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Chemistry 17 Online
OpenStudy (anonymous):

Determine the empirical formula of a compound with the following composition by mass... 62.1% Carbon 5.21% Hydrogen 12.1% Nitrogen 20.59% Oxygen

OpenStudy (anonymous):

please! im not just copying and pasting my question, I actually really need help and an explanation

OpenStudy (anonymous):

If you would have 100 grams of this compound how many grams would you have of each element?

OpenStudy (anonymous):

the percentages, but just as numbers @LeibyStrauss

OpenStudy (anonymous):

Correct. So can you convert the grams to moles?

OpenStudy (anonymous):

just multiply by 6.02 x 10^21?

OpenStudy (anonymous):

6.02 x 10^23 is Avogadro's number, it will give you the number of atoms in a mole. We need to do something else, how many moles are in Carbon, Hydrogen etc. Do you know how many moles are in 63.1 g of Carbon?

OpenStudy (anonymous):

there are 12 grams per mole of Carbon, so do I take 63.1 times 12?

OpenStudy (anonymous):

*in carbon

OpenStudy (anonymous):

How many grams would be in 5 moles of carbon?

OpenStudy (anonymous):

60

OpenStudy (anonymous):

right?

OpenStudy (anonymous):

You have 62.1 g of carbon, (i made a typo earlier it's not 63.1) which will be ~ 5 moles. If you multiply it by 12 you'll get a large number, not 5. Do you know how to perform dimensional analysis?

OpenStudy (anonymous):

i don't think so:/ it's 745.2

OpenStudy (anonymous):

Were you taught how to convert from grams to moles?

OpenStudy (anonymous):

okay I found the moles for all of them and it's C: 745.2 H: 5.21 N: 169.4 O: 392.44

OpenStudy (anonymous):

yes, i'm just not very good at it

OpenStudy (anonymous):

sorry, I meant grams for all of them

OpenStudy (anonymous):

You can't have 745.2 mole of C. 1 mole is 12.01 g. 5 mole is 60.05 g. You have 62.1 g so you have around 5 mole. You multiplied and got a high number.

OpenStudy (anonymous):

i don't understand. why is that significant?

OpenStudy (anonymous):

To find the empirical formula you need to know the number of moles of each element. The number needs to be correct. If you start with 62.1 g of carbon you need the number of moles of carbon to be around 5 (based on the reasoning of the previous post). If you have the wrong number you can't solve the problem.

OpenStudy (anonymous):

okay so, it'd be 5.175

OpenStudy (anonymous):

OHHHHH I THINK I GET IT

OpenStudy (anonymous):

Good. So now calculate the moles of the rest of the elements.

OpenStudy (anonymous):

C: 5.2 H.5.2 N: .86 o: 1.3

OpenStudy (anonymous):

okay now i don't know what to do

OpenStudy (anonymous):

According to your calculations you have \[C _{5.2 mol}H _{5.2mol}N _{0.86 mol}O _{1.3mol}\] Divide each one by the smallest mole

OpenStudy (anonymous):

okay so

OpenStudy (anonymous):

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