Question

Haemoglobin contains \[0.25\% \] iron ($Fe$ ) by mass. The molar mass of haemoglobin is $89600g/mol$ . The number of $Fe$atoms present in each molecule of haemoglobin is _______.

Answer 1

Hint: The mass percent of any element in a compound is the amount of substance present in the compound or simply the ratio of the total mass of the element divided by the sum of masses of all the elements in the compound or mixture and the ratio multiplied by 100. The number of moles is the ratio of total number of atoms of the element and the Avogadro number.

Complete step by step answer:
As it has been given that haemoglobin contains \[0.25\% \]iron by mass. This means that in every $100g$ of haemoglobin, $0.25g$ of iron is present. Thus,
$100g$ haemoglobin contains = $0.25g$ of iron
$89600g/mol$ haemoglobin will contain = $\dfrac{{0.25}}{{100}} \times 89600 = 224g/mol$ of iron
As we have found that the given weight of iron is $224g/mol$ , we can find the number of moles of iron from the mathematical representation:
$n = \dfrac{w}{{G.A.W}}$
Where, $n = $ number of moles
$w = $ Given weight $ = 224g$
$G.A.W = $ Gram atomic weight of iron $ = 56g$
Substituting the values in the above equation, we have:
$n = \dfrac{{224}}{{56}}$ = 4 $moles$
Thus, in order to determine the number of atoms of iron in 4 $moles$ of it, we need to apply the relation between the number of moles and number of particles (atoms/ ions/ molecules).
$n = \dfrac{N}{{{N_A}}}$
Where, $n = $ number of moles = $4$
$N = $ Number of atoms = ?
${N_A} = $ Avogadro number = $6.023 \times {10^{23}}$ atoms
Thus, substituting the values, we have:
$4 = \dfrac{N}{{6.023 \times {{10}^{23}}}} \Rightarrow N = 24.092 \times {10^{23}}$ atoms of iron
Thus, the number of atoms in each molecule of haemoglobin is equal to $24.092 \times {10^{23}}$.

Note:
The relation between the number of moles of a substance, its given weight, its given volume in the solution and the number of particles that it is composed of is given by the following relation:
$n = \dfrac{w}{{{M_w}}} = \dfrac{N}{{{N_A}}} = \dfrac{V}{{22.4(l)}}$
Where, $w = $ given weight
${M_w} = $ gram atomic weight/ gram molecular weight
$N = $ Number of particles
${N_A} = $ Avogadro number
$V = $ Given volume of the substance