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

Explanation:

$.Percentage composition of Hydrogen in the compound is 4.07 %</p><p></p><p>Percentage composition of Carbon in the compound is 24.47 %</p><p></p><p>Percentage composition of Chlorine in the compound is 71.65 %</p><p></p><p>Molar mass = 99 g/mol</p><p></p><p>\LARGE{\underline{\underline{\bf{To \: find:}}}} </p><p>Tofind:</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Molecular formula of the compound</p><p></p><p>{\LARGE{\underline{\underline{\bf{Solution:}}}}} </p><p>Solution:</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Strep 1 : Calculating relative number of atoms</p><p></p><p>a. Hydrogen</p><p></p><p>Percentage composition = 4.07 %</p><p></p><p>Atomic weight of hydrogen = 1</p><p></p><p>Relative number of atoms = \sf{\dfrac{4.07}{1}} </p><p>1</p><p>4.07</p><p> </p><p> = 4.07</p><p></p><p>b. Carbon</p><p></p><p>Percentage comppsition = 24.47 %</p><p></p><p>Aomic weight of carbon = 12</p><p></p><p>Relative number of atoms = \sf{\dfrac{24.47}{12}} </p><p>12</p><p>24.47</p><p> </p><p> = 2.02</p><p></p><p>c. Chlorine</p><p></p><p>Percentage comppsition = 71.65%</p><p></p><p>Atomic weight of chlorine = 35.5</p><p></p><p>Relative number of atoms = \sf{\dfrac{71.65}{35.5}} </p><p>35.5</p><p>71.65</p><p> </p><p> = 2.01</p><p></p><p>Step 2 : Calculating Simplest ratio for each element</p><p></p><p>Simplest ratio is calculted by dividing all the values we got by the smallest value we got.</p><p></p><p>The smallest value we got is 2.01</p><p></p><p>a. Hydrogen</p><p></p><p>Simplest ratio = \sf{\dfrac{4.07}{2.01}} </p><p>2.01</p><p>4.07</p><p> </p><p> \approx≈ 2</p><p></p><p>b. Carbon</p><p></p><p>Simplest ratio = \sf{\dfrac{2.02}{2.01}} </p><p>2.01</p><p>2.02</p><p> </p><p> \approx≈ 1</p><p></p><p>c. Chlorine</p><p></p><p>Simplest ratio = \sf{\dfrac{2.01}{2.01}} </p><p>2.01</p><p>2.01</p><p> </p><p> = 1</p><p></p><p>The empirical formula becomes CH₂Cl</p><p></p><p>Molecular weight of Empirical formula = 12 + 2(1) + 35.5 = 49.5</p><p></p><p>We are already given that molecular weight as 99.</p><p></p><p>▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬</p><p></p><p>Relation between Empirical formula and Molecular formula is given by ,</p><p></p><p>\star\large{\boxed{ \sf{ \purple{Molecular\:Formula=Empirical\:formula\times\:n}}}}⋆ </p><p>MolecularFormula=Empiricalformula×n</p><p> </p><p> </p><p></p><p>n is some integer which is given by ,</p><p></p><p>\star\large{\boxed{\purple{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}⋆ </p><p>n= </p><p>Empiricalformulaweight</p><p>MolecularFormulaweight</p><p> </p><p> </p><p> </p><p> </p><p></p><p>We know the values for calculating the value of n . So , by substituting we get ;</p><p></p><p>\begin{gathered} \\ : \implies \sf \: n = \frac{99}{49.5} \\ \\ \\ : \implies \sf {\boxed {\underline{ \sf{n = 2}}}}\end{gathered} </p><p>:⟹n= </p><p>49.5</p><p>99</p><p> </p><p> </p><p>:⟹ </p><p>n=2</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬</p><p></p><p>Applying the first formulae , (i.e , relation between empirical formulae and Molecular formulae) we get ;</p><p></p><p>\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} </p><p>⟹MolecularFormula=Empiricalformula×n</p><p>⟹MolecularFormula=CH </p><p>2</p><p> </p><p> Cl×2</p><p>⟹MolecularFormula=C </p><p>2</p><p> </p><p> H </p><p>4</p><p> </p><p> Cl </p><p>2</p><p> </p><p> </p><p> </p><p> </p><p></p><p>Hence , The molecular formula of the given compound is C₂H₄Cl₂$

Answer 2

Find an answer to your 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 ...