Question

An insecticide has the following percentage composition by mass:47.5% C, 2.54% H and 50.0%CI. Determine its empirical formula and molecular formulae. Molar mass of the substance is 354.5gmol-1​

Answer 1

Question:

Answer:

$\bigstar{\bold{Empirical\:formula=C_3H_2Cl}}$

$\bigstar{\bold{Molecular\:formula=C_{15}H_{10}Cl_5}}$

Explanation:

$\Large{\underline{\rm{Given:}}}$

• Mass of Carbon = 47.5 g
• Mass of Hydrogen = 2.54 g
• Mass of Chlorine = 50 g

$\Large{\underline{\rm{To\:Find:}}}$

• Empirical formula
• Molecular formula

$\Large{\underline{\rm{Solution:}}}$

↬ First we have to find the number of moles of each element presesnt in the compund.

↬ We know that,

    Number of moles = Given mass/Molar mass

↬ Hence,

   Number of moles of carbon = 47.5/12

  Number of moles of carbon = 4 moles

   Number of moles of hydrogen = 2.54/1

  Number of moles of hydrogen = 2.54 moles

   Number of moles of chlorine = 50/35.5

  Number of moles of chlorine = 1.4 moles

↬ Next finding the simplest whole number ratio of all the elements.

    Whole number ratio of carbon = 4/1.4 = 3

    Whole number ratio of hydrogen = 2.54/1.4 = 2

    Whole number ratio of chlorine = 1.4/1.4 = 1

↬ Hence the empirical formula of the compound is C₃H₂Cl.

    $\boxed{\bold{Empirical\:formula=C_3H_2Cl}}$

↬ Now we have to find the molecular formula of the compund.

↬ Molecular mass of the compound = 12 × 3 + 2 × 1 + 35.5

   Molecular mass of the compound = 36 + 2 + 35.5 = 73.5

↬ Now,

   Molar mass/Molecular mass = n

    n = 354.5/73.5

    n = 5

↬ Hence,

   Molecular formula = n × Empirical formula

   Molecular formula = 5 × (C₃H₂Cl)

   Molecular formula = C₁₅H₁₀Cl₅

↬ Hence molecular formula of the compund is C₁₅H₁₀Cl₅.

    $\boxed{\bold{Molecular\:formula=C_{15}H_{10}Cl_5}}$