Question

How do you find the molecular formula of nicotine\[\left( {{C_x}{H_y}{N_z}} \right)\], when \[0.438\] grams of nicotine burns to make \[1.188g\] $C{O_2}$ and \[0.341g\] ${H_2}O$? The molecular weight of the nicotine is \[162.2g/mole\].

Answer 1

Hint: We need to know that the molecular formula specifies the numbers of atoms present in the molecule of a chemical substance. And the molecular formula is the same as that of the empirical formula. The molecular formula mainly contains the chemical symbols of the corresponding element and it has numeric subscripts to express the number of atoms present in a molecule. Therefore, the molecular formula will show the actual number of atoms present in a molecule.

Complete answer:
The molecular formula of the nicotine is equal to \[{C_{10}}{H_{14}}{N_2}\]. By using the percentage composition, we can find out the amount of carbon we would get from \[1.188g\] sample of \[C{O_2}\]. Every mole of \[C{O_2}\] contains one mole of C atom. The molecular weight of carbon is equal to \[12.011g/mol\] and molecular weight carbon dioxide is equal to \[44.01g/mol\]. Hence,
\[\dfrac{{1 \times 12.011g/mol}}{{44.01g/mol}} \times 100 = 27.29\% C\]
From this, we can say that every \[100g\] of \[C{O_2}\] consists of \[27.29g\] carbon. Therefore,
\[1.188gC{O_2} = \dfrac{{27.29gC{O_2}}}{{100gC{O_2}}} = 0.324gC\]
And every mole of water consists of two moles of hydrogen atom. Thus,
\[\dfrac{{2 \times 1.00794g/mol}}{{18.015g/mol}} \times 100 = 11.19\% H\]
Hence, \[0.341g\] ${H_2}O$ contains,
\[0.341g{H_2}O = \dfrac{{27.29gH}}{{100g{H_2}O}} = 0.0382gH\]
Therefore, \[{m_{sample}} = {m_N} + {m_H} + {m_C}\]
Substitute the values and rearranging the equation,
\[{m_N} = 0.438g - 0.324g - 0.0382g\]
On simplification we get,
\[{m_N} = 0.0758gN\]
By using the molar mass of hydrogen, carbon and nitrogen, we can find out the number of moles of each and that is,
\[For C:\dfrac{{0.324g}}{{12.0110\dfrac{g}{{mol}}}} = 0.02698 molesC\]
\[For H:\dfrac{{0.382g}}{{1.00784\dfrac{g}{{mol}}}} = 0.03790molesH\]
\[For N:\dfrac{{0.0758}}{{14.007\dfrac{g}{{mol}}}} = 0.005412molesN\]
Divide the number of moles with the smallest value to get,
\[For C:\dfrac{{0.0758moles}}{{0.005412moles}} = 4.985 \approx 5\]
\[For H:\dfrac{{0.03790moles}}{{0.005412moles}} = 7.003 \approx 7\]
\[For N:\dfrac{{0.005412moles}}{{0.005412moles}} = 1\]
Thus, the empirical formula of nicotine is \[{C_5}{H_7}{N_1}\]. And the molar mass of the empirical formula is equal to,
\[5 \times 12.011g/mol + 7 \times 1.00794g/mol + 1 \times 14.007g/mol = 81.12g/mol\]
Which means, \[81.12g/mol \times n = 162.2g/mol\]
Therefore, \[n = \dfrac{{162.2}}{{81.12}} = 1.995 \approx 2\]
Hence the molecular mass of the nicotine is equal to \[{C_{10}}{H_{14}}{N_2}\]

Note:
Nicotine is a chemical compound which is mainly used as an anxiolytic and stimulant. And the molecular mass of the nicotine is equal to \[{C_{10}}{H_{14}}{N_2}\] and the empirical formula is \[{C_5}{H_7}{N_1}\]. The components present in the nicotine are carbon, hydrogen and nitrogen. When the nicotine is burning, there is a formation of a definite amount of carbon dioxide and water.