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
please! im not just copying and pasting my question, I actually really need help and an explanation
If you would have 100 grams of this compound how many grams would you have of each element?
the percentages, but just as numbers @LeibyStrauss
Correct. So can you convert the grams to moles?
just multiply by 6.02 x 10^21?
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?
there are 12 grams per mole of Carbon, so do I take 63.1 times 12?
*in carbon
How many grams would be in 5 moles of carbon?
60
right?
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?
i don't think so:/ it's 745.2
Were you taught how to convert from grams to moles?
okay I found the moles for all of them and it's C: 745.2 H: 5.21 N: 169.4 O: 392.44
yes, i'm just not very good at it
sorry, I meant grams for all of them
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.
i don't understand. why is that significant?
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.
okay so, it'd be 5.175
OHHHHH I THINK I GET IT
Good. So now calculate the moles of the rest of the elements.
C: 5.2 H.5.2 N: .86 o: 1.3
okay now i don't know what to do
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
okay so
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