solve the attachment
Answers
Explanation:
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{\dfrac{4.07}{1}}
1
4.07
= 4.07
b. Carbon
Percentage comppsition = 24.47 %
Aomic weight of carbon = 12
Relative number of atoms = \sf{\dfrac{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{\dfrac{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{\dfrac{4.07}{2.01}}
2.01
4.07
\approx≈ 2
b. Carbon
Simplest ratio = \sf{\dfrac{2.02}{2.01}}
2.01
2.02
\approx≈ 1
c. Chlorine
Simplest ratio = \sf{\dfrac{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}}}}⋆
MolecularFormula=Empiricalformula×n
n is some integer which is given by ,
\star\large{\boxed{\purple{ \sf{ n = \dfrac{Molecular\:Formula \: weight}{Empirical\:formula \: weight}}}}}⋆
n=
Empiricalformulaweight
MolecularFormulaweight
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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Answer:
Calculate moles of each element - divide
% by atomic mass
C=19.998/12.011 = 1.665
H= 3.331/ 1.008 = 3.305
N= 23.320 / 14.007 = 1.665
O= 53.302 / 15.999 = 3.332
Divide through by smallest
C=1
H=2
N=1
O = 2 ( small rounding is accepted)
Empirical formula = CH2NO2
Formula mass = 12.011 +1.008'2 + 14.007 + 15.999:2 = 60.03 g/formula unit
Formula units in 1 molecule = 120.07 /
60.03 = 2
Molecular formula = ( CH2NO2)2 = C2H4N204 (1,2-dinitroethane.)