Respuesta :
Answer: The empirical formula is [tex]C_3H_3O[/tex].
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2[/tex] = 12.24 g
Mass of [tex]H_2O[/tex] = 2.505 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 12.24 g of carbon dioxide, =[tex]\frac{12}{44}\times 12.24=3.338g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 2.505 g of water, =[tex]\frac{2}{18}\times 2.505=0.278g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g
Step 1 : convert given masses into moles.
Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{0.278}{0.104}=3[/tex]
For H =[tex]\frac{0.278}{0.104}=3[/tex]
For O =[tex]\frac{0.104}{0.104}=1[/tex]
The ratio of C : H : O = 3: 3: 1
Hence the empirical formula is [tex]C_3H_3O[/tex].
