Question

Calculate the number of atoms of carbon and oxygen in 72.5 gram isopropanol.

Answer 1

Given that

$Isopropanol\left(C_3{H}_{7}OH\right)=72.5g$

So, molecular mass of $C_3{H}_{7}OH=36+7+16+1=60$

In $72.5gis\frac{72.5}{60}=1.208moles$

Now,

$1moleof{C}_3{H}_{7}OH$ Contains$3N(N=6.023\times {10}^{23})$ atoms of Carbon

1.208 moles contains $1.208\times 3Natoms=3.624Natoms$ of Carbon

So,

$=3.624\times 6.023\times {10}^{23}=2.183\times {10}^{24}atoms$ Of Carbon

Similarly,

Number of Oxygen atoms$=1.208N$

$\Rightarrow 1.208\times 6.023\times {10}^{23}$

$\Rightarrow 7.276\times {10}^{23}atoms$

Hence,

Number of Carbon $=2.183\times {10}^{24}atoms$

Number of Oxygen $=7.276\times {10}^{23}atoms$

This is the required solution.