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I dont want the answer but rather how to find out the answer so i can do it on my own.....

2007-01-15 13:29:39 · 9 answers · asked by just_curious 1 in Science & Mathematics Chemistry

9 answers

There are 8 carbon atoms in each molecule of C8H10N4O2. If you have 3 moles of this molecule you will have 3 x 8 = 24 moles of carbon atoms.

2007-01-15 13:34:53 · answer #1 · answered by physandchemteach 7 · 0 0

24.

A mole is a quantity of molecules. So you have 3 moles of C8H10N4O2. Each molecule with 8 Carbons. Multiply.

2007-01-15 21:32:04 · answer #2 · answered by DT 4 · 0 0

8 Carbons in C8H10N4O2, so if there are 3 moles of C8H10N4O2 you of course multiply them together to get 24.

2007-01-15 21:53:29 · answer #3 · answered by Anonymous · 0 0

I believe from my days of knowing organic chemistry that there would be 8
drop your #'s below the letter so...C=carbon 8 is the amount of atoms. Then H=hydrogen, 10 is the amount of atoms, then N=nitrogen,that would be 4 atoms , and the last is oxygen , which has 2 atoms. I cant remember how many molecules are in a mole?

2007-01-15 21:53:09 · answer #4 · answered by KIM O 1 · 0 0

8 carbons per molecule, 3 moles of molecules...

8 * 3 = 24

2007-01-15 21:33:28 · answer #5 · answered by Richard B 4 · 0 0

Okay, everyone here has it wrong so far. THE ANSWER IS NOT 24!!!!

Find the formula weight of your molecule. Then set up a proportion of 3/(f.w. of molecule)=x(your variable)/12(f.w. of carbon)
Then cross multiply and solve for x.

2007-01-15 21:46:44 · answer #6 · answered by Lizrd 3 · 0 0

There are 24

8x3=24

2007-01-15 21:33:24 · answer #7 · answered by -Eugenious- 3 · 0 0

ok its like you have to add up all the molecular weights of all the atoms

then divide

then figure the percentage of carbon

2007-01-15 21:33:54 · answer #8 · answered by kurticus1024 7 · 0 1

http://webnt.calhoun.cc.al.us/distance/internet/Natural/Chm111/CHM%20111_UNIT02.pdf

try this PDF document it should help you figure it out. it explains alot on page 6

2007-01-15 21:41:37 · answer #9 · answered by Anonymous · 0 0

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