Question

Hemoglobin contains 0.25% iron/weight the molecular mass of hemoglobin is 89600 calculate the no. Of helium atoms present per molecule of hemoglobin​

Answer 1

Correct Question:-

Haemoglobin contains 0.25% iron/weight the molecular mass of hemoglobin is 89600 calculate the no. of Iron (Fe) atoms present per molecule of hemoglobin$?$

Required Answer:-

Molecular mass of Haemoglobin = 89600

Percentage of Iron = 0.25% / weight. This means the total mass of Iron out of the total molecular mass of Haemoglobin is 0.25% of 89600.

GiveN mass of Iron:

= 0.25% × 89600

= 25/10000 × 89600

= 1/400 × 89600

= 224 gm of Iron.

Moles of Iron:

= Given Mass / Atomic mass

= 224 gm / 56 gm

= 4 moles.

Now here, 1 mole of Haemoglobin contains 4 moles of Iron (Fe). Then 1 molecule of Haemoglobin will have 4 atoms of Fe.

Hence:-

• No. of Iron atoms present per molecule of hemoglobin = 4

Answer 2

Answer:

Question?

Haemoglobin contains 0.25% iron/weight the molecular mass of hemoglobin is 89600 calculate the no. of Iron (Fe) atoms present per molecule of hemoglobin?

Answer :-

Molecular mass of hemoglobin = 89600

Percentage of Iron (Fe) = 0.25

Haemoglobin is 0.25% of 89600.

So, for this we first find the mass of iron

$\sf \implies \: 0.25\% \: of \: 89600$

$\sf \implies \: \dfrac{25}{10000} \times 89600$

$\sf \implies \: \dfrac{1}{4} \times 89600$

$\huge \bf Iron \: = 224 \: gm$

Now,

Finding mole of iron

$\sf \implies \: mole \: = \dfrac{ given \: mass}{atomic \: mass}$

$\sf \implies \: mole = \dfrac{224}{56}$

$\huge \bf \: \: = 4 \: mole$