Question

Calculate the molar mass of Fe2(SO4)3. Then Calculate the mass of iron(III) sulfate of 4.05x10^23 formula units. Work must be shown.​

Answer 1

Answer:

Molar mass: 399.9 g/mol

The mass of iron(III) sulfate is 269 grams.

Explanation:

First of all, you have to find the molar mass of Fe2(SO4)3.

So you find the molar first for each element

Fe = 55.8 amu

S = 32.1 amu

O = 16.0 amu

However, there are 2 iron atoms, 3 sulfur atoms, and 12 oxygen atoms.(Don't forget the 3 outside the parenthesis of SO4).

So you do this

55.8(2) + 32.1(3) + 16.0(12) = 399.9

So the final answer is 399.9 g/mol

Second of all, if you want to find the mass of Fe2(SO4)3, you have to convert from formula units to moles. After that, then you convert from moles to grams.

If you want to convert from molecules to molecules, you have to use Avogadro's number. Avogadro's number is 6.02 x 10^23. Then if you want to convert from moles to grams, you use the molar mass. We already know that the molar mass is 399.9 g/mol.

Therefore, now use dimensional analysis to show your work.

(4.05 x 10^23) formula units of Fe2(SO4)3 * 1 mol/ 6.02 x 10^23 formula units * 399.9 g/mol / mol

Formula units in the beginning and the moles will cancel out.

So [(4.05 x 10^23) / (6.02 x 10^23)] x 399.9 = 269.0357143

But we need to use significant figures with the least digits(which is 4.05) So we round to the nearest whole number.

269.0357143 = 269

So the final answer for the mass of iron(III) sulfate is 269 grams(don't forget the units).

Hope it helped!