Haemoglobin contains $0.33 %$ Fe by weight, the molecular weight of haemoglobin is approximately $672200$ , then calculate the number of Fe atoms present in the molecule of haemoglobin.

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Solution

Total weight $= 672200$

Percentage of $F e = 0.33 %$ molecular weight of haemoglobin
Weight of $F e$ present $= \frac{0.33}{100} \times 672200 = 2218.26$

Atomic weight of $F e = 55.84$ $g$

$∴$ No. of atoms of $F e = \frac{2218.26}{55.84} = 40$ atoms

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Haemoglobin contains 0.33% of iron by weight . The molecular weight of haemoglobin is approximately 67200 . The number of iron atoms ( at . wt. of Fe=56) present in one molecule of haemoglobin is

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Haemoglobin contains 0.33% of iron by weight . The molecular weight of haemoglobin is approximately 67200 . The number of iron atoms ( at . wt. of Fe=56) present in one molecule of haemoglobin is

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is 56) present in one molecule of haemoglobin is:

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Haemoglobin contains 0.33% Fe by weight, the molecular weight of haemoglobin is approximately 672200, then calculate the number of Fe atoms present in the molecule of haemoglobin.

Total weight =672200Percentage of Fe=0.33% molecular weight of haemoglobinWeight of Fe present =0.33100×672200=2218.26Atomic weight of Fe=55.84 g∴ No. of atoms of Fe=2218.2655.84=40 atoms

Haemoglobin contains $0.33 %$ Fe by weight, the molecular weight of haemoglobin is approximately $672200$ , then calculate the number of Fe atoms present in the molecule of haemoglobin.

Total weight $= 672200$

Percentage of $F e = 0.33 %$ molecular weight of haemoglobin
Weight of $F e$ present $= \frac{0.33}{100} \times 672200 = 2218.26$

Atomic weight of $F e = 55.84$ $g$

$∴$ No. of atoms of $F e = \frac{2218.26}{55.84} = 40$ atoms