Question

A compound contains H=4.07%, c=24.47%and Ci=71.65% its molar mass is 99gram/mol find the molecular formula of its compound.......​

Answer 1

Answer:

$\LARGE{\bf {\underline{\underline{Given : }}}}$

Percentage composition of Hydrogen in the compound is 4.07 %

Percentage composition of Carbon in the compound is 24.47 %

Percentage composition of Chlorine in the compound is 71.65 %

Molar mass = 99 g/mol

$\LARGE{\underline{\underline{\bf{To \: find:}}}}$

Molecular formula of the compound

${\LARGE{\underline{\underline{\bf{Solution:}}}}}$

Strep 1 : Calculating relative number of atoms

a. Hydrogen

Percentage composition = 4.07 %

Atomic weight of hydrogen = 1

Relative number of atoms = $\sf{\dfrac{4.07}{1}}$ = 4.07

b. Carbon

Percentage comppsition = 24.47 %

Aomic weight of carbon = 12

Relative number of atoms = $\sf{\dfrac{24.47}{12}}$ = 2.02

c. Chlorine

Percentage comppsition = 71.65%

Atomic weight of chlorine = 35.5

Relative number of atoms = $\sf{\dfrac{71.65}{35.5}}$ = 2.01

Step 2 : Calculating Simplest ratio for each element

Simplest ratio is calculted by dividing all the values we got by the smallest value we got.

The smallest value we got is 2.01

a. Hydrogen

Simplest ratio = $\sf{\dfrac{4.07}{2.01}}$ $\approx$ 2

b. Carbon

Simplest ratio = $\sf{\dfrac{2.02}{2.01}}$ $\approx$ 1

c. Chlorine

Simplest ratio = $\sf{\dfrac{2.01}{2.01}}$ = 1

The empirical formula becomes CH₂Cl

Molecular weight of Empirical formula = 12 + 2(1) + 35.5 = 49.5

We are already given that molecular weight as 99.

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Relation between Empirical formula and Molecular formula is given by ,

$\star\large{\boxed{ \sf{ \purple{Molecular\:Formula=Empirical\:formula\times\:n}}}}$

n is some integer which is given by ,

$\star\large{\boxed{\purple{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}$

We know the values for calculating the value of n . So , by substituting we get ;

$\\ : \implies \sf \: n = \frac{99}{49.5} \\ \\ \\ : \implies \sf {\boxed {\underline{ \sf{n = 2}}}}$

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Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get ;

$\\ \implies \sf Molecular\:Formula=Empirical\:formula \times \: n \\ \\ \\ \implies \sf Molecular\:Formula=CH_2Cl \times 2 \\ \\ \\ \implies \sf \: Molecular\:Formula=C_2H_4Cl_2$

Hence , The molecular formula of the given compound is C₂H₄Cl₂

Answer 2

$\LARGE{\bf {\underline{\underline{Given : }}}}$

Percentage composition of Hydrogen in the compound is 4.07 %

Percentage composition of Carbon in the compound is 24.47 %

Percentage composition of Chlorine in the compound is 71.65 %

Molar mass = 99 g/mol

$\LARGE{\underline{\underline{\bf{To \: find:}}}}$

Molecular formula of the compound

${\LARGE{\underline{\underline{\bf{Solution:}}}}}$

Step 1 : Calculating relative number of atoms

a. Hydrogen

Percentage composition = 4.07 %

Atomic weight of hydrogen = 1

Relative number of atoms = $\sf{\dfrac{4.07}{1}}= 4.07$

b. Carbon

Percentage comppsition = 24.47 %

Aomic weight of carbon = 12

Relative number of atoms = $\sf{\dfrac{24.47}{12}} = 2.02$

c. Chlorine

Percentage comppsition = 71.65%

Atomic weight of chlorine = 35.5

Relative number of atoms = $\sf{\dfrac{71.65}{35.5}} = 2.01$

Step 2 : Calculating Simplest ratio for each element

Simplest ratio is calculted by dividing all the values we got by the smallest value we got.

The smallest value we got is 2.01

a. Hydrogen

Simplest ratio = $\sf{\dfrac{4.07}{2.01}}$

$\approx$ 2

b. Carbon

Simplest ratio = $\sf{\dfrac{2.02}{2.01}}$

$\approx$ 1

c. Chlorine

Simplest ratio = $\sf{\dfrac{2.01}{2.01}} = 1$

The empirical formula becomes CH₂Cl

Molecular weight of Empirical formula = 12 + 2(1) + 35.5 = 49.5

We are already given that molecular weight as 99.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Relation between Empirical formula and Molecular formula is given by ,

$\star\large{\boxed{ \sf{ \pink{Molecular\:Formula=Empirical\:formula\times\:n}}}}$

n is some integer which is given by ,

$\star\large{\boxed{\pink{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}$

We know the values for calculating the value of n . So , by substituting we get ;

$\begin{gathered} \\ : \implies \sf \: n = \frac{99}{49.5} \\ \\ \\ : \implies \sf {\boxed {\underline{ \sf{n = 2}}}}\end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get ;

$\begin{gathered} \\ \implies \sf Molecular\:Formula=Empirical\:formula \times \: n \\ \\ \\ \implies \sf Molecular\:Formula=CH_2Cl \times 2 \\ \\ \\ \implies \sf \: Molecular\:Formula=C_2H_4Cl_2\end{gathered}$

Hence , The molecular formula of the given compound is C₂H₄Cl₂