Question

The molar mass of the reagent, Fe(NH4)2(SO4)2.6H2O is 392.14 g/mol. Now suppose the measured mass of Fe(NH4)2(SO4)2.6H2O during the practical was 0.0218 g and this was dissolved in 50ml of deionized water that contains 1 to 2ml of concentrated sulphuric acid and transfer quantitatively to a 250ml volumetric flask using a funnel and a glass rod and dilute to the mark. Determine the iron concentration in the standard solution in ppm (mg L-1) units, showing full details of your working.

Answer 1

Convert 0.218 g of the compound to g Fe. That is done by 0.218 x (atomic [mass Fe/molar mass Fe(NH4)2(SO4)2.6H2O]*100 = ?

Convert g Fe to mg = ?

Convert 250 mL to L.

Then mg Fe/L = your answer.

Answer 2

DRBob222 what do you mean when you say That is done by 0.218 x (atomic [mass Fe/molar mass Fe(NH4)2(SO4)2.6H2O]*100 = ?
Could you please explain further?

Answer 3

OK, but first I made a typo. That is supposed to be 0.0218 g.

You are weighing Fe(NH4)2(SO4)2.6H2O but you don't want to calculate mg/L of that stuff, you want to calculate mg/L of Fe. Not all of that stuff is Fe.
What fraction is Fe? That is [atomic mass Fe/molar mass Fe(NH4)2(SO4)2.6H2O] = fraction that is Fe.
So 0.0218 x fraction that is Fe = grams Fe in the mass of Fe(NH4)2(SO4)2.

Answer 4

The mass of the compound is 0.0218g and if I calculate the mass of Fe I get a mass of 0.31g which is more than the mass of the whole compound.

Answer 5

Thank you for all your help.

Answer 6

I was gone for awhile. I think you just misplaced the decimal.

fraction = atomic mass Fe/molar mass Fe(NH4)2(SO4)2 = (55.85/392.14) = 0.142
Then mass Fe = mass Fe(NH4)2*(SO4)2 x fraction = 0.0218 x 0.142 = 0.00318 g Fe which is not larger than 0.0218.
Convert that to mg which is 3.18 mg and divide by L (0.250) which comes out to be 3.18/0.25 = ?