Question

An analysis of nicotine (mw = 162 g/mol) gives 74.0% carbon, 8.65% hydrogen, and 17.3% nitrogen. what is the molecular formula of nicotine

Answer 1

C=74÷12 =6.1 6.1÷1.23=4.95

H=8.65÷1=8.65 8.65÷1.23=7.03

N=17.3÷14=1.23 1.23÷1.23=1
so emperical formula C5H7N

E.F weight=81
M.F weight =162
MF=C10H14N2

Answer 2

Answer: The molecular formula of nicotine is $C_{10}H_{14}N_2$

Explanation:

To know the molecular formula of the compound, we will follow some steps:

Step 1: Converting all these percentages into mass.

We take the total mass of the compound to be 100 grams, so, the percentages given for each element becomes its mass.

Mass of Carbon = $\frac{74}{100}\times 100g=74 g$

Mass of Hydrogen = $\frac{8.65}{100}\times 100g=8.65g$

Mass of Nitrogen = $\frac{17.3}{100}\times 100g=17.3g$

Molar mass of Carbon = 12 g/mole

Molar mass of Hydrogen = 1 g/mole

Molar mass of Nitrogen = 14 g/mol

Step 2: Converting the given masses into their respective moles.

Moles of Carbon = $\frac{\text{Given mass of C}}{\text{Molar mass of C}}= \frac{74g}{12g/mole}=6.16moles$

Moles of Hydrogen = $\frac{\text{Given mass of H}}{\text{Molar mass of H}}= \frac{8.65g}{1g/mole}=8.65moles$

Moles of Nitrogen = $\frac{\text{Given mass of N}}{\text{Molar mass of N}}= \frac{17.3g}{14g/mole}=1.23moles$

Step 3: Now, calculating mole ratio, we divide each number of moles by the smallest number of moles calculated.

For Carbon = $\frac{6.16}{1.23}=5$  

For Hydrogen = $\frac{8.65}{1.23}=7$  

For Nitrogen = $\frac{1.23}{1.23}=1$

The ratio of C : H : N = 5 : 7 : 1

Hence the empirical formula is $C_{5}H_{7}N_1$  

The empirical weight of$C_{5}H_{7}N_1$ = 5(12) + 7(1) + 1(14)= 81 g

The molecular weight = 162 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight of metal}}{\text{Equivalent of metal}}=\frac{162}{81}=2$

Thus Molecular formula of nicotine is =$2\times C_5H_7N_1=C_{10}H_{14}N_2$