Question

Calculate the number of atoms of $C$ and $O$ in 72.5 kg isopropanol, $C_2{H}_{7}OH$(molar mass 60).

Answer 1

Isopropanol $C_3{H}_{7}OH=60g/mol$

$1$ mole contains $3$ moles of carbon & $1$ mole of oxygen

$\therefore 60g$ of isopropanol contains $3$ moles of carbon & $1$

mole of oxygen

$\therefore$ In $72.5kg$ of isopropanol,

Moles of $C=\frac{72.5\times {10}^3\times 3}{60}=3.625\times {10}^3$

$=3.625\times 6\times {10}^{23}\times {10}^3$ atoms

$=2.1\times {10}^{27}$ atoms

Moles of $O=\frac{72.5\times {10}^3\times 1}{60}$

$=1.208\times {10}^3=1.208\times 6\times 10623\times {10}^3$

$=7.248\times {10}^{26}$ atoms