Avogradro's Constant

[Mr. Study]

Hey guys!

Just an example question: Copper Sulfate solution is electrolysed, 0.175g of copper  is deposited by a current of 0.863A, in 10 minutes. If the charge on one electron is 1.602 x 10^-19 and the charge on a copper ion is 2+, Calculate a value for Avogradro's constant

What I did was:






Avogradro's constant=

The answer is that but the book (Checkpoints) does it differently.

I've done it this way for a number of other questions and I always arrive at the correct answer.

:S, would anyone be able to explain to me, conceptually, what it is that I am actually doing (In regards to my working out)?

(Tell me if my working out is clear! D: )

[Lasercookie]

You're finding the number of electrons via mole ratios.

Then you're finding the value of Faraday's constant from your data.

Faraday's constant is the electric charge on one mole of electrons.



So we know that the charge on one electron is

So if we have the charge for one mole, and the charge for one electron, then we can figure out how many electrons in a mole. Which in other words will be Avogadro's constant.



which gets a value that is reasonably close to the actual value.

As an aside, I think it's interesting that:


which is reasonably close seeing that we're using rounded values.

[FlorianK]

Why does everybody calculate Faradays constant first?






Avo's constant=Q/n(e)/Charge of an electron
=517.8/0.0055/1.602*10^19=5.88*10^23

[Lasercookie]

Why does everybody calculate Faradays constant first?

Avo's constant=Q/n(e)/Charge of an electron

Isn't that calculating Faraday's constant right there?

[FlorianK]

Why does everybody calculate Faradays constant first?

Avo's constant=Q/n(e)/Charge of an electron

Isn't that calculating Faraday's constant right there?

Yes, but in my formula you do it in one step instead of Mr.Study's 2 steps and therefore you don't have any rounding errors.

[Lasercookie]

Yes, but in my formula you do it in one step instead of Mr.Study's 2 steps and therefore you don't have any rounding errors.

What do you mean?

For any calculation in chemistry you'd be carrying the values over in your calculator regardless, so what you've said about rounding errors is irrelevant.
 
The level of error in the final answer would come from the actual data you're given at the start (sigfigs).