What is the mass in grams of a sample of Fe2(SO4)3 that contains 3.59x10^23 sulfate ions, SO4 with negative two at the top ?

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Every mole of any compound has a fixed mass in grams and a fixed number of molecules, namely 6.02x10^23. What we need to do here is:

find out how many moles of the compound we have

find out how many grams in a mole of the compound

multiply moles times grams per mole.

Each molecule of Fe2(SO4)3 has 3 sulfate ions SO4. So a mole of iron sulfate would contain 3 x 6.02x10^23=18.06 x 10^23 sulfate ions. We have 3.59 x 10^23 sulfate ions so the number of moles we have is 3.59/18.06 = 0.19878 moles of iron sulfate.

each mole of iron sulfate has an atomic weight which is the sum of the stomic weights of all the atoms that make up a molecule.

at.weight = 2Fe + 3S +3x4 O= 2 (55.847) + 3 (32.06) + 12 (15.9994) = 399.8668. The mass in grams of one mole of iron sulfate is 399.8668 g

thhe mass of .19878 moles = .19878 X 399.8668g = 79.486 g

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