Answer: The empirical formula for the given compound is CH
Explanation:
The chemical equation for the combustion of hydrocarbon follows:
[tex]C_xH_y+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', and 'y' are the subscripts of Carbon and hydrogen respectively.
We are given:
Mass of [tex]CO_2=6.785g[/tex]
Mass of [tex]H_2O=1.389g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44 g of carbon dioxide, 12 g of carbon is contained.
So, in 21.33 g of carbon dioxide, [tex]\frac{12}{44}\times 6.785=1.853g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18 g of water, 2 g of hydrogen is contained.
So, in 4.366 g of water, [tex]\frac{2}{18}\times 1.389=0.154g[/tex] of hydrogen will be contained.
To formulate the empirical formula, we need to follow some steps:
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.853g}{12g/mole}=0.154moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.154g}{1g/mole}=0.154moles[/tex]
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.154 moles.
For Carbon = [tex]\frac{0.154}{0.154}=1[/tex]
For Hydrogen = [tex]\frac{0.154}{0.154}=1[/tex]
The ratio of C : H = 1 : 1
Hence, the empirical formula for the given compound is CH