Question

Hemoglobin is a protein having a molar mass of 64500g/mol and containing four atoms of iron(II). When 73.5g of a sample containing an unknown amount of hemoglobin is analyzed by precipitating all the iron(II) as iron(II) phosphate, 0.500g of precipitation is collected. What is percentage of hemoglobin contained in the sample?

Answer 1

Molar Mass of Iron (II) Phosphate i.e. Fe₃(PO₄)₂ = 357.48 g/mol

Moles of Fe₃(PO₄)₂ = Mass / Molar Mass = 0.500 g / (357.48 g/mol) = 0.0013987 mol

1 mol of Fe₃(PO₄)₂ contains 3 mol of Fe²⁺

Thus, mol of Fe²⁺ = 3 x 0.0013987 mol = 0.0041961 mol

1 mol Hemoglobin contains 4 mol of Fe²⁺

Thus, mol of Hemoglobin = Moles of Fe²⁺ / 4 = 0.0041961 mol / 4 =0.0010490 mol

Mass of Hemoglobin = Moles x Molar Mass = 0.0010490 mol x 64500 g/mol = 67.66 g

Mass of Sample = 73.5 g

Mass Percent of Hemoglobin = 67.66 g / 73.5 g x 100 = 92.1 %