[L4Z0]

How do you work out:

the number of moles of iron (II) ions in a conicle flask?

the number of moles of iron (II) ions in the 250cm3 of solution in the volumetric flask

the mass of iron in the 250cm3 of solution in the volumetric flask

the iron that must was in the mass of hydrated iron (II) ammonium sulphate

the percentage of iron in the hydrated iron (II) ammonium sulphate crystals

[clownfish]

You need to work out the balanced equation for the reaction between permaganate and iron, this will then give you the mole ratio. use the mole ratio to find the number of moles of iron in the conical flask.
Then multiply this up to the size of the volumetric flask (eg if you used 25cm3 samples in the conical flask, then the no of moles in the volumetric flask would be 10x more).
Then use mass = moles x atomic mass to find the mass of iron
and divide by the mass of the hydrated crystals and x100 to get percentage



I'm pretty sure that's the method, but I'm not 100% clear on some of your steps above. I hope that helps.

[L4Z0]

Thanks I figured out the balanced equation 5 Fe2+(aq) + MnO4-(aq) + 8H+(aq)=5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
Sorry but what is the mole ratio and how do you use the mole ratio to find the number of moles of iron in the conicle flask?

[clownfish]

In your balanced equation you have written that 5 moles of iron react with 1 mole of permanganate - this is the mole ratio, therefore multiply the no of moles of permanganate by 5 to get the number of moles in the conical flask.

[L4Z0]

Ok, thanks so much for your help

But if I got 1.50 mol for example in the conicle flask would that mean I would have to devide 250 by 1.50 to work out the number of moles in the 250cm3 solution?

[clownfish]

(Original post by L4Z0)
Ok, thanks so much for your help

But if I got 1.50 mol for example in the conicle flask would that mean I would have to devide 250 by 1.50 to work out the number of moles in the 250cm3 solution?

No.
At this point you just scale up the volumes, so if the conical flask contains 25cm3 of your solution, then the volumetric flask would have contained 10x more, so the number of moles in the volumetric flask would be 1.5 x10 = 15 moles.

[L4Z0]

Yeah, I realised it was wrong.

Is the part for the mass of iron just 15x55.845?

[clownfish]

(Original post by L4Z0)
Yeah, I realised it was wrong.

Is the part for the mass of iron just 15x55.845?

Yes it would be. Are you sure your number is 15? It's normally less than 1 when done on a laboratory scale (the examples when they make you work in tonnes naturally are much bigger!). Remember when working out number of moles in a solution we have to change the volume into dm3 - you do this by dividing the cm3 by 1000.

[L4Z0]

Yeah thats what I was thinking.

I'm using a similar figures to my real one cause I want to do it the working out by myself.

But when I used 15 it came out as 837.675. So maybe I should divide that by a thousand because I didnt do it before?

[clownfish]

(Original post by L4Z0)
Yeah thats what I was thinking.

I'm using a similar figures to my real one cause I want to do it the working out by myself.

But when I used 15 it came out as 837.675. So maybe I should divide that by a thousand because I didnt do it before?

Yes you will need to divide by 1000, but you want to be in the habit of doing that in the first step moving forwards.

[L4Z0]

Hmmmm.

Is the iron used in the mass of hydrated iron (II) aluminium sulphate......

0.837675 divided by the atomic mass of hydrated iron (II) ammonium suplate?

[clownfish]

no it's divided by the mass of salt that you were given in the question (ie mass in grams, not RMM)

[L4Z0]

Oh right so 0.837675 divided by 5g

[clownfish]

(Original post by L4Z0)
Oh right so 0.837675 divided by 5g

Yep, then x100 to get a percentage.

[L4Z0]

Thanks so much sorry for being such a pain lool

[clownfish]

(Original post by L4Z0)
Thanks so much sorry for being such a pain lool

Happy to help, hope you get how to attack this kind of question now.