Respuesta :
Answer: The empirical formula and the molecular formula of the hydrocarbon is [tex]C_2H_3[/tex] and [tex]C_4H_6[/tex] respectivley.
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', 'y' are the subscripts of Carbon, hydrogen
We are given:
Mass of [tex]CO_2[/tex] = 12.74 g
Mass of [tex]H_2O[/tex]= 3.913 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 12.74 g of carbon dioxide, =[tex]\frac{12}{44}\times 12.74=3.474g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 3.913 g of water, =[tex]\frac{2}{18}\times 3.913=0.435g[/tex] of hydrogen will be contained.
Mass of C = 3.474 g
Mass of H = 0.435 g
Step 1 : convert given masses into moles.
Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.474g}{12g/mole}=0.289moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.435g}{1g/mole}=0.435moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{0.289}{0.289}=1[/tex]
For H =[tex]\frac{0.435}{0.289}=1.5[/tex]
The ratio of C : H = 1: 1.5
The whole number ratio will be = 2: 3
Hence the empirical formula is [tex]C_2H_3[/tex].
The empirical weight of [tex]C_2H_3[/tex] = 2(12.01)+3(1.008)= 27.04 g.
The molecular weight = 54.09 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{54.09}{27.04}=2[/tex]
The molecular formula will be=[tex]2\times C_2H_3=C_4H_6[/tex]
