# Theoretical yield of a precipitate called lead(II) iodide

Submitted by Gumber on Fri, 04/04/2008 - 11:40

Hello again, I need some help ONCE AGAIN...

The quesiton is as follows..

In a lab experiment when 20ml of Lead (II) Nitrate is mixed with 40ml of Potassium Iodide and then strained through a buncher funnel there is 9.112 grams of Lead(II) Iodide precipitate, with 0.108g of Lead(II) Iodide left over along with 4.043796g of Potassium Nitrate and 60g of Water.

The question is

A) what is the molar mass of Lead(II) Iodide ( I got 253.8g/mol)
B)What is the theoretical yield of lead (II) Iodide using the following equation...

Theo yield = 20ml X 1L/1000mL X 1.00 Mol/L X molar mass of PbI2

What I have so far is

Theo Yield = 20ml PbI2 X 1.00 mol PbI2/253.8ml PbI2
= 0.08mol/mL PbI2

Now where I am unsure is the point where 1.00mol/L.  If mol are measured in grams than how can a liquid unit of a mol be measure in Litres?  Anyways its probably something silly.

C)calculate the percentage yield of solid lead(II) iodide.
I know that the percentage yield = actual yield/thoretical yield  but because I am unsure as to whether or not my theoretical yield is correct i cannot be sure the percentage yield is correct

Please help me in anyway you can, this site has been a great help in the past and I hope to hear form somebody soon.

To get the theoretical yield we will need the molarity of the two reacting solutions

Not exactly...

You see my course hasn't covered molarity yet just mol mass.

an example would be....

2 KClO3 -> 2KCl + 3O2

calculate percentage yield of this reaction if 250g of potassium chlorate is reacted and found to produce 80.3g of oxygen gas

step 1:

nKClO3 = 250g KClO3 x 1mol KClO/122.6 KClO3
= 2.04 mol KClO3
nO2    = 2.04 mol KClO3 x 3mol O2/2mol KClO3
= 3.06mol O2
MO2    = 3.06mol O2 x 32.0g O2/ 1mol O2
= 97.9g O2

step 2:

% yield = actual yield/theoretical yield
= 80.3g O2/97.9g O2
= 82.0%
the percentage yield of oxygen in this reaction is 82%

For what you have written above, to calculate the theoretical yield, you need 1 of the following:

1 )molarity

then the Volume of the solution (which you have) x Molarity (which you don't have, yet you used it in your cacluations already at 1 mol / L)

so rereading that, is 1 M the molarity of the solution?

2) the density of the solution (not the solvent),  would give you grams which would give you moles which you can use to find the theoretical yield

so yes you do need it to do the problem,

now the way you have it written, it appears he has already given you a molarity of 1 M.

to get the theoretical yield,

he gave you an equation that already appears to take molarity into account.

Theo yield = 20ml X 1L/1000mL X 1.00 Mol/L X molar mass of PbI2

this is the volume of the solution, converted to mL, multiplied by the molarity and then multipled by the molar mass.

ok then,  my mistake "where 1.00mol/L is the concentrarion of the solution" would make sense now after your explanation.  So is molarity the amount of moles per litre?  and if so how am i supposed to compare 9.11g of precipitate with a liquid unit?  i am so very confused, i appreciate your help immensly but i am still very lost.

also i dont know if this is correct but sample weight of 20ml plus 40ml is now 63.152g (not including the 9.112g of precipitate) so would the concentraion (molarity) of the solution be 63.152g?

so the equation would look like this...

Theo Yield = 20ml X 1L/1000mL x 1mol/.06352L x 253.8g PbI2

is this how the equation would now look?

ok i think i see your confusion,

molarity is a unit of concentration, how much solute is in a solution.

Your teachers theoretical yield equation, gives an answer in grams. (do unit cancellation on the problem and you will see)

So you aren't comparing two different things.

the theoretical yield is what you should get according to the balanced chemical.

equation.

you compare this number to amoutn of PbI2 you collected.

Don't worry, it will always be less than 100%.

Does that help

i wrote this earlier but u might have missed it so.... ::)

also i dont know if this is correct but sample weight of 20ml plus 40ml is now 63.152g (not including the 9.112g of precipitate) so would the concentraion (molarity) of the solution be 63.152g?

so the equation would look like this...

Theo Yield = 20ml X 1L/1000mL x 1mol/.06352L x 253.8g PbI2

is this how the equation would now look?

also could you show me (in the equation) how cancellation of units leaves the answer in grams?

Theo yield = 20ml X 1L/1000mL X 1.00 Mol/L X molar mass of PbI2

i am just going to do units

mL  x  L/mL  =  L  (yes?)

L  x mol / L  = mol

mol x g / mol = g

Are you jesus? haha just kidding but seriously i definatly will tell all my friends (the nerdy ones atleast  :P) about this fabulous site.  just....one...last...QUESTION!!

if the concentraion of PbI2 is 0.108g in a 64.152 g solution ( the rest consists of 4.042796g KNO3 and 60g of water) than what would the molarity be...0.000108L/Mol?

and also what would thr final answer be? i know its kinda cheating but i work backwards better than forwards:D

once again thank you for saving me 2 days as oppose to a couple of hours.

Anything....? ;D ::) ???

did i do this right?...

5.08g / 9.11g x 100%

= 56%????

teaching class atm. will have to wait a bit

if the concentraion of PbI2 is 0.108g in a 64.152 g solution ( the rest consists of 4.042796g KNO3 and 60g of water) than what would the molarity be...0.000108L/Mol?

Molarity = mol of solute divided by liters of solution

so you need to convert 0.108 g PbI2 to moles and then divide by the liters of solution

0.108 g PbI2 x MW PbI2 =  moles PbI2

the problem you have is that you need Liters of solution  , you have grams of solution

you really need the density of the solution to get the actual volume,

usually though if it is aqueous you can assume the mass of the solute is negligible and just use the mass of water.

60 g of water would be 60 mL of water.....

you would need to convert it to liters

OK this is the answer (atleast from what i got) and yes the molarity was simply 1.00mol/L

Theo Yield PbI2 = 20mL x 1L/1000mL x 1.00mol/L x 461g/mol
(Eliminate matching units ;))
= 20/1 x 1/1000 x 1/1 x 461g/1
= 9220g/1000
= 9.22

So the percentage yield would be..

%Yield = Theoretical yield/ actual yield

= 9.11g/9.22g
= 98.8%

The percentage yield of PbI2 in this reaction is 98.8%

Thank you VERY VERY much,  I know I left important things out and put non-important things in, but in the end you helped me realize the answer without telling me, which I think is the goal here.

THANK YOU!

1. Add 40ml of Potassium Iodide to 100ml beaker.
2. Add 20 ml of Lead (II) Nitrate into the beaker with potassium iodide.
3. Stir the contents of the beaker with the stirring rod.