Question

Haemoglobin contains $0.33\% {\text{ Fe}}$ by weight, the molecular weight of haemoglobin is approximately 67200, then calculate the number of ${\text{Fe}}$ atoms present in the molecule of haemoglobin.

Answer 1

We are given that haemoglobin contains $0.33\% {\text{ Fe}}$ by weight. When any substance is present in a compound in percentage by weight, we consider the mass of the compound as $100{\text{ g}}$ and percentage as the mass of the substance.

Formula Used:
${\text{Number of atoms}} = \dfrac{{{\text{Mass of substance}}}}{{{\text{Molar mass of the substance}}}}$

Complete step by step answer:
We know that haemoglobin is a red coloured pigment which is present in the blood. Haemoglobin carries oxygen from the lungs to the rest of the body. Haemoglobin contains iron.
-We are given that haemoglobin contains $0.33\% {\text{ Fe}}$ by weight. This means that $100{\text{ g}}$ of haemoglobin contains $0.33{\text{ g}}$ of iron or $\left( {{\text{Fe}}} \right)$. And $1{\text{ g}}$ of haemoglobin contains $0.0033{\text{ g}}$ of iron $\left( {{\text{Fe}}} \right)$.
We are given that the molecular weight of haemoglobin is approximately 67200.

-Now, calculate the amount of iron present in $67200{\text{ g}}$ of haemoglobin. Thus,
Amount of iron $= 67200{\text{ g haemoglobin}} \times \dfrac{{{\text{0}}{\text{.0033 g Fe}}}}{{1{\text{ g haemoglobin}}}}$
Amount of iron $= 221.76{\text{ g Fe}}$

Thus, the amount of iron present in $67200{\text{ g}}$ of haemoglobin is $221.76{\text{ g}}$.
Thus, one molecule of haemoglobin contains $221.76{\text{ g Fe}}$.
-Calculate the number of atoms of ${\text{Fe}}$ present in one molecule of haemoglobin as follows:
${\text{Number of atoms}} = \dfrac{{{\text{Mass of substance}}}}{{{\text{Molar mass of the substance}}}}$
Substitute $221.76{\text{ g}}$ for the mass of ${\text{Fe}}$, $56{\text{ g/mol}}$ for the molar mass of ${\text{Fe}}$. Thus,
${\text{Number of atoms}} = \dfrac{{221.76{\text{ g}}}}{{56{\text{ g/mol}}}}$
${\text{Number of atoms}} = 3.96 \approx 4$

Thus, the number of ${\text{Fe}}$ atoms present in the molecule of haemoglobin are 4.

Note: The amount of ${\text{Fe}}$ can also be calculated by multiplying the percentage by weight of ${\text{Fe}}$ with the molecular weight of haemoglobin. Thus,
Amount of iron $= 0.33\% \times 67200 = \dfrac{{0.33}}{{100}} \times 67200 = 221.76{\text{ g}}$