Respuesta :
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:
Conversion factor: 1 g = 1000 mg
Mass of [tex]CO_2=6.32mg=0.00632g[/tex]
Mass of [tex]H_2O=2.58g=0.00258g[/tex]
Mass of compound = 2.78 mg = 0.00278 g
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 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.00632 g of carbon dioxide, [tex]\frac{12}{44}\times 0.00632=0.00172g[/tex] of carbon will be contained.
- For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 0.00258 g of water, [tex]\frac{2}{18}\times 0.00258=0.000286g[/tex] of hydrogen will be contained.
- Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g
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{0.00172g}{12g/mole}=1.43\times 10^{-4}moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\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]4.83\times 10^{-5}mol[/tex]
For Carbon = [tex]\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3[/tex]
For Hydrogen = [tex]\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6[/tex]
For Oxygen = [tex]\frac{4.83\times 10^{-5}}{4.83\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_{6}O_1=C_3H_6O[/tex]
