Respuesta :
Answer: The empirical formula and the molecular formula of the organic compound is [tex]CHO[/tex] and [tex]C_4H_4O_4[/tex] respectively.
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[/tex] = 21.71 g
Mass of [tex]H_2O[/tex]= 5.926 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 21.71 g of carbon dioxide, =[tex]\frac{12}{44}\times 21.71=5.921g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 5.926 g of water, =[tex]\frac{2}{18}\times 5.926=0.658g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g
Mass of C = 5.921 g
Mass of H = 0.658 g
Mass of O = 10.53 g
Step 1 : convert given masses into moles.
Moles of C =[tex] \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles[/tex]
Moles of H=[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles[/tex]
Mass of O=[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =[tex]\frac{0.493}{0.493}=1[/tex]
For H =[tex]\frac{0.658}{0.493}=1[/tex]
For O=[tex]\frac{0.658}{0.493}=1[/tex]
The ratio of C : H: O = 1: 1: 1
Hence the empirical formula is [tex]CHO[/tex].
empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29
Molecular mass = 104.1 g/mol
[tex]n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4[/tex]
Thus molecular formula = [tex]n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4[/tex]
