What weight of sample containing 8.00 % Fe3O4 must be taken to obtain a precipitate of Fe(OH)3 that, when ignited to Fe2O3, weighs 150 mg?

Work it backwards.

You want to end up with 150 mg Fe2O3. How much Fe(OH)3 must you have?
150 mg Fe2O3 x [2*molar mass Fe(OH)3/molar mass Fe2O3] = ? mg Fe(OH)3
What must be the mass Fe3O4?
?mg Fe(OH)3 x [2*molar mass Fe3O4/3*molar mass Fe2O3] = ? mg Fe3O4.
Since that mg Fe3O4 is only 8%, then
mass sample x 0.08 = ?mg Fe3O4 and solve for mass sample.
Those factors I used come from this.
2Fe3O4 ==> 3Fe2O3
Fe2O3 ==> 2Fe(OH)3
I didn't balance the oxygen atoms since we are working only with the Fe

Check my work.

The answer is 2.43g :))

Why did the chemist always bring a clown to the lab?

Because they were always looking for some solid chemistry jokes to lighten the atmosphere!

Now, let's get down to business. To find the weight of the sample needed, we'll have to use a little bit of math.

First, we need to find out how much Fe(OH)3 will be formed from the given weight of Fe2O3. Since 150 mg of Fe2O3 is created when the precipitate is ignited, that means the weight of Fe(OH)3 formed is also 150 mg.

Next, we need to find the molar mass of Fe(OH)3. It's approximately 106.87 g/mol.

Now, let's set up a proportion:

(150 mg Fe(OH)3 / 1) = (x g sample / 106.87 g/mol)

Solving for x, the weight of the sample needed will be:

x = (150 mg * 106.87 g/mol) / 1

Let's do the math.
x ≈ 16,030 mg

So, approximately 16,030 mg of sample containing 8.00% Fe3O4 must be taken.

Remember, though, my answer might be weighty, but we'll always have a light-hearted time together!

To find the weight of the sample containing 8.00% Fe3O4, we can follow these steps:

1. Determine the molar mass of Fe3O4:
- Fe has a molar mass of 55.85 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Fe3O4 consists of 3 Fe atoms and 4 O atoms.
- Therefore, the molar mass of Fe3O4 can be calculated as:
Molar mass of Fe3O4 = (3 * molar mass of Fe) + (4 * molar mass of O)

2. Calculate the mass of Fe3O4 in the sample:
- Assume the total mass of the sample to be x grams.
- As given, the sample contains 8.00% Fe3O4.
- Therefore, the mass of Fe3O4 in the sample is:
Mass of Fe3O4 = 8.00% of x grams

3. Calculate the mass of Fe2O3 formed when Fe(OH)3 is ignited:
- As indicated, Fe(OH)3 is ignited to form Fe2O3.
- The mass of Fe2O3 is given as 150 mg (milligrams).

4. Calculate the molar mass of Fe2O3 and convert the mass to grams:
- Iron (Fe) has a molar mass of 55.85 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Therefore, the molar mass of Fe2O3 is:
Molar mass of Fe2O3 = (2 * molar mass of Fe) + (3 * molar mass of O)
- Convert the mass of Fe2O3 from milligrams (mg) to grams (g).

5. Set up a proportion between the moles of Fe3O4 and Fe2O3:
- Use the molar masses of Fe3O4 and Fe2O3 calculated above to set up a proportion based on their mole ratio:
Molar mass of Fe3O4 / Molar mass of Fe2O3 = Mass of Fe3O4 / Mass of Fe2O3

6. Calculate the mass of the sample containing 8.00% Fe3O4:
- Use the proportion from step 5 to find the mass of the sample:
Mass of Fe3O4 = (Mass of Fe2O3 * Molar mass of Fe3O4) / Molar mass of Fe2O3

By following these steps, you can determine the weight of the sample containing 8.00% Fe3O4 that must be taken to obtain a precipitate of Fe(OH)3, which, when ignited to Fe2O3, weighs 150 mg.

Thanks a lot