An organic compound on analysis gave the following results C=54.5%, O=36.4%, H=91%. The Empirical formula of the compound is
I assume you made a typo and that is 9.1% H.
Take a 100 g sample to give you 54.5 g C, 36.4 g O, 9.1 g H
mols C = 54.5/12 = approx 4.5
mols O = 36.4/16 = approx 2.3
mols H = 9.1/1 = approx 9.1
Divide all by the smallest to get
4.5/2.3 = about 2
2.3/2.3 = 1
9.1/2.3 = about 4
ratio is C2H4O
To find the empirical formula of the compound, we need to determine the ratios of the different elements present in the compound.
First, assume that we have 100 grams of the compound. This assumption allows us to convert the percentages into grams.
Given:
C = 54.5%
O = 36.4%
H = 91%
The mass of carbon (C) in 100 grams of the compound would be:
(54.5/100) x 100 = 54.5 grams
The mass of oxygen (O) in 100 grams of the compound would be:
(36.4/100) x 100 = 36.4 grams
The mass of hydrogen (H) in 100 grams of the compound would be:
(91/100) x 100 = 91 grams
Next, we need to convert these masses into moles. To do this, we divide the mass of each element by its molar mass. The molar mass for C is approximately 12 g/mol, for O it is approximately 16 g/mol, and for H it is approximately 1 g/mol.
The number of moles of carbon (C) would be:
54.5 grams / 12 g/mol = 4.54 moles
The number of moles of oxygen (O) would be:
36.4 grams / 16 g/mol = 2.275 moles
The number of moles of hydrogen (H) would be:
91 grams / 1 g/mol = 91 moles
Now, we need to find the simplest whole number ratio of the moles of each element. To do this, we can divide the moles by the smallest number of moles obtained.
In this case, the smallest number of moles is 2.275 moles (from the oxygen calculation).
The ratio of moles becomes:
Carbon: 4.54 moles / 2.275 moles = 2
Oxygen: 2.275 moles / 2.275 moles = 1
Hydrogen: 91 moles / 2.275 moles = 40
Therefore, the empirical formula of the compound is C2O(H2O)40.