How Do You Calculate the Mass Percent of Fe3+ in Iron Ore Using Stoichiometry?

[Illuvitar]

Hello, I am in desperate need of help with setting up this problem and I would very much appreciate being
steered in the right direction.

1. Homework Statement
Find the mass percent of Fe³⁺ in a 0.0293g sample of iron ore if 13.25 ml of a .200 M stannous chloride solution is required to react completely with Fe³⁺.

Homework Equations

2 Fe³⁺(aq) + Sn²⁺(aq) ⇒2Fe²⁺(aq)+Sn⁴⁺

is the net ionic equation provided

The Attempt at a Solution

So I am trying to find M of Fe³⁺ so

g Fe³⁺→mol Fe/g Fe→mol Fe/mol SnCl²→.200molSnCl²/1L ...etc

Sorry I don't really know how to represent a dimensional analysis problem. Basically I am trying to end up with mol/L to calculate the molarity for Fe³⁺ but I have a strong suspicion that I am attempting this problem in the wrong way. I have tried to rework this problem several times sometimes starting with the 13.25ml but I am ending up at a dead end every time.

Any help or advice is sincerely appreciated.

 

[Bystander]

One step at a time. You have 13.25 ml of what, and what information does that give you?

 

[Illuvitar]

One step at a time. You have 13.25 ml of what, and what information does that give you?

Its 13.25 ml of stannous chloride solution which contains 0.200 moles per liter. It is the amount of solution that is required to completely react with 0.0293g Fe³⁺.

 

[Bystander]

Not two steps at a time --- ONE.

Its 13.25 ml of stannous chloride solution which contains 0.200 moles per liter.

What information?

 

[Chestermiller]

How many moles of stannous ions are present in 13.25 ml of 0.200 molar stannous chloride?

Chet

 

[Illuvitar]

Not two steps at a time --- ONE.
What information?

Thanks for the response. It gives me the information about the number of moles contained in 13.25 ml solution.

How many moles of stannous ions are present in 13.25 ml of 0.200 molar stannous chloride?

Chet

2.65x10^-3 moles. I've tried to use this: 13.25ml x 1L/10³ml x.200mole SnCl₂/ 1 L

but I keep having trouble ending up with mol/L to find the molarity of Fe³⁺. Ill keep trying to play with the equation to see what I get. Thank you.

 

[Bystander]

Moles of what?

 

[Illuvitar]

Chet[/QUOTE]

Moles of what?

Oh sorry, moles of stannous chloride.

 

[Bystander]

So, now you know how many moles of stannous chloride reacted with Fe³⁺, and you've been given a reaction that tells you the proportional relationship between Sn²⁺ and Fe³⁺. Step 2 is using the number you just found with that relationship.

 

[Illuvitar]

So, now you know how many moles of stannous chloride reacted with Fe³⁺, and you've been given a reaction that tells you the proportional relationship between Sn²⁺ and Fe³⁺. Step 2 is using the number you just found with that relationship.

Yes the problem I am having is utilizing this relationship in dimensional analysis to end with the units I am looking for. Here is one of the strategies I have come up with (the main problem I am having is fitting in the 0.0293g Fe³⁺ )

2.65x10-3 moles SnCl₂ x 2 mol Fe³⁺/1mol SnCl₂ x 55.85g Fe³⁺/1 mol Fe³⁺

I keep trying to cancel out units to end up with mol Fe3+/L but I feel as though I am approaching the dimensional analysis set up in the wrong way.

 

[Bystander]

I keep trying to cancel out units to end up with mol Fe3+/L but I feel as though I am approaching the dimensional analysis set up in the wrong way.

What is the point of looking for mol Fe/L

Find the mass percent of Fe3+ in a 0.0293g sample of iron ore

when this is the question you're trying to answer?

Ill keep trying to play with the equation to see what I get

You have yet to write an equation.

 

[Illuvitar]

What is the point of looking for mol Fe/Lwhen this is the question you're trying to answer?
You have yet to write an equation.

Well mol/L is the mass percent of a solution right? I am guessing this is wrong.

Yeah I tried to use this to represent a dimensional analysis chain : 2.65x10-3 moles SnCl2 x 2 mol Fe3+/1mol SnCl2 x 55.85g Fe3+/1 mol Fe3+...

 

[Bystander]

You are interested in solving a pair of proportions, "a is to b as c is to d;" you've seen such proportions written as, " a:b::c:d," or as an equation (includes an equal sign) "a/b = c/d." For the first proportion, you have been given "a" (moles of Fe), "b" moles of Sn, "d" moles of Sn to titrate an unknown quantity of Fe. For the second, you are calculating "a" (mass of Fe³⁺ from the first proportion), given "b" (mass of sample), and calculating a numerical value for "c" (number of %), and know "d" (100).

Do these proportions ONE step at a time, and once you understand the steps, you can solve the proportions for individual values of a, b, c, or d as necessary and substitute them into previous or subsequent proportions to generate a lumped dimensional equation.

 

[Illuvitar]

Im sorry I am such an idiot for not understanding...but I just don't understand this problem, I am even more confused now. Thank you so much for your patience.

 

[Chestermiller]

You are not trying to find the amount of Fe³⁺ is any solution. You are trying to find it in the solid ore, that is, the mass percent in the solid ore. You have the right idea in trying to calculate the number of grams of Fe³⁺, but you made a mistake when you said there are 2 moles of Fe³⁺ for every mole of stannous. There is only 1/2 mole of Fe³⁺ for every mole of stannous.

