# If one molecule of hemoglobin contains 4 iron atoms, and they constitute 0.335% by mass of hemoglobin, what is the molar mass of hemoglobin?

Apr 25, 2016

$66686.57 g m o {l}^{-} 1$, rounded to two decimal places.

#### Explanation:

Let molecular mass of heamoglobin molecule be $= M$

One molecule of heamoglobin contains $4 \text{Fe}$ atoms of atomic mass

$= 55.85 \text{ each}$,

Total mass of iron in one molecule of heamoglobin

$= 4 \times 55.85 = 223.4$

To calculate the percentage of iron:

$M$ amu of heamoglobin has iron$= 223.4$

$100$ amu of heamoglobin has iron$= \frac{223.4}{M} \times 100$

Equating to the given value,

$\frac{223.4}{M} \times 100 = 0.335$

Solving for $M$,

$M = \frac{223.4}{0.335} \times 100$

$= 66686.57 g m o {l}^{-} 1$, rounded to two decimal places.
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At some places the iron percentage in heomoglobin is given as 0.3335%. Using the published average atomic mass of iron as $55.847$,

$M = 66982.9085 g m o {l}^{-} 1$