Question

calculate the empirical and molecular formula of the compound having t
following percentage composition:
C=26.59%. H=2.22% 0=71.19%, its molecular mass is 90.​

Answer 1

Given :

Per. composition of C = 26.59%

Per. composition of H = 2.22%

Per. composition of O = 71.19%

To Find :

Molecular and empirical formula of the compound.

Solution :

Atomic mass of C = 12g

Atomic mass of H = 1g

Atomic mass of O = 16g

$\bullet\sf \ No. \ of \ moles \ of \ C = \dfrac{26.59}{12} = 2.22$

$\bullet\sf \ No. \ of \ moles \ of \ H= \dfrac{2.22}{1} = 2.22$

$\bullet\sf\ No \ of \ moles \ of \ O = \dfrac{71.79}{16} = 4.44$

■ Simplest ratio :

$\implies\sf \sf \dfrac{2.22}{2.22} \ : \ \dfrac{2.22}{2.22} \ : \ \dfrac{4.44}{2.22}$

$\implies\bf 1 \ : \ 1 \ : \ 2$

$\bigstar\:\underline{\boxed{\bf{\purple{Empirical\:Formula=CHO_2}}}}$

Empirical formula mass :

• C + H + 2(O)
• 12 + 1 + 2(16)
• 45

We know that,

$\dag\:\bf Molecular \ formula = n \times Empirical\ formula$

➠ n = Molar mass / Formula mass

➠ n = 90/45

➠ n = 2

$\bullet\sf\:Molecular\ formula=2\times CHO_2$

$\dag\:\underline{\boxed{\bf{\pink{Molecular\ Formula=C_2H_2O_4}}}}$

Answer 2

$\huge{\mathtt{\red{\underline {Given : }}}}$

Percentage composition of C = 26.59%

Percentage composition of H = 2.22%

Percentage composition of O = 71.19%

$\huge{\mathtt{\purple{\underline {To\: Find : }}}}$

Molecular and empirical formula of the compound.

$\huge{\mathtt{\pink{\underline {Solution : }}}}$

Atomic mass of C = 12g

Atomic mass of H = 1g

Atomic mass of O = 16g

$\bullet\sf \ No. \ of \ moles \ of \ C = \dfrac{26.59}{12} = 2.22.$

$\bullet\sf \ No. \ of \ moles \ of \ H= \dfrac{2.22}{1} = 2.22∙$

$\bullet\sf\ No \ of \ moles \ of \ O = \dfrac{71.79}{16} = 4.44∙$

Simplest ratio :

$:\implies\sf \sf \dfrac{2.22}{2.22} \ : \ \dfrac{2.22}{2.22} \ : \ \dfrac{4.44}{2.22}$

$:\implies\bf 1 \ : \ 1 \ : \ 2⟹1 : 1 : 2$

${\boxed{\boxed{\bf{\orange{Empirical\:Formula=CHO_2}}}}}$

We know that,

$\dag\:\bf Molecular \ formula = n \times Empirical\ formula$

$\sf n = \dfrac {Molar\: mass }{Formula \:mass}$

$\sf n = \dfrac {90}{45}$

★ n = 2 ★

$\bullet\sf\:Molecular\ formula=2\times CHO_2.$

${\boxed{\boxed{\bf{\blue{Molecular\ Formula=C_2H_2O_4}}}}}$