Answer: The empirical formula and the molecular formula of the organic compound is
and
respectively.
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of
= 21.71 g
Mass of
= 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, =
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, =
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 =
Moles of H=![\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20H%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20H%7D%7D%3D%20%5Cfrac%7B0.658g%7D%7B1g%2Fmole%7D%3D0.658moles)
Mass of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =![\frac{0.658}{0.493}=1](https://tex.z-dn.net/?f=%5Cfrac%7B0.658%7D%7B0.493%7D%3D1)
For O=
The ratio of C : H: O = 1: 1: 1
Hence the empirical formula is
.
empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29
Molecular mass = 104.1 g/mol
![n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%5Ctext%20%7BMolecular%20mass%7D%7D%7B%5Ctext%20%7BEquivalent%20mass%7D%7D%3D%5Cfrac%7B104.1%7D%7B29%7D%3D4)
Thus molecular formula = ![n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4](https://tex.z-dn.net/?f=n%5Ctimes%20%7B%5Ctext%20%7BEmpirical%20formula%7D%7D%3D4%5Ctimes%20CHO%3DC_4H_4O_4)