Empirical formula from "experiment" data

[wanabe]

 

So I'm stuck. I don't know what to do if I don't have the formula for the copper chloride... The table of information just makes it more confusing, because I don't know if i should average the values out or what...it all seems so ambiguous to me. Help please.

Edited September 23, 2010 by wanabe

[Mr Skeptic]

What they give you is the concentration of the copper halide, and the weight of the copper in a given volume. So you can find the mass of your halide, which would give you a pretty good idea of the formula. You also should know the charges of the copper ion and halide ions.

[wanabe]

Thats all the information I'm given(I can't assume the charge?)...In any case i came up with: (46.62gCuCl/1L)(1L/1mol)(1000mL/1L)(0.786gCu/49.6mL)= a totally bogus 738g Cu??? the 1mol was never used...

 

Then I tried 0.786gCu(1L/42.62)(1000mL/1L)(?/49.6mL)=? I can't just assume some kind of Cu_xCl_y combo can I; and use that atomic mass?

[Mr Skeptic]

Well for your first formula you aren't conserving units.

 

What you want is to use the mass of copper precipitate to find the concentration of copper (grams Cu/L). Then compare that to the concentration of your copper halide.

[wanabe]

Here is what I have.

 

0.786gCu(1L/42.62)(1000mL/1L)=18.442mL Cu/42.62*100=43.2707% Cu-100=56.729%Cl

 

0.432707 Cu / 63.546amu Cu*42.62=0.29021=1 mol Cu

0.56729 Cl / 35.453 amu Cl *42.62=0.68197 =2 mol Cl

 

 

(i chose to multiply by 42.62 instead of 100 because thats the grams of the sample not 100g)

CuCl_2, but thats not right!?

 

Well i got desperate (time constraint)and put CuCl_3 as a guess, and got it right, but i still want to know whats up. Thanks.

Edited September 23, 2010 by wanabe