Respuesta :
Answer: The empirical formula for the given compound is [tex]C_2H_4O[/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=8.59g[/tex]
Mass of [tex]H_2O=3.52g[/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 8.59 g of carbon dioxide, [tex]\frac{12}{44}\times 8.59=2.34g[/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 3.52 g of water, [tex]\frac{2}{18}\times 3.52=0.391g[/tex] of hydrogen will be contained.
For calculating the mass of oxygen:
Mass of oxygen in the compound = [tex](4.30g)-[(2.34g)+(0.391g)]=1.57g[/tex]
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{2.34g}{12g/mole}=0.195moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.391g}{1g/mole}=0.391moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.57g}{16g/mole}=0.0981moles[/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]0.0981[/tex] moles.
For Carbon = [tex]\frac{0.195}{0.0981}=1.98\approx 2[/tex]
For Hydrogen = [tex]\frac{0.391}{0.0981}=3.98\approx 4[/tex]
For Oxygen = [tex]\frac{0.0981}{0.0981}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 4 : 1
Hence, the empirical formula for the given compound is [tex]C_2H_4O_1=C_2H_4O[/tex]
