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Following are the possible isomers of secondary alcohol and ketones for six carbon molecules. In order to distinguish between sec. alcohol and ketone we can simply treat the unknown compound with acidified Potassium Dichromate (VI) in the presence of acid. If with treatment with unknown compound the colour of K2Cr2O7 (potassium dichromate VI) changes from orange to green then it is confirmed that the unknown compound is sec. alcohol, or if no change in colour is detected then ketone is confirmed. This is because ketone can not be further oxidized while, sec. alcohol can be oxidized to ketones as shown below,
Answer: The empirical formula is .
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 = 12.24 g
Mass of = 2.505 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 12.24 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 2.505 g of water, = of hydrogen will be contained.
Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H=
Moles of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For O =
The ratio of C : H : O = 3: 3: 1
Hence the empirical formula is .