# Homework Help: Redox titration and mass percent composition calculations.

1. Oct 27, 2014

### kirsten_2009

1. The problem statement, all variables and given/known data

Hello everyone,

I have completed a lab and I’m working on my report. I’m not sure if my work is correct but I would most definitely appreciate it if anyone could check it for me? I’m particularly having trouble finishing question #3. It’s worth a lot of my grade so I want to make sure I get it right! Thanks a lot in advance for all the time and help :)

Basically, the lab is a set of 2 redox titrations. For the first part, we standardized a KMnO₄ solution and in the second we titrated K₃[Fe(C₂O₄)₃]∙3H₂O

2. Relevant equations

2MnO₄²¯(aq) + 5C₂O₄²¯(aq) + 16H⁺(aq) à 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)

3. The attempt at a solution

1. Given a mass of 0.2349 g of Na₂C₂O₄ in 100 mL, what is the concentration of Na₂C₂O₄?
A:// 0.2349 g Na₂C₂O₄ = 1 mol Na₂C₂O₄ / 0.1 L = 0.0175 M
2. You used 6.9933 mL of KMnO₄, what is the concentration of the KMnO₄ solution?
A:// 0.00175 mol Na₂C₂O₄ x 2 mol KMnO₄ / 5 mol Na₂C₂O₄ = 0.0007 mol KMnO₄ / 0.0069933 L =
0.1001 M

3. You dissolved 0.2196 g of K₃[Fe(C₂O₄)₃]∙3H₂O in 100 mL. Given the concentration of KMnO₄ you just calculated and assuming that you used 7.1667 mL of it: determine the Mass Percent Composition of C₂O₄²¯ in K₃[Fe(C₂O₄)₃]∙3H₂O.
A:// 7.1667 mL x 1L / 1000 mL = 0.0071667 L x 0.1001mol / 1L = 0.000717386 mol KMnO₄

0.000717386 mol KMnO₄ x 5 mol Na₂C₂O₄ / 2 mol KMnO₄ = 0.0017935 mol Na₂C₂O₄ x 491.243 g/mol =
0.08810 g Na₂C₂O₄

And then I’m stuck…
4. Compare your experimentally determined % oxalate in your product to theoretically mass percentage of C₂O₄²¯ in K₃[Fe(C₂O₄)₃]∙3H₂O. Use the following equation to aid in your discussion: Experimentally determined % oxalate / Theoretically determined % oxalate x 100%
This seems fairly straight forward if I could correctly figure out #3…however, what would my theoretical
mass percent composition come from? Is it from the 0.2349 g of Na₂C₂O₄?

I know I'm not suppose to post more than one question per post but they are all related and if I made a mistake in the previous one then I carry it forward to the next; so pardon my breach of protocol.

2. Oct 27, 2014

### Staff: Mentor

Try to write balanced reaction equation of K3[Fe(C2O4)3]∙3H2O with permanganate.

3. Oct 27, 2014

### kirsten_2009

Allow me to restate question 3 since it was a 25 mL aliquot of K3[Fe(C2O4)3]∙3H2O that was titrated.

So the question should say:

"You dissolved 0.2196 g of K₃[Fe(C₂O₄)₃]∙3H₂O in 100 mL. Given the concentration of KMnO₄ you just calculated and assuming that you used 7.1667 mL of it to titrate a 25 mL aliquot of K₃[Fe(C₂O₄)₃]∙3H₂O : determine the Mass Percent Composition of C₂O₄²¯ in K₃[Fe(C₂O₄)₃]∙3H₂O".

