Chemistry Help

[The Cheat]

I know this is a weird topic but i am in desperate need of help. Im sure there is some chem genious out there.

 

I have to do some calculations for this chemistry experiment.

 

I have to find the actual and theoretical yield of copper in this chemical reaction.

 

I mixed cupric sulfate crystals (CuSO4) and Iron Fillings (Fe) in deionized water to get solid copper (Cu)

 

2.2 g of Iron fillings was the limiting reactant.

 

i am in desperate need of help to find the actual and theoretical yield

 

oh and the amount of solid copper at the end was 2.85 grams, im thinking this is the actual yield but im not sure

[ET Warrior]

I got a B in my college chemistry class last semester...but...yeah...I really didn't understand what I was doing the whole semester...chemistry sucks.

[matt--]

Originally posted by The Cheat

I know this is a weird topic but i am in desperate need of help. Im sure there is some chem genious out there.

 

I have to do some calculations for this chemistry experiment.

 

I have to find the actual and theoretical yield of copper in this chemical reaction.

 

I mixed cupric sulfate crystals (CuSO4) and Iron Fillings (Fe) in deionized water to get solid copper (Cu)

 

2.2 g of Iron fillings was the limiting reactant.

 

i am in desperate need of help to find the actual and theoretical yield

 

Actual yield is what would be produced in a reaction you did.

 

Theoretical yield is calculated.

 

I'm assuming the Fe is Iron (II) and not Iron (III).

the reaction equation: CuSO4 + Fe = FeSO4 + Cu

 

Fe has a molar mass of 55.845 g/mol, and you have 2.2 g of it as a limit. 2.2 g / 55.845 g/mol = 0.039395 mol Fe

 

This will produce an equal number of moles of Cu.

.039395 mol Cu * 63.546 g/mol = 2.5034 g Cu

 

Your theoretical yield is 2.5 g Cu.

[MuRaSaMuNe]

I did the EXACT same experiment two years ago my memory may be a little rusty so bear with me if I have a few things wrong .

 

The Actual Yield is also called Experimental Yield and it is what you find in the experiment. So the amount of Cu that was produced should be that. Keep in mind that if you measured the weight of the Cu that was produced while it was on a paper tray you gotta subtract the weight of the tray first.

 

The Theoretical Yield is what you should have had idealy and you find this with calculations. Now this can be found numerous ways based on the information you have. The thing I usually do first is make an overall equation. Once you have that, you can use mole ratios (assuming you know how to do this) in order to figure out the moles of Cu. So... you find the molar mass of CuSO4 by adding up each elements molar mass and use your molar mass/mass/mole triangle. If you forget it's;

 

__m_

mm| n

 

 

So nCuSO4 = mCuSO4/mmCuSO4

 

This is assuming you have the mass of CuSO4!!

 

Remember I told you to derive the general reaction equation? Now you gotta use this. Use the coefficients in the equation as the moles of the substance so.... for example if you have

 

4CuSO4

 

You'd go

 

nCuSO4 = 4

 

And if you had

 

4Cu

 

Then you'd have

 

nCu = 4

 

Now you can set up your mole ratio.

 

nCuSO4/nCuSO4 = nCu/nCu

 

So using the fake coefficients I came up with and the mole of CuSO4 we calculated earlier;

 

nCuSO4/nCUSO4(calculated CuSO4 from above) = Xmol Cu/nCu

 

4/(calculated CuSO4 from above) = XmolCu/4

 

 

Then you just solve for XmolCu

 

XmolCu = 4 * 4/(Calculated SuSO4 from above)

 

***Keep in mind that the value of four I am using is made up. Use the co efficient for that number from the equation.***

 

Now you have the moles of Cu.

 

Now, to get the mass you go;

 

mCu = mmCu * nCu

 

And mCu is your theoretical yield.

 

The procedure I explained is called mass to mass I don't know if your teacher calls it something else or if you even did it but I remember doing it this way. Good luck

[MuRaSaMuNe]

Or that... lol

[The Cheat]

i was getting 2.5 g for the theoretical yield but with the actual yield was 2.8 g then that would make the percent yield over 100% which lost me because it should be below 100%

[matt--]

Originally posted by The Cheat

i was getting 2.5 g for the theoretical yield but with the actual yield was 2.8 g then that would make the percent yield over 100% which lost me because it should be below 100%

 

It's possible to get over 100% yield, but percent error is calculated as follows.

 

| theoretical yield - actual yield |

theoretical yield

 

So, although your percent yield is 112%, your percent error is only 12%.