[Generic Username]

The description about the table below reads: '125mg of sodium oxalate is weighed, dissolved in warm acid and titrated with 18.3 ml of permanganate solution. The results are shown in the table here:'

I can work out the right hand column (sodium oxalate) from the information given, but not the left (MnO4^-). I can't work out the concentration using c = n/V because I don't have the amount of substance for MnO4^-. I tried converting n in the formula to m/M, but that didn't help because I don't have the mass for MnO4^-. How on earth did they work out c and n for MnO4^-?

[Borek]

Stoichiometry.

[mikasaur]

I tried converting n in the formula to m/M, but that didn't help because I don't have the mass for MnO4^-

Do you need the mass?

Borek's hint was subtle, but a darn good one.

[Generic Username]

I tried converting n in the formula to m/M, but that didn't help because I don't have the mass for MnO4^-

Do you need the mass?

Borek's hint was subtle, but a darn good one.

Hi mikasaur,
I took the hint (assuming I understood it correctly) and have been playing about with Vm = V/n, [Vm = L/mol], where I took the number of moles to be 2, resulting in  [Vm = 0.0183/2 = 0.009150000000, which when I use with n = V/Vm gives me a result of 2. Correct - if only I didn't have to divide it by 100 for it to be so.

How would I then work out the mass?

I now have the correct answer for n = 0.02, but only if I write [Vm = 0.0183*1000/2 = 0.009150000000]. I'm not sure if I should do that at all for Vm.

Using m = M * n, I've just worked out the mass to be m = 118.9356*2 = 237.8712, which just seems way off. Converting the formula gives me an n of 2 - again. 

Chemistry, chemistry thou art but a mystery!

Perhaps if I'd looked, I would've noticed that 0.02 is the figure for c and not n - as I previously stated. It's been a long day. Back to the drawing board.


Ah, I think I've got it. The ratio is 2:5 - not 2:1. Take the value for n for sodium oxalate: 0.933, divide by 5 and then multiply by 2 to give n for MnO4 = 0.373. Thanks guys!

[Borek]

How many moles of permanganate will react with 0.125 g of sodium oxalate?

[Generic Username]

How many moles of permanganate will react with 0.125 g of sodium oxalate?

As far as I can tell, 2 moles of permanganate for every 5 of sodium oxalate.

[mikasaur]

How many moles of permanganate will react with 0.125 g of sodium oxalate?

As far as I can tell, 2 moles of permanganate for every 5 of sodium oxalate.

You made your edit seconds before Borek wrote his response, so I'm guessing he didn't see it.

It looks like you've figured this one out. You can use your 2:5 ratio to calculate n and then c.

[Generic Username]

How many moles of permanganate will react with 0.125 g of sodium oxalate?

As far as I can tell, 2 moles of permanganate for every 5 of sodium oxalate.

You made your edit seconds before Borek wrote his response, so I'm guessing he didn't see it.

It looks like you've figured this one out. You can use your 2:5 ratio to calculate n and then c.

It seems that stoichiometric ratios really are the key to solving these sorts of problems - something easy for the uninitiated (like me) to forget. Thanks again, guys.