What is the ppm of standard Fe that is made by dissolving 0.1405g of reagent grade Fe(NH4)2(SO4)2 6H2O (FM 392.14 g/mol) in DI water in a 500mL volumetric flask containing 0.5mL of 98 mass% H2SO4
Parts Per Million (ppm) are equivalent to milligrams/liter = mg/L.
The % of Fe in Fe(NH4)2(SO4)2 6H2O is:
(100)(55.845 g)/(392.14g) = 14.24%
•Find 14.24% of 0.1405g of the iron containing sample to get the grams of Fe in it.
•Convert the grams of Fe to milligrams.
•Divide the milligrams of Fe by 0.500 L of solution.
The last step will give an answer in mg/L or ppm.