Question

a compound contains 4.07% hydrogen, 24 look.27% of carbon and 71.65% of chlorine.
its molecular mass is 98.96g. what are its empirical and molecular formula?​

Answer 1

Answer:

$\bigstar{\bold{Empirical\:formula=CH_2Cl}}$

$\bigstar{\bold{Molecular\:formula=C_2H_4Cl_2}}$

Explanation:

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

• Mass of Hydrogen = 4.07 g
• Mass of carbon = 24.27 g
• Mass of chlorine = 71.75 g
• Molecular mass = 98.96 g

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

• Empirical formula
• Molecular formula

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

➔ Here we have to first find the simplest whole number ratio of all the elements.

➔ For that, we have to find the number of moles of each element.

➔ Number of moles of an element = Give mass/Molar mass

➔ Number of moles of hydrogen = given mass of hydrogen/molar mass

    Number of moles of hydrogen = 4.07/1

    Number of moles of hydrogen = 4.07 moles

➔ Number of moles of carbon = 24.27/12

    Number of moles of carbon = 2.02 moles

➔ Number of moles of chlorine = 71.65/35.5

   Number of moles of chlorine = 2.02 moles

➔ Now finding the whole number ratio of all the elements

➔ Whole number ratio of hydrogen = 4.07/2.02

    Whole number ratio of hydrogen = 2

➔ Whole number ratio of carbon = 2.02/2.02

    Whole number ratio of carbon = 1

➔ Whole number ratio of chlorine = 2.02/2.02

    Whole number ratio of chlorine = 1

➔ Hence the empirical formula of the compund is CH₂Cl

    $\boxed{\bold{Empirical\:formula=CH_2Cl}}$

➔ Now we have to find the molar mass of the compound

    Molar mass = 12 + 2 × 1 + 35.5

    Molar mass = 12 + 2 + 35.5

    Molar mass = 49.5

➔ Now we know that,

    Molecular mass/Molar mass = n

➔ Hence,

    98.96/49.5 = n

    n = 2

➔ Molecular formula is given by,

   Molecular formula = n × Empirical formula

    Molecular formula =  2 × (CH₂Cl)

    Molecular formula = C₂H₄Cl₂

➔ Hence molecular formula of the compund is  C₂H₄Cl₂

    $\boxed{\bold{Molecular\:formula=C_2H_4Cl_2}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

➛ Number of moles of an element = Give mass/Molar mass

➛  Molecular mass/Molar mass = n

➛   Molecular formula = n × Empirical formula