Respuesta :
Answer: The empirical formula for the given compound is [tex]CH_2[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen 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=1.80g[/tex]
Mass of [tex]H_2O=0.738g[/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 [tex]1.80g[/tex] of carbon dioxide, [tex]\frac{12}{44}\times 1.80=0.491g[/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 [tex]0.738g[/tex] of water, [tex]\frac{2}{18}\times 0.738=0.082g[/tex] of hydrogen will be contained.
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.491g}{12g/mole}=0.0409moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.082g}{1g/mole}=0.082moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is [tex]0.0409[/tex] moles.
For Carbon = [tex]\frac{0.0409}{0.0409}=1[/tex]
For Hydrogen = [tex]\frac{0.082}{0.0409}=2.00\approx 2[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 2
Hence, the empirical formula for the given compound is [tex]C_1H_2=CH_2[/tex]
