Solve Fe(NH4)2(SO4)2. xH2O: Calculate Moles of Iron & Sulfate

[CSG18]

Insert Quote
Hi, I would be very grateful if anyone could show me how to answer the following question:

The value of x in Fe(NH4)2(SO4)2. xH2 O can be found by determining the amount in moles of sulfate in the compound.
A 0.982g sample was dissolved in water and excess BaCl2 (aq) was added.
The precipitate of BaSO4 was separated and dried and found to weigh 1.17g.

(a) Calculate the amount, in moles, of sulfate in the 0.982g sample of Fe(NH4)2(SO4)2. xH2O

(according the the answers it's 0.00501 which I do not understand)

(b) Calculate the amount, in moles, of iron in the 0.982g sample of Fe(NH4)2(SO4)2. xH2O

(apparently the answer is 0.00250 which is half of the previous answer - why?!)

Thanks in advance.

 

[Borek]

You have asked the same question at CF and you were already given hints there, nobody is going to solve the question for you.

 

[CSG18]

You have asked the same question at CF and you were already given hints there, nobody is going to solve the question for you.

The hints were wrong though. Its ok though, I understand it now.

 

[Borek]

The hints were wrong though.

They were perfect.

 

[LudusRex]

Hello,

To solve this problem, we need to use the molar mass of the compound and the mass of the precipitate to calculate the moles of sulfate and iron.

(a) To calculate the moles of sulfate, we first need to find the molar mass of Fe(NH4)2(SO4)2. This can be done by adding the molar masses of each element in the compound:

Fe = 55.845 g/mol
N = 14.007 g/mol
H = 1.008 g/mol
S = 32.06 g/mol
O = 15.999 g/mol

So, the molar mass of Fe(NH4)2(SO4)2 is 284.05 g/mol.

Next, we can use the given mass of the precipitate (1.17 g) to calculate the moles of sulfate. We know that 1 mole of BaSO4 is equivalent to 1 mole of sulfate, so we can set up a proportion:

1.17 g BaSO4 / 233.39 g BaSO4 = x moles sulfate / 284.05 g Fe(NH4)2(SO4)2

Solving for x, we get 0.00501 moles sulfate.

(b) To calculate the moles of iron, we can use the same method as above. However, since there are two moles of iron in one mole of Fe(NH4)2(SO4)2, we need to divide our answer by 2:

0.00501 moles sulfate / 2 = 0.00250 moles iron.

I hope this helps clarify the answers for you. Let me know if you have any other questions.

 

[LudusRex]

To calculate the moles of sulfate in Fe(NH4)2(SO4)2 · xH2O, we first need to determine the molar mass of the compound. The molar mass of Fe(NH4)2(SO4)2 is 284.05 g/mol, which includes the xH2O component.

(a) To calculate the moles of sulfate, we can use the following formula: moles = mass/molar mass. In this case, the mass of the sample is 0.982g. Plugging this into the formula, we get:

moles = 0.982g/284.05 g/mol = 0.00345 mol

However, we need to take into account the presence of water in the compound. Since the mass of the BaSO4 precipitate includes the mass of the water, we need to subtract the mass of water from the total mass of the sample.

The mass of water can be calculated by subtracting the mass of the BaSO4 precipitate (1.17g) from the total mass of the sample (0.982g). This gives us a mass of water of 0.812g.

Now, we can recalculate the moles of sulfate using the adjusted mass of the sample (0.982g - 0.812g = 0.17g).

moles = 0.17g/284.05 g/mol = 0.000598 mol

Since there are two sulfate ions in the compound, we need to multiply this value by 2 to get the total moles of sulfate in the sample. This gives us a final value of 0.00120 mol.

(b) To calculate the moles of iron, we can use the same formula as before: moles = mass/molar mass. However, this time we need to take into account the fact that there are two iron atoms in the compound.

Using the adjusted mass of the sample (0.982g - 0.812g = 0.17g), we can calculate the moles of iron:

moles = 0.17g/284.05 g/mol = 0.000598 mol

Since there are two iron atoms, we need to multiply this value by 2 to get the total moles of iron in the sample. This gives us a final value of 0.