A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Fe^2+ in acid and then titrating the Fe^2+ with MnO4^-. A 1.3510 g sample was dissolved in acid and then titrated with 39.32 mL of 0.03190 M of KMnO4. The balanced equation is:

8H^+(aq)+5Fe^2+(aq)+MnO4^-(aq)->5Fe^2+(aq)+Mn^2+(aq)+4H2O(l)
Calculate the mass percent of iron in the ore.

moles MnO4^- = M x L

moles Fe = moles MnO4^- x (5 moles Fe/1 mole MnO4^-) = ??
g Fe = moles Fe x atomic mass Fe.
%Fe = (mass Fe/mass sample)*100 = ??

I know this is just a problem but in real life I doubt a chemical engineer did any of this. In the U. S. at least, I think MOST would have taken the sample to the lab and told a chemist, "Here, analyze this and tell me the percent Fe."

Well, it seems like our chemical engineer is on a mission to unlock the iron's secrets in that ore sample! Let's help them out with some calculations.

First, let's calculate the number of moles of MnO4^- used in the titration. We can use the equation:

Moles of MnO4^- = (volume of KMnO4 solution in liters) * (molarity of KMnO4 solution)

For our given values:

Moles of MnO4^- = (39.32 mL * 0.001 L/1 mL) * (0.03190 M)

Let me crunch the numbers real quick...

Okay, so we end up with moles of MnO4^- = 0.001255 moles.

Now let's use the stoichiometry of the balanced equation to figure out the moles of Fe^2+ that reacted. According to the equation, for every 5 moles of Fe^2+, we need 1 mole of MnO4^-.

So, moles of Fe^2+ = (moles of MnO4^-) * (5 moles of Fe^2+/1 mole of MnO4^-)

Let me calculate that for you...

Moles of Fe^2+ = 0.001255 moles * 5 moles/1 mole = 0.006275 moles.

Since we know the mass of the sample (1.3510 g), we can calculate the mass of iron in the sample. According to the stoichiometry, for every 5 moles of Fe^2+, we have 1 mole of Fe. The molar mass of Fe is approximately 55.845 g/mol.

Mass of iron = (moles of Fe^2+) * (1 mole of Fe/5 moles of Fe^2+) * (molar mass of Fe)

Let me crunch that last calculation...

The mass of iron in the sample is approximately 0.006275 moles * (1 mole/5 moles) * 55.845 g/mol = 0.0068727 g.

Finally, to find the mass percent of iron in the ore, we use the following formula:

Mass percent of iron = (mass of iron/mass of sample) * 100

Now, let me plug in the values and find the percentage for you...

The mass percent of iron in the ore sample is approximately (0.0068727 g/1.3510 g) * 100 = 0.508%

So, the clownish conclusion is that the ore sample contains approximately 0.508% of iron. Remember, this answer comes with a side dish of humor!

To calculate the mass percent of iron in the ore, we need to determine the amount of iron present in the sample and then divide it by the mass of the entire sample.

1. Calculate the moles of KMnO4 used:
Molarity of KMnO4 (M) = 0.03190 M
Volume of KMnO4 used (L) = 39.32 mL = 0.03932 L

Moles of KMnO4 used = Molarity * Volume = 0.03190 M * 0.03932 L = 0.001254 moles KMnO4

2. Use stoichiometry to find the moles of Fe in the sample:
From the balanced equation, we can see that the ratio of Fe to KMnO4 is 5:1.

Moles of Fe = 0.001254 moles KMnO4 * (5 moles Fe / 1 mole KMnO4) = 0.00627 moles Fe

3. Calculate the molar mass of Fe:
The molar mass of Fe is 55.845 g/mol.

4. Calculate the mass of Fe in the sample:
Mass of Fe = Moles of Fe * Molar mass of Fe = 0.00627 moles Fe * 55.845 g/mol = 0.3498 g Fe

5. Calculate the mass percent of Fe in the ore:
Mass percent of Fe = (Mass of Fe / Mass of sample) * 100
Mass of sample = 1.3510 g

Mass percent of Fe = (0.3498 g / 1.3510 g) * 100 = 25.89%

Therefore, the mass percent of iron in the ore sample is approximately 25.89%.

To calculate the mass percent of iron in the ore, we need to know the number of moles of iron and the total mass of the sample.

Step 1: Calculate the number of moles of KMnO4 used in the titration.
The given volume of KMnO4 solution is 39.32 mL, and the molarity is 0.03190 M.
Using the formula: moles = volume (L) × concentration (M), we can calculate the number of moles of KMnO4.
moles of KMnO4 = (39.32 mL / 1000 mL/L) × 0.03190 M

Step 2: Calculate the number of moles of Fe^2+ reacted.
From the balanced equation, we can see that 1 mole of KMnO4 reacts with 5 moles of Fe^2+. Therefore, the number of moles of Fe^2+ is equal to 5 times the number of moles of KMnO4:
moles of Fe^2+ = 5 × moles of KMnO4

Step 3: Calculate the mass of Fe in the sample.
To calculate the mass of Fe, we need to know the number of moles of Fe and the molar mass of Fe.
The molar mass of Fe is given as 55.845 g/mol.
mass of Fe = moles of Fe^2+ × molar mass of Fe

Step 4: Calculate the mass percent of iron in the ore.
The mass percent is defined as the mass of Fe divided by the total mass of the sample, multiplied by 100.
mass percent of iron = (mass of Fe / total mass of the sample) × 100

Plug in the values from the given data into the above equations to calculate the mass percent of iron in the ore.