Given :
Moles of Na : 1.06
Moles of C : 0.528
Moles of O : 1.59
To Find :
The empirical formula of the compound.
Solution :
Dividing moles of each atom with the smallest one i.e 0.528 .
So,
Na : 1.06/0.528 = 2.007 ≈ 2
C : 0.528/0.528 = 1
O : 1.59/0.528 = 3.011 ≈ 3
Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.
So, empirical formula is .
Hence, this is the required solution.
Explanation:
Haemoglobin consists of heme unit which is comprised of an <u></u> and porphyrin ring. The ring has four pyrrole molecules which are linked to the iron ion. In oxyhaemoglobin, the iron has coordinates with four nitrogen atoms and one to the F8 histidine residue and the sixth one to the oxygen. In deoxyhaemoglobin, the ion is displaced out of the ring by 0.4 Å.
The prosthetic group of hemoglobin and myoglobin is - <u>Heme</u>
The organic ring component of heme is - <u>Porphyrin</u>
Under normal conditions, the central atom of heme is - <u></u>
In <u>deoxyhemoglobin</u> , the central iron atom is displaced 0.4 Å out of the plane of the porphyrin ring system.
The central atom has <u>six</u> bonds: <u>four</u> to nitrogen atoms in the porphyrin, one to a <u>histidine</u> residue, and one to oxygen.