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The unknowns are a mixture of ferrous ammonium sulfate and ammonium sulfate. calculate the amount of ferrous ammonium sulfate and the amount of ammonium sulfate needed to prepare a sample that is 10% fe with a total mass of 0.2 g

User AaronBaker
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Answer: -

Molar mass of ferrous ammonium sulfate =392 g / mol

Mass of iron present in ferrous ammonium sulfate = 56 g/mol

Total mass required = 0.2 g

Total mass of iron Fe required = 10% of 0.2 g

=
(10)/(100)x0.2

= 0.02 g

Only ferrous ammonium sulfate has iron. So all iron must be present only as ferrous ammonium sulfate.

56 g of iron is present in 392 g of ferrous ammonium sulfate

0.02 g of iron is present in
(392)/(56) x 0.02

= 0.14 g

Mass of ammonium sulfate = 0.2 - 0.14 = 0.06 g

Thus the required composition must be 0.06 g ammonium sulfate and 0.14 g ferrous ammonium sulfate.

User Bealer
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