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~ Chemistry Question ~

[From: ] [author: ] [Date: 11-08-15] [Hit: ]
mol C (6.My answer was 3.197x10^21, but the answer key says 1.06x10^23.Im confused because I used the same approach when solving a very similar problem,......
I'm working on a AP chemistry summer-assignment and have a few questions:

1) How many carbon atoms are in a 5.00 gram sample of sucrose, C12H22011?

-This is how I solved it:

5.00g/342.34g (molar mass) = .0146...mol (12molC/33mol H/O) = .005...mol C (6.02E23) [avogadro's #]

My answer was 3.197x10^21, but the answer key says 1.06x10^23.

I'm confused because I used the same approach when solving a very similar problem, and it worked that time. I assume my error came from the stoichiometric conversion (12mol/33mol H/O) but perhaps I'm just doing the entire thing wrong... -_______-

I COULD solve this problem if it asked for the number of atoms in the sample for the entire molecule, but not for a specific element. Please explain how to approach this problem.

on an unrelated note ~ is chemAP hard because I'm sort of nervous :l

-
5.00 g sucrose x 1 mole sucrose/342.34 g sucrose x 6.022 x 10^23 molecules sucrose/ mole sucrose x 12 atoms C / molecule sucrose = 1.06 x 10^23 atoms of C. If you set the problem up correctly, and do the math correctly, you'll always get the right answer.

-
5.00g(1mol/342.34g) = 0.0146 mol C12H22O11
0.0146 mol(12 mol C/1 mol C12H22O11) = 0.175 mol C
0.175 mol C(6.023x10^23 atom/1 mol) = 1.06x10^23

i hope you understand this setup, not just getting the right answer but knowing how it works.. 6.023x10^23atom/mol is avogadros # thats the only way to get the # of atom in a substance.
GOOD LUCK
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