I believe the balanced net ionic equation is:

2MnO₄²¯(aq) + 5C₂O₄²¯(aq) + 16H⁺(aq) à 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)

and I know the moles of MnO₄²¯ that react is 0.000717386 mol and based on the stoichiometric ratio then 0.0017935 mol of C₂O₄²¯ react

but i still don't see the relationship between the moles of C₂O₄²¯ and the grams of C₂O₄²¯ present in K₃[Fe(C₂O₄)₃]∙3H₂O

I know the theoretical % mass composition would be 53.75% oxalate in K₃[Fe(C₂O₄)₃]∙3H₂O just by the ratio of their molecular weights but I don't know how to find the experimental % mass composition of my sample. Thanks for replying and for your time.

4. Oct 27, 2014

### Staff: Mentor

To be honest, I don't understand your statement

What is molar mass of C2O42-?

Note: please format your posts using tags: [SUB]4[/SUB][SUP]2-[/SUP], or using subscript and superscript buttons (x2 and x2) in the editor, as of now, they don't always display correctly.

5. Oct 27, 2014

### kirsten_2009

Hi Mr.Borek,

I think I'm not understanding myself either. I'm sorry for the confusion. I'm just going to start from the beginning so as to get my thoughts in order so please ignore the above posts and if this doesn't work then I'll leave everyone alone :D

Question Is: What is the % mass composition of oxalate (C2O42-) in K3[Fe(C2O4)3]*3H2O

Background Information: 0.2196 g K3[Fe(C2O4)3]*3H2O was dissolved in 100 mL and 25 mL aliquot of this solution was titrated with 7.1667 mL of 0.1001 M KMnO4

2MnO42- + 5C2O42- + 16H+ --> 2Mn2+ + 10CO2 + 8H2O

Attempt at a Solution:

1. Find the moles of KMnO4 that were present in 7.1667 mL or 0.0071667 L of the solution :
0.0071667 L * 0.1001 mol / 1 L KMnO4 = 0.00071739 mol KMnO4

2. Using the stoichiometric ratio from the balanced net ionic equation above; find the moles of K3[Fe(C2O4)3]*3H2O that reacted with the KMnO4
0.00071739 mol KMnO4 * 5 mol K3[Fe(C2O4)3]*3H2O / 2 mol KMnO4 = 0.0017935 mol K3[Fe(C2O4)3]*3H2O

My Question: Where do I go from here? I could convert the moles of potassium trioxalatoferrate trihydrate to grams and that would be how many grams were present in a 25 mL aliquot but I don't know how to determine the % mass composition of just the oxalate

6. Oct 27, 2014

### Staff: Mentor

Why not just

$$\frac {mass~of~the~oxalate~anion}{mas~of~the~sample} 100%$$

I must admit I find the wording confusing - it is not entirely clear to me what they mean by "oxalate". It can either mean just the oxalate anion, or the potassium iron(III) oxalate trihydrate (or whatever systematic name of the compound is).

7. Oct 27, 2014

### kirsten_2009

I agree, and that is what I did originally....but when I convert the moles of K3[Fe(C2O4)3]*3H2O into grams (0.0017935 mol * 491.243 g/mol = 0.8810 g) then it becomes greater than the initial sample of 0.2196 so that doesn't make sense.

Alternatively, If i convert the moles of K3[Fe(C2O4)3]*3H2O to grams using the molar mass of oxalate (C2O4 ) then it gives me a smaller number (0.0017935 mol * 88.019 g/mol = 0.15786 g) which makes more sense ...but still wouldn't be correct because that would give me 0.15786/0.2196 * 100 = 72% and the theoretical % mass composition of C2O4 in K3[Fe(C2O4)3]*3H2O is 53.75 %
Anyways, I appreciate all the time and effort to try and help me...I just don't get it though. Thanks!

8. Oct 27, 2014

### Staff: Mentor

Have you titrated whole 100 mL, or just an aliquot?

9. Oct 27, 2014

### kirsten_2009

a 25 mL aliquot (three 25 mL aliquot actually....but we averaged the tree titration, so just one?)

10. Oct 28, 2014

### Staff: Mentor

If it was an aliquote, what was mass of sodium oxalate per an aliquot? And if so, what was the concentration of the permanganate?