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An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives NH_{3} alongwith a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound is

Option: 1

C\! H_{3}N\! C\! O\;


Option: 2

\; \; CH_{3}CONH_{2}\;


Option: 3

\; (NH_{2})_{2}CO\;


Option: 4

\; CH_{3}CH_{2}CONH_{2}


Answers (1)

 

Element Percentage \frac{Percentage}{At. wt. } Simple ratio
C 20.0 \frac{20.0}{12 } = 1.66 1
H 6.67 \frac{6.67}{1} = 6.67

\frac{6.67}{1.66} =4

N 46.67 \frac{46.67}{14} = 3.33

\frac{3.33}{1.66} =2

O 26.66 \frac{26.66}{16} =1.66

\frac{3.33}{1.66} =1

 

Empirical formula = CH4N2O

Empirical formula wt. = 12 + (4 X 1 ) + (2 X 14) + 16 = 60

n = \frac{Mol. \;formula\; wt.}{Empirical\; formula\; wt.} = \frac{60}{60} = 1

\therefore , molecular formula = CH4N2O

Therefore, Option(3) is correct

Posted by

Sumit Saini

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