Question

A compound has haemoglobin like structure. It has one Fe. It contains $4.5\%$ of Fe. Determine its molar mass.

Answer 1

We will take the mass of the compound to be 100 g. It contains 1 atom of Fe and we know the mass of the Fe atom is 56 g/mol. The percentage of Fe in the compound is given as $4.6\%$. Let the molar mass of the compound be ‘M’, so, the percentage of Fe will be equal to its atomic mass by molar mass multiplied by 100.
Complete step by step answer:
Suppose 100 g is the mass of the compound and let M be the molecular mass of the compound having Haemoglobin like structure which contains one Fe atom.
Percentage of Fe in the compound $= 4.5\%$ (given in the question)
Number of moles of the compound $= \dfrac{{100}}{{molar\,mass}} = \dfrac{{100}}{M}$
Atomic mass of $Fe = 56$
So, if 56 amu of Fe is present in 100 g of the compound
Now, $\dfrac{{100}}{m} = 1 \times \dfrac{{4.5}}{{56}}$
$\Rightarrow$ $M = \dfrac{{100 \times 56}}{{4.5}} = 1244.4gm$

Hence, the molar mass of the compound is 1244.4 gm

Note: If mass of Fe is 4.5 g, the molecular mass of the compound is 100 g and if the mass of Fe is 56 g, then the molecular mass of the compound will be equal to $\dfrac{{100}}{{4.5}} \times 56$.
The average quantity of iron in the human body is about 4.5 grams of which almost $65\%$ is in the form of the haemoglobin which helps to transport the molecular oxygen from the lungs throughout the body and about $1\%$ in the various enzymes that control the intracellular oxidation and most of the rest are stored in the body.