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, 201015 yr by wanabe

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.

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?

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.

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, 201015 yr by wanabe