# Chemistry A, real simple for most people. Mol problems.

1. Nov 19, 2013

### Fishingaxe

Chemistry A, real simple for most people. "Mol" problems.

1. The problem statement, all variables and given/known data

Hello, first I'd like to say thank you to this community as I have gotten a lot of help in the physics department before and is basically a big reason of me passing the class.

Now chemistry is here and I am having a little bit of trouble so I thought I'd put up a few problems here and you can look over my attempt of solving it and maybe help me along the way.

Problem A) A copper test shows to contain 0.126 mol of copper, the whole test weighs 8.562 g. Does the test contain of 100% copper?

Attempt to solve: n=m/M

n= 8.562g / 0.126*6.02*10^23
n = 8.562 / 7.585*10^22
n = 4.09*10^25.

^--- this is how far I've managed to come on my own. I know the solution is more or less done by now but I am having trouble with what to write, in terms of grams etc to finish off the question. I know that if the answer is not 8.562g the test doesn't contain of 100% copper. As said, don't know how to take it from here.

Problem B) A sugarcube has the mass 2.1g. The sugar is made of sackaros with the formula C11H22O11

a) calculate "n" sackaros in the sugarcube.
b) calculate "n" carbonatoms in the sugarcube.

Attempt to solve: <---- This is where I am completely lost. I know that I have to use the formula n=m/M and I understand what the formula means. What I don't understand is how to get those numbers from "C11H22O11". I am a complete novice when it comes to this. Any help is very much appreciated.

PS: Posted this in another chemistry section in this forum but I've realised that this is the homework section so I apologize.

2. Nov 20, 2013

### Saitama

To calculate the moles of sackaros, you need its molar mass. Can you calculate that?

3. Nov 20, 2013

### Fishingaxe

Thank you for helping, I had another thread up here that contained the exact same problems and already solved them with the help from borek and one other person as well.

I am much more familiar with these types of things now :)

4. Nov 20, 2013