Question

A certain alkaloid has $70.8$% carbon, $6.2$% hydrogen, $4.1$% nitrogen and the rest oxygen. What is its empirical formula?

• A. $C_{20}{H}_{21}N{O}_4$
• B. $C_{20}{H}_{20}N{O}_4$
• C. $C_{21}{H}_{20}N{O}_3$
• D. $C_{20}{H}_{19}N{O}_3$

Answer 1

The correct option is A $C_{20}{H}_{21}N{O}_4$

A certain alkaloid has $70.8$% carbon, $6.2$% hydrogen, $4.1$% nitrogen and $100-70.8-6.2-4.1=18.9$ oxygen
$100$ g of alkaloid has $70.8$ g carbon, $6.2$g hydrogen, $4.1$ g nitrogen and $18.9$ g oxygen.
The number of moles of $C$ present in $100$ g of alkaloid is$\frac{70.8}{12}=5.9mol$
The number of moles of $H$ present in $100$ g of alkaloid is $\frac{6.2}{1}=6.2mol$
The number of moles of $N$ present in $100$ g of alkaloid is $\frac{4.1}{14}=0.29mol$
The number of moles of $O$ present in $100$ g of alkaloid is $\frac{18.9}{16}=1.18mol$
The molar ratio is $C:H:N:O=5.9:6.2:0.29:1.18$
To get the whole number ratio, we divide the ratio with $0.29$.
The molar ratio is $C:H:N:O=20.3:21.4:1:4$. This ratio is approximately equal to $C:H:N:O=20:21:1:4$
Hence, the empirical formula is $C_{20}{H}_{21}N{O}_4$.