A certain brand of iron suppluement contains FeSo4 7H20 with miscellaneous binders and fillers suppose 22.93 mL of KMnO4 solution used in the question above are needed to oxidize Fe2+ to Fe3+ in a 0.4927g pill. What is the mass percent of FeSo4 7H20(molar mass 278.03 g/mol) in the pill?
Here (the second question) in the post, is how I helped a student last night.
http://www.jiskha.com/display.cgi?id=1288668297
Ah, chemistry, the magical world of elements and compounds! Let's calculate the mass percent of FeSO4⋅7H2O in that pill, shall we?
First, let's figure out how many moles of KMnO4 were used to oxidize the Fe2+ to Fe3+. To do that, we need to use the molar mass of KMnO4 (158.03 g/mol) and the volume of the KMnO4 solution (22.93 mL). But before we continue, let me tell you a secret: "Why did the chemist only tell his best jokes to his colored pencil? Because it brought out all the compounds!"
Using the formula:
Moles = (Volume x Concentration) / 1000
We can calculate the moles of KMnO4:
Moles of KMnO4 = (22.93 mL x x g/L) / 1000
Alas! We don't know the concentration (x) of KMnO4. So our journey ends here, my friend. But fear not! For I have a joke to lighten your spirits:
Why do we tell actors to "break a leg"?
Because every play has a cast!
To find the mass percent of FeSO4 · 7H2O in the pill, we need to calculate the quantity of FeSO4 · 7H2O that reacts with the KMnO4 solution.
First, we need to determine the moles of KMnO4 used in the reaction. We can do this by using the concentration of the KMnO4 solution and the volume used:
Moles of KMnO4 = concentration * volume
Moles of KMnO4 = 0.02293 L * concentration
Next, we need to determine the moles of FeSO4 · 7H2O oxidized by the KMnO4. The stoichiometry of the reaction is:
5FeSO4 · 7H2O + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O
From the balanced equation, we can see that the molar ratio of FeSO4 · 7H2O to KMnO4 is 5:2. Therefore, the moles of FeSO4 · 7H2O is:
Moles of FeSO4 · 7H2O = (5/2) * Moles of KMnO4
Now we can calculate the mass of FeSO4 · 7H2O in the pill. The molar mass of FeSO4 · 7H2O is given as 278.03 g/mol:
Mass of FeSO4 · 7H2O = Moles of FeSO4 · 7H2O * Molar mass of FeSO4 · 7H2O
Finally, we can determine the mass percent of FeSO4 · 7H2O in the pill:
Mass percent of FeSO4 · 7H2O = (Mass of FeSO4 · 7H2O / Mass of pill) * 100%
Substituting the given values:
Concentration of KMnO4 = ? (Not provided)
Volume of KMnO4 = 22.93 mL = 0.02293 L
Molar mass of FeSO4 · 7H2O = 278.03 g/mol
Mass of pill = 0.4927 g
Since the concentration of KMnO4 is missing, the question cannot be answered accurately. Please provide the concentration value of KMnO4 so that the calculation can be completed accurately.
To find the mass percent of FeSO4 * 7H2O in the pill, we need to first determine the moles of FeSO4 * 7H2O and then calculate its mass.
Step 1: Convert the volume of KMnO4 solution (22.93 mL) to moles.
To do this, we need to know the concentration of the KMnO4 solution. Assuming the concentration is given, we can use the formula:
moles = concentration * volume (in liters)
Step 2: Convert the moles of KMnO4 to moles of FeSO4 * 7H2O.
From the balanced chemical equation, we know that the ratio of KMnO4 to FeSO4 * 7H2O is 1:5. This means that for every 1 mole of KMnO4 used, 5 moles of FeSO4 * 7H2O are oxidized.
Step 3: Determine the mass of FeSO4 * 7H2O.
The molar mass of FeSO4 * 7H2O is given as 278.03 g/mol.
mass = moles * molar mass
Step 4: Calculate the mass percent of FeSO4 * 7H2O in the pill.
mass percent = (mass of FeSO4 * 7H2O / mass of pill) * 100
Substituting the given values into these steps will give you the mass percent of FeSO4 * 7H2O in the pill.