an organic compound was found to contain the following constituents:C=33%,H=4.7%,N=130%,Cl=33.4% . Determine the Empirical formula of the compound. Hint: Calculate the percentage of oxygen since the sum of percentages of elements is not 100. The percentage of oxygen=15.7% .
Answers
Answer:
Answer =C3H5N9ClO
Explanation: HERE'S YOUR ANSWER
Step -1 .. Percentage of oxygen=100--(33+4.7+13.2+33.4) = 100--84.3 = 15.7 Step -2 .. Emperical formula of the organic compound.
Element Percentage AM GAM Atomic ratio Simpleratio
C 33.0 12 33.0÷12 2.75÷0.94 3
=2.75 =2.93
H 4.7 1 4.7÷1 4.70÷0.94 5
=4.70 =5.0
N 130 14 130÷14 9.28÷0.94 9
=9.28 =9.87
Cl 33.4 35.5 33.4÷35.5 0.94÷0.94 1
=0.94 =1
O 15.7 16 15.7÷16 0.98÷0.94 1
=0.98 =1.04
EMPERICAL FORMULA=C3H5N9ClO
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