Chet

 

[Borek]

Well mol/L is the mass percent of a solution right?

No, it is a concentration, but not mass percent.

Besides, you don't need concentration - you need amount of Fe. That can be easily calculated from teh stoichiometry of the reaction and known amount of tin reaction.

I have a feeling you are paralyzed by trying to blindly apply dimensional analysis and without understanding what is happening first. Bystander suggestion of doing calculations a step at a time is perfect.

 

[Illuvitar]

You are not trying to find the amount of Fe³⁺ is any solution. You are trying to find it in the solid ore, that is, the mass percent in the solid ore. You have the right idea in trying to calculate the number of grams of Fe³⁺, but you made a mistake when you said there are 2 moles of Fe³⁺ for every mole of stannous. There is only 1/2 mole of Fe³⁺ for every mole of stannous.

Chet

Why are there 1/2 mole Fe for every 1 mole of stannous if the balanced equation is : 2 Fe3+(aq) + Sn2+(aq) ⇒2Fe2+(aq)+Sn4+ ?

No, it is a concentration, but not mass percent.

Besides, you don't need concentration - you need amount of Fe. That can be easily calculated from teh stoichiometry of the reaction and known amount of tin reaction.

I have a feeling you are paralyzed by trying to blindly apply dimensional analysis and without understanding what is happening first. Bystander suggestion of doing calculations a step at a time is perfect.

So I am trying to find grams of Fe? And yes I I am blindly trying to apply dimensional analysis to this problem because I don't understand what to do. There are no example problems like this in the book. When I created the thread I thought I was heading in the right direction but now I don't even know.

 

[Bystander]

Okay, let's clarify some background for purposes of us (Bystander, Chet, Borek) helping you. What's the highest level of math you've completed?

 

[Illuvitar]

Okay, let's clarify some background for purposes of us (Bystander, Chet, Borek) helping you. What's the highest level of math you've completed?

Algebra 2 :/

 

[Bystander]

Thank you --- give me a couple minutes to double check myself regarding what all I'm taking for granted and rethink how to talk you through this. "I shall return."

 

[Chestermiller]

Why are there 1/2 mole Fe for every 1 mole of stannous if the balanced equation is : 2 Fe3+(aq) + Sn2+(aq) ⇒2Fe2+(aq)+Sn4+ ?

Oops. You're right. I must have misread the equation.

So I am trying to find grams of Fe?

Yes, and you've already shown how to do that. So, what number do you get? What fraction is this of the mass of the ore? What percent?

Chet

 

[Illuvitar]

Oops. You're right. I must have misread the equation.Yes, and you've already shown how to do that. So, what number do you get? What fraction is this of the mass of the ore? What percent?

Chet

Okay so in 13.25 ml of a .200 M stannous there are 2.65x10^-3 moles. So I convert moles SnCl to moles Fe to grams Fe?

2.65x10^-3moles SnCl₂ x 2 moles Fe/1 mol SnCl₂ x 55.85g Fe/1 mole Fe to end up with 0.296 grams Fe? So this is 29.6% of the mass of Fe? Is this what I am looking for?

 

[Bystander]

2.65x10-3moles SnCl2 x 2 moles Fe/1 mol SnCl2 x 55.85g Fe/1 mole Fe to end up with 0.296 grams Fe?

Good to here. Now, what's the mass of the sample?

 

[Illuvitar]

the originally mass of the sample was 0.0293g Fe. So am I supposed to divide the actual by the theoretical yield and multiply by 100% to get the mass percentage?

 

[Bystander]

... and, who wrote this problem? Please tell me you transcribed 0.0293 g from 0.0293 kg. And why didn't I work it through first and catch this disaster?

"Houston, we have a problem." If there's a larger mass of ferric iron in the analysis than there was in the original sample, it's about par for textbook writing and proofreading these days.

Please double check the units in the original problem statement for us. One of us should have caught that earlier --- me particularly.

 

[Illuvitar]

... and, who wrote this problem? Please tell me you transcribed 0.0293 g from 0.0293 kg. And why didn't I work it through first and catch this disaster?

"Houston, we have a problem." If there's a larger mass of ferric iron in the analysis than there was in the original sample, it's about par for textbook writing and proofreading these days.

Please double check the units in the original problem statement for us. One of us should have caught that earlier --- me particularly.

My professor writes out our homework problem sets but it is g not kg. This has kinda happened before. On our first test he gave everyone 10 points because he messed up the units and phrasing on two of the questions.

 

[Bystander]

Sorry I did not spot that earlier --- so, you have found 0.296 g Fe³⁺ in a 0.0293 g sample; 0.0293 g is 100%, and 0.296 g is then how much? It ain't possible in the real universe, but it's the solution to the problem as written.

 

[Illuvitar]

Sorry I did not spot that earlier --- so, you have found 0.296 g Fe³⁺ in a 0.0293 g sample; 0.0293 g is 100%, and 0.296 g is then how much? It ain't possible in the real universe, but it's the solution to the problem as written.

Ill talk to my professor about it, like I said he has made some mistakes before because english is not his first language. Hopefully I will get partial credit even though the problem doesn't make sense. Thank you so much for helping me through it though, I was very confused.

 

[Chestermiller]

Yeah. I agree with Bystander. It must've been kg, not g.

Chet