[Guilo]

Hello,

I have no idea how to solve this...

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO₄ is required to titrate the unknown to the endpoint.

a) How many moles of MnO₄^- were added?
b) How many moles of iron (II) must be present in the sample?
c) How many grams of iron (II) must be present in the sample?
d) What is the percent of iron present in the sample?


I would know b, c, d, if I would know how to solve a. b, c, d, questions are simple, but I totally don't understand how to solve a. I wrote the rest so people who don't know similar exercise could get something more out of that topic.
I know I shouldn't ask such a basics here, but I spent like an hour on YouTube and internet with no success. My professor is very unclear on every topic.

Thank you so much for helping.

Guilo

[sjb]

Hello,

I have no idea how to solve this...

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO₄ is required to titrate the unknown to the endpoint.

a) How many moles of MnO₄^- were added?
b) How many moles of iron (II) must be present in the sample?
c) How many grams of iron (II) must be present in the sample?
d) What is the percent of iron present in the sample?


I would know b, c, d, if I would know how to solve a. b, c, d, questions are simple, but I totally don't understand how to solve a. I wrote the rest so people who don't know similar exercise could get something more out of that topic.
I know I shouldn't ask such a basics here, but I spent like an hour on YouTube and internet with no success. My professor is very unclear on every topic.

Thank you so much for helping.

Guilo

What does 0.01522 M KMnO₄ mean?

[Guilo]

What does 0.01522 M KMnO₄ mean?

The molarity of KMnO₄ solution is 0.01522
Molarity = moles/1L

[Borek]

Molarity = moles/1L

Yes and no. What if you have just 0.5 L, or 12.6 mL of the solution?

[Guilo]

Molarity = moles/1L

Yes and no. What if you have just 0.5 L, or 12.6 mL of the solution?

Yes, I understand what you mean. Molarity of KMnO₄ solution solution in this exercise is 0.01522, so it is 0.01522 moles in 0.0126 L.


But, I don't have any idea how to start solving a)


^{0.01522 moles KMnO4}/_{0.0126 L KMnO4}=^{158 g KMnO4}/_{1 mol KMnO4}=^2.40476/_0.0126≈191^g/_L


And I have no idea what to do next...

[Arkcon]

Yes, I understand what you mean. Molarity of KMnO4 solution solution in this exercise is 0.01522, so it is 0.01522 moles in 0.0126 L.

Incorrect.  The molarity of the solution was given as 0.01522 moles per 1 L.  The experiment uses less than one liter, you have to convert to determine how many moles were used.

You will also have to write a balanced chemical equation for the reaction between iron (II) and permanganate.  You will the relate the moles of permanganate to moles of iron.

[Guilo]

the two half-reactions are

MnO₄^- + 8H⁺ + 5e^- = Mn²⁺ + 4H₂O
Fe²⁺ = Fe³⁺ + 1e^-   // x5

the balanced equation is

MnO₄^- + 5Fe²⁺ + 8H⁺ = Mn²⁺ + 5Fe³⁺ + 4H₂O


I don't understand why there is KMnO₄ not just MnO₄ (In the exercise question).


Is that right?
^{0.01522 moles}/_{0.0126 L} x 0.0126L ≈ 0.0002 = 2x10^-4 moles

Still no idea what to do next...

[Borek]

I don't understand why there is KMnO₄ not just MnO₄ (In the exercise question).

How are you going to prepare solution containing just MnO₄^-? It has to contain some counterion.

Is that right?
^{0.01522 moles}/_{0.0126 L} x 0.0126L ≈ 0.0002 = 2x10^-4 moles

No. Basically you multiplied and divided by the same number, so it canceled out.

To speed things up, see here:

http://www.titrations.info/titration-calculation

[Guilo]

Hey Borek,

I made mistake. Please confirm if everything is right for each answer:

The initial mass of an unknown sample containing iron (II) is 0.512 g.
12.6 mL of 0.01522 M KMnO4 is required to titrate the unknown to the endpoint.

a) How many moles of MnO4- were added?
^{0.01522 moles}/_{1 L} x 0.0126L ≈ 0.0002 = 2x10^-4 moles

b) How many moles of iron (II) must be present in the sample?
2x10^-4 x 5 = 0.001

c) How many grams of iron (II) must be present in the sample?
0.001 moles of Fe²⁺ x ^{56g Fe}/_{1 mole Fe} = 0.056g Fe

d) What is the percent of iron present in the sample?
^{0.056g Fe}/_{0.512 g unknown sample} x 100 = 10.9%

[Borek]

Never round down intermediate results - that is, report them rounded, but use fulls precision for calculations.

You are off by almost 2% (or almost 20%, depending on how you look at it) because of roundings, but the approach is correct.