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Topic: Chemistry Mole Question involving atoms  (Read 3626 times)

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Offline artfuldodger23

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Chemistry Mole Question involving atoms
« on: February 01, 2009, 08:36:05 PM »
How many copper atoms are in 2.50 grams of copper (I) chloride?

If possible set up equation

Offline Astrokel

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Re: Chemistry Mole Question involving atoms
« Reply #1 on: February 02, 2009, 04:21:39 AM »
Where's your equation?
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Hello12

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Re: Chemistry Mole Question involving atoms
« Reply #2 on: February 02, 2009, 04:10:48 PM »
I would like to know how to work this out to? Would you find the moles of the copper chloride then use the avogadro constant. so it would be 0.252*6.023*10(23). Or would you find out the percentage of the Copper in copper chloride then work out that percentage of the avogadro constant

Offline ARGOS++

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Re: Chemistry Mole Question involving atoms
« Reply #3 on: February 02, 2009, 04:59:42 PM »

Dear daniel01;

Your first method is the better/more economic one, because the structure formula tells you that  1.0 mole copper(1)chloride contains 1.0 mole copper.

Good Luck!
                    ARGOS++


Offline Blitzkrieg

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Re: Chemistry Mole Question involving atoms
« Reply #4 on: February 03, 2009, 03:10:28 AM »
First you need the molecular formula for copper (I) chloride. Since it's copper (I), the charge on copper is +1. Since Cl is in Group 7A (or 17, depending on your periodic table), its charge is -1. The empirical formula is CuCl, because since they both have a charge of +/- 1, only one of each is needed to cancel out. After a quick google search, the molecular formula is also CuCl.

The molecular mass of CuCl is 99 g/mole (63.55 + 35.45).

Like, ARGOS++ said, the ratio of Cu to CuCl is 1:1.

Now you just need to find the moles of CuCl (grams of CuCl divided by the molar mass), and then multiply that by Avagadro's number.


Offline Hello12

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Re: Chemistry Mole Question involving atoms
« Reply #5 on: February 03, 2009, 04:36:34 AM »
Ahh thanks  :)

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