A compound with haemoglobin like structure contains one atom of iron per molecule. If it contains 4.6 % iron, its approximate molecular mass is

Question

A compound with haemoglobin like structure contains one atom of iron per molecule. If it contains 4.6 % iron, its approximate molecular mass is (A) 100 g mol-1 (B) 1200 g mol-1 (C) 1600 g mol-1 (D) 1400 g mol-1

Tardigrade Admin · Accepted Answer

If 4.6 of Fe is present mass of compound =100 g If 56 a.m.u. of Fe is present mass of compound =(100/46) ×56 a.m.u =1217.39 a.m.u. Molar mass of compound =1217.39 g mol-1

Q. A compound with haemoglobin like structure contains one atom of iron per molecule. If it contains 4.6 % iron, its approximate molecular mass is

100 g $m o l^{- 1}$

1200 g $m o l^{- 1}$

1600 g $m o l^{- 1}$

1400 g $m o l^{- 1}$

If $4.6$ of $F e$ is present mass of compound $= 100 g$
If $56$ a.m.u. of $F e$ is present mass of compound
$= \frac{100}{46} \times 56$ a.m.u
$= 1217.39$ a.m.u.
Molar mass of compound $= 1217.39 g m o l^{- 1}$

Q. A compound with haemoglobin like structure contains one atom of iron per molecule. If it contains 4.6 % iron, its approximate molecular mass is

100 g mol−1

1200 g mol−1

1600 g mol−1

1400 g mol−1

If 4.6 of Fe is present mass of compound =100g If 56 a.m.u. of Fe is present mass of compound =46100​×56 a.m.u =1217.39 a.m.u. Molar mass of compound =1217.39gmol−1

100 g $m o l^{- 1}$

1200 g $m o l^{- 1}$

1600 g $m o l^{- 1}$

1400 g $m o l^{- 1}$

If $4.6$ of $F e$ is present mass of compound $= 100 g$
If $56$ a.m.u. of $F e$ is present mass of compound
$= \frac{100}{46} \times 56$ a.m.u
$= 1217.39$ a.m.u.
Molar mass of compound $= 1217.39 g m o l^{- 1}$