Respuesta :
Answer:
The empirical formula is [tex]C_5H_5O[/tex].
Explanation:
Given that, the combustion of 29.5 mg produced 80.2 mg CO₂ and 16.4 mg of H₂O.
Mass of C [tex]=80.1\times 10^{-3} g CO_2\times \frac{\textrm{1 mol C}O_2}{44 g}\times \frac{12 g CO_2}{\textrm{1 mol C}O_2}[/tex]
[tex]=0.022[/tex] g
Mass of H [tex]=16.4\times 10^{-3} g HO_2\times \frac{\textrm{1 mol }H_2O}{18 g}\times \frac{2 g H}{\textrm{1 mol }H_2O}[/tex]
[tex]=1.8 \times 10^{-3}[/tex] g
Mass of O = [tex]29.5\times 10^{-3}g-(0.022 g+1.8 \times 10^{-3} g)[/tex]
[tex]=5.68 \times 10^{-3}[/tex] g
Number of mole of C [tex]=\frac{Mass}{\textrm{ Atomic mass}}[/tex]
[tex]=\frac{0.022g}{12 g/mol}[/tex]
[tex]=1.83\times 10^{-3}[/tex] mol
Number of mole of H [tex]=\frac{Mass}{\textrm{ Atomic mass}}[/tex]
[tex]=\frac{1.8\times 10^{-3}g}{1g/mol}[/tex]
[tex]=1.8 \times 10^{-3}[/tex] mol
Number of mole O [tex]=\frac{Mass}{\textrm{ Atomic mass}}[/tex]
[tex]=\frac{5.68 \times 10^{-3}g}{16 g/mol}[/tex]
[tex]=0.355\times 10^{-3}[/tex] mol
Dividing by lowest integer
[tex]C:\frac{1.83\times 10^{-3}}{0.355\times 10^{-3}}=5[/tex]
[tex]H:\frac{1.8\times 10^{-3}}{0.355\times 10^{-3}}=5[/tex]
The empirical formula is [tex]C_5H_5O[/tex].
