Answer: The empirical formula is [tex]C_3H_3O_2[/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] = 0.8635
Mass of [tex]H_2O[/tex]= 0.1767 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 0.8635 g of carbon dioxide, =[tex]\frac{12}{44}\times 0.8635=0.2355g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 0.1767 g of water, =[tex]\frac{2}{18}\times 0.1767=0.0196g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (0.4647) - (0.2355+0.0196) = 0.2096 g
Mass of C = 0.2355 g
Mass of H = 0.0196 g
Step 1 : convert given masses into moles.
Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{0.2355g}{12g/mole}=0.0196moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0196g}{1g/mole}=0.0196moles[/tex]
Moles of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.2096g}{16g/mole}=0.0131moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{0.0196}{0.0131}=1.5[/tex]
For H =[tex]\frac{0.0196}{0.0131}=1.5[/tex]
For O = [tex]\frac{0.0131}{0.0131}=1[/tex]
The ratio of C : H: O = 1.5 : 1 .5 : 1
Converting to simple whole number ratio:
The ratio of C : H: O = 3 : 3: 2
Hence the empirical formula is [tex]C_3H_3O_2[/tex].
