Solveeit Logo

Question

Question: The elemental mass percent composition of succinic acid is \( C \) \( 40.68\% \) , \( H \) \( 5.12\%...

The elemental mass percent composition of succinic acid is CC 40.68%40.68\% , HH 5.12%5.12\% and OO 54.19%54.19\% . What is the empirical formula of succinic acid?

Explanation

Solution

The molecular formula gives the exact number of each different atom present in a molecule, whereas the empirical formula gives the simplest ratio of the number of various atoms present. It is an empirical formula if the formula is simplified.

Complete answer:
In order to be able to determine the empirical formula of succinic acid, we must first determine what the mole ratio between the atoms that make up the molecule is.
The mass percent composition of carbon, hydrogen, and oxygen are given
From that we can know the mass of each element present,
Mass of C=40.68gC = 40.68g
Mass of H=5.12gH = 5.12g
Mass of O=54.19gO = 54.19g
Now we need to find the number of moles of each element,
Number of moles of C=40.6812=3.39C = \dfrac{{40.68}}{{12}} = 3.39
Number of moles of H=5.121=5.12H = \dfrac{{5.12}}{1} = 5.12
Number of moles of O=54.1916=3.38O = \dfrac{{54.19}}{{16}} = 3.38
The mole ratio these elements in the compound can be determined by dividing these values by the smallest value
Mole Ratio of C=3.393.38=1.002C = \dfrac{{3.39}}{{3.38}} = 1.002
Mole ratio of H=5.123.38=1.51H = \dfrac{{5.12}}{{3.38}} = 1.51
Mole ratio of O=3.383.38=1O = \dfrac{{3.38}}{{3.38}} = 1
Hence, we get the formula C1H1.5O1{C_1}{H_{1.5}}{O_1}
Since, we cannot have fractional subscripts, multiply every subscript by 22 to get the empirical formula of succinic acid
Hence, the empirical formula of succinic acid C2H3O2{C_2}{H_3}{O_2} .

Note:
The formula of a substance expressed with the smallest integer subscript is called an empirical formula. The empirical formula specifies the number of atoms in the compound in a given ratio. The empirical formula of a compound is directly proportional to its percent composition.