Respuesta :
The mass sample of nicotine combusted = 2.625 mg (given)
Mass of [tex]CO_2[/tex] produced = 7.1210 mg (given)
Mass of [tex]H_2O[/tex] produced = 2.042 mg (given)
Molar mass of [tex]CO_2[/tex] = 44 g/mol
Molar mass of [tex]H_2O[/tex] = 18 g/mol
Percentage of Carbon = [tex]\frac{12 g}{44 g/mol}\times \frac{7.120 mg of CO_2}{2.625 mg sample}\times 100 = 74.04[/tex]%
Percentage of hydrogen = [tex]\frac{2 g}{18 g/mol}\times \frac{2.04 mg of H_2O}{2.625 mg sample}\times 100 = 8.62[/tex]%
Now, for percentage of nitrogen = [tex]100 - (74.04+8.62) = 17.34[/tex]%
Calculating the moles of each element:
[tex]Number of moles = \frac{given mass}{Molar mass}[/tex]
- For [tex]C[/tex]
[tex]\frac{74.04 g}{12 g/mol} = 6.17 mol[/tex]
- For [tex]H[/tex]
[tex]\frac{8.62 g}{1 g/mol} = 8.62 mol[/tex]
- For [tex]N[/tex]
[tex]\frac{17.34 g}{14 g/mol} = 1.24 mol[/tex]
Dividing with the smallest mole value to calculate the molar ratio of each element:
[tex]C_{\frac{6.17}{1.24}} = C_{4.9} = C_5[/tex]
[tex]H_{\frac{8.62}{1.24}} = H_{6.9} = H_7[/tex]
[tex]N_{\frac{1.24}{1.24}} = N_1[/tex]
Hence, the empirical formula for nicotine is [tex]C_5H_7N[/tex].
