Question

Calculate the concentration of Fe in an iron (II) solution in units of mg/L.

Solution is FeH₁₄O₁₁S, 0.10g of FeH₁₄O₁₁S is used and dissolved in 100ml of distilled water. FeH₁₄O₁₁S has a molecular weight of 278.006g/mol. Fe has a molecular weight of 55.85g/mol.

1. What is concentration of Iron in this 100ml solution in mg/L?

This solution is then diluted, this is done by using a pipette to draw 10ml from the iron (II) solution and is then added to a seperate 100mL flask. This seperate 100mL flask is then filled to 100mL with distilled water.

2. What is the concentration of Iron in this diluted 100mL solution in mg/L?

Hope someone can help me with an answer soon, thank you!

Answer 1

Ans. #1. Moles of FeH₁₄O₁₁S = Mass / Molar mass

                                                = 0.10 g / (278.006 g/ mol)

                                                = 3.597045 x 10^-4 mol

# 1 mol FeH₁₄O₁₁S consists of 1 mol Fe-atom.

So,

Moles of Fe-atom = Moles of Moles of FeH₁₄O₁₁S

Hence,

            Moles of Fe-atom = 3.597045 x 10^-4 mol

            Mass of Fe in sample taken = Moles of Fe x its Molar mass

                                                = 3.597045 x 10^-4 mol x (55.85 g/ mol)

                                                = 0.02009 g

                                                = 20.09 mg

# Given, volume of solution = 100.0 mL = 0.100 L

Now,

            [Fe] = Mass of Fe in mg / Volume of solution in liters

                        = 20.09 mg/ 0.100 L

                        = 200.9 mg/ L

#2. Using       C1V1 (stock solution) = C2V2 (diluted solution)

                        Or, 200.9 mg L^-1 x 10.0 mL = C2 x 100.0 mL

                        Or, C2 = (200.9 mg L^-1 x 10.0 mL) / 100.0 mL

                        Hence, C2 = 20.09 mg L^-1

Therefore, [Fe] in diluted solution = 20.09 mg / L