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6. Explain how the collection of 159 g of Fe,o, from the combustion of 4.824x10* g of hemoglobin can be used to determine the
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Answer #1

Moles of Fe2O3 = weight of Fe2O3/molecular weight of Fe2O3 = 159 g/(159.69 g/mol) = 0.995679 mol

It's a fact that hemoglobin contains 4 iron atoms, i.e. one mole of hemoglobin on combustion produces two moles of Fe2O3.

Hence, the moles of hemoglobin used in the combustion reaction = (0.995679/2) mol = 0.4978395 mol

Therefore, 0.49784 mol = 48240 g/molecular weight of hemoglobin

i.e. The molecular weight of hemoglobin = 48240 g/0.4978395 mol = 96898.7 g/mol

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