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........
Answers
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:}}}}
Tofind:
Molecular formula of the compound
{\LARGE{\underline{\underline{\bf{Solution:}}}}}
Solution:
Strep 1 : Calculating relative number of atoms
a. Hydrogen
Percentage composition = 4.07 %
Atomic weight of hydrogen = 1
Relative number of atoms = \sf{\description{4.07}{1}}
1
4.07
= 4.07
b. Carbon
Percentage comppsition = 24.47 %
Aomic weight of carbon = 12
Relative number of atoms = \sf{\description{24.47}{12}}
12
24.47
= 2.02
c. Chlorine
Percentage comppsition = 71.65%
Atomic weight of chlorine = 35.5
Relative number of atoms = \sf{\description{71.65}{35.5}}
35.5
71.65
= 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{\description{4.07}{2.01}}
2.01
4.07
\approx≈ 2
b. Carbon
Simplest ratio = \sf{\description{2.02}{2.01}}
2.01
2.02
\approx≈ 1
c. Chlorine
Simplest ratio = \sf{\description{2.01}{2.01}}
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}}}}⋆
Molecular Formula=Empirical Formula×n
n is some integer which is given by ,
\star\large{\boxed{\purple{ \sf{ n = \description{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}⋆
n=
Empirical Formula Weight
Molecular 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}
:⟹n=
49.5
99
:⟹
n=2
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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}
⟹Molecular Formula=Empirical Formula×n
⟹Molecular Formula=CH
2
Cl×2
⟹Molecular Formula=C
2
H
4
Cl
2
Hence , The molecular formula of the given compound is C₂H₄Cl
Divide the given percentages of atoms with their molecular masses
H---4.07/1 =4.07
C---24.27/12 =2.02
Cl--71.65/35.5=2.01
step 2
divide all values with the lowest value obtained.
H---4.07/2.01=2
C---2.02/2.01=1
Cl---2.01/2.01=1
therefore the empirical formula is
CH
2
Cl
WEIGHT OF EMPIRICAL FORMULA=12+2+35.5=49.5
Given molecular weight =98.96 which is double of empirical weight .therefore molecular formula is C
2
H
4
Cl
2
.