Respuesta :
Answer: The molecular of the compound is, [tex]C_2H_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=3.095g[/tex]
Mass of [tex]H_2O=1.902g[/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]3.095g[/tex] of carbon dioxide, [tex]\frac{12}{44}\times 3.095=0.844g[/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]1.902g[/tex] of water, [tex]\frac{2}{18}\times 1.092=0.121g[/tex] of hydrogen will be contained.
For calculating the mass of oxygen:
Mass of oxygen in the compound = [tex](1.621)-[(0.844)+(0.121)]=0.656g[/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{0.844g}{12g/mole}=0.0703moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles[/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]0.041[/tex] moles.
For Carbon = [tex]\frac{0.0703}{0.041}=1.71\approx 2[/tex]
For Hydrogen = [tex]\frac{0.121}{0.041}=2.95\approx 3[/tex]
For Oxygen = [tex]\frac{0.041}{0.041}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 3 : 1
Hence, the empirical formula for the given compound is [tex]C_2H_3O_1=C_2H_3O[/tex]
The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]
[tex]n=\frac{46.06}{43}=1[/tex]
Molecular formula = [tex](C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O[/tex]
Therefore, the molecular of the compound is, [tex]C_2H_3O[/tex]
