Question

The percentage of Se in peroxidase anhydrous enzyme is 0.5% by weight (atomic weight=78.4). Then minimum molecular weight of peroxidase anhydrous enzyme is:
a.) $1.568\times {10}^4$
b.) $1.568\times {10}^3$
c.) $15.68$
d.) $3.136\times {10}^4$

Answer 1

Hint: In peroxidase anhydrous enzyme, 0.5% Se is present by weight, it means that 0.5 g Se is present in 100 g of enzyme. Minimum molecular formula is defined as the molecular weight divided by the number of atoms of an element present in a molecule.

Complete answer:
As we know, the minimum molecular formula is the molecular weight which is divided by the number of atoms of an element present in a molecule.
$Minimummolecularweight=\frac{Molecularweight}{No.ofatomsofanelement}$

In this question, we will be able to find the minimum molecular weight if we consider that in 1 molecule of the peroxidase anhydrous enzyme only 1 Se atom is present. Therefore, the percentage of Se in peroxidase anhydrous enzyme can be found as:

$PercentageofSe=\frac{AtomicweightofSe}{Min.molecularweight(m)}\times 100$
$⟹m=\frac{Atomicweight}{PercentageofSe}\times 100$

Now, given that Atomic weight of Se = 78.4 and 5 of Se = 0.5. Substituting these values in the above equation, we get,
$m=\frac{78.4}{0.5}\times 100$
$⟹m=156.8\times 100=1.568\times {10}^{4}g/mol$

Therefore, the minimum molecular weight of Se in peroxidase anhydrous enzymes is $1.586\times {10}^4$.
So, the correct answer is “Option A”.

Note: Don't confuse between the molecular weight and the minimum molecular weight. Molecular weight is the sum of the atomic weights of individual atoms in a molecule. Whereas, minimum molecular weight is the molecular weight divided by the number of atoms of an element present in a molecule.