Question

An organic substance containing carbon hydrogen and oxygen gave the percentage composition carbon 40.687./. oxygen 54.228./. calculate the emperical formula of the compound.​ ( chemistry )

Answer 1

Given :-

• % C =40.68
• % O = 54.22
• % H = 100 - (40.68 + 54.22) = 5.1
• Atomic mass of C = 12 g
• Atomic mass of O = 16 g
• Atomic mass of H = 1 g

Solution :-

➥ Relative no of atoms

• Divide the % composition of the elements with their atomic masses

$\longrightarrow\mathtt{C =\dfrac{ 40.68}{12}=3.39}$

$\longrightarrow\mathtt{O=\dfrac{ 54.28}{16}=3.39}$

$\longrightarrow\mathtt{H=\dfrac{ 5.1}{1}=5.1}$

➥Simple ratio

• Divide all the values obtained above with the least value (i.e. 3.39 in this case)

$\longrightarrow\mathtt{C =\dfrac{ 3.39}{3.39}=1}$

$\longrightarrow\mathtt{O =\dfrac{ 3.39}{3.39}=1}$

$\longrightarrow\mathtt{H =\dfrac{ 5.1}{3.39}=1.5}$

Hence ,

$\longrightarrow\mathtt{1 :1:1.5}$

$\longrightarrow\mathtt{2:2:3}$

➽ Empirical formula of the compound = C2H3O2

The empirical formula of the compound is C2H3O2