Respuesta :
By stoichiometry and assume that:
CxH2xOy + zO2 -> xCO2 + xH2O
CO2: 9.48/44 = 0.215 mmol
H2O: 3.87/18 = 0.215 mmol
mass of C = 0.215 * 12 = 2.58 mg
mass of H = 0.215 * 2 * 1 = 0.43 mg
mass of O in ethylbutyrate = 4.17 - 2.58 - 0.43 = 1.11 mg
So C/O = 2.58/1.11 ≈ 3
Thus we have C3H6O
Answer: The empirical formula for the given compound is [tex]C_3H_6O[/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=9.48mg=9.48\times 10^{-3}g[/tex]
Mass of [tex]H_2O=3.87mg=3.87\times 10^{-3}g[/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]9.48\times 10^{-3}g[/tex] of carbon dioxide, [tex]\frac{12}{44}\times 9.48\times 10^{-3}=2.58\times 10^{-3}g[/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]3.87\times 10^{-3}g[/tex] of water, [tex]\frac{2}{18}\times 3.87\times 10^{-3}=4.30\times 10^{-4}g[/tex] of hydrogen will be contained.
For calculating the mass of oxygen:
Mass of oxygen in the compound = [tex](4.17\times 10^{-3})-[(2.58\times 10^{-3})+(4.30\times 10^{-4})]=1.16\times 10^{-3}g[/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{2.58\times 10^{-3}g}{12g/mole}=2.15\times 10^{-4}moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{4.30\times 10^{-4}g}{1g/mole}=4.30\times 10^{-4}moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.16\times 10^{-3}g}{16g/mole}=7.25\times 10^{-5}moles[/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]7.25\times 10^{-5}[/tex] moles.
For Carbon = [tex]\frac{2.15\times 10^{-4}}{7.25\times 10^{-5}}=2.96\approx 3[/tex]
For Hydrogen = [tex]\frac{4.30\times 10^{-4}}{7.25\times 10^{-5}}=5.93\approx 6[/tex]
For Oxygen = [tex]\frac{7.25\times 10^{-5}}{7.25\times 10^{-5}}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 3 : 6 : 1
Hence, the empirical formula for the given compound is [tex]C_3H_6O_1=C_3H_6O[/tex]
