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

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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

TO FIND:

Molecular formula of the compound

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

Answer:

compound contains H=4.07%, c=24.47%and Ci=71.65% its molar mass is