, I didn't do the 100g method and therefore got different mol of X atoms. Is my method correct and will it work every time?

Andrew Nguyen 3G
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

Re: 2013 Midterm Q2A

Postby Andrew Nguyen 3G » Tue Nov 01, 2016 5:32 pm

The method is correct and it will work every time. The "100g" method doesn't REALLY tell you the weight distribution in your sample (how can 8.00 grams of a sample suddenly transform to 100g?), but rather shows you in an easy to comprehend way the distribution of your grams IF you had 100 grams of it. Thinking of your sample as 100g might be easier to keep track mentally. Although your method and the 100g both use different intermediate processes, they both achieve the same answer. Personally, I've been doing it your way because it is faster.


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