Quinine,${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$, is a component of tonic water and bitter lemon.
a. How many moles of $\mathrm{H}$are in$1.5$ moles of quinine?
b. How many moles of $\mathrm{C}$ are in$5.0$ moles of quinine?
c. How many moles of $\mathrm{N}$are in$0.020$ mole of quinine?

We need to find moles of$\mathrm{H}$in$1.5$moles of quinine

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$= localid="1652692050146" $24$moles of $\mathrm{H}$

Conversion Factor=localid="1652857636012" $\frac{24moles\mathrm{H}}{1mole{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}$

Moles of $\mathrm{H}$in $1.5$moles of quinine = $$\begin{gathered} \frac{24mole\mathrm{H}}{1\mathrm{mole}{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}\times 1.5 \\ =36 \end{gathered}$$

Hence,$36$moles of$\mathrm{H}$in$1.5$mole quinine.

We need to find moles of $\mathrm{C}$in $5.00$moles of quinine

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$= $20$moles of $\mathrm{C}$

Conversion Factor=$\frac{20mole\mathrm{C}}{1mole{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}$

Moles of $\mathrm{C}$in $5.00$moles of quinine =$$\begin{gathered} \frac{20mole\mathrm{C}}{1mole{C}_{20}{H}_{24}{N}_2{O}_2}\times 5mole{C}_{20}{H}_{24}{N}_2{O}_2 \\ =100mole\mathrm{C} \end{gathered}$$

Hence, $100$moles of $\mathrm{C}$in $5$mole quinine.

We need to find moles of $\mathrm{N}$in $0.020$moles of quinine

Using equalities and conversion factor using subscripts:

One mole of ${\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2$= $2$moles of $\mathrm{N}$

Conversion Factor=$\frac{2mole\mathrm{N}}{1\mathrm{mole}{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}$

Moles of $\mathrm{N}$ in $0.020$moles of quinine =$$\begin{gathered} \frac{2mole\mathrm{N}}{1\mathrm{mole}{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2}\times 0.020\mathrm{mole}{\mathrm{C}}_{20}{\mathrm{H}}_{24}{\mathrm{N}}_2{\mathrm{O}}_2 \\ =0.040mole\mathrm{N} \end{gathered}$$

Hence, $0.040$moles of $\mathrm{N}$in $0.020$mole quinine.