Organic Chemistry – Some Basic Principles and Techniques

Question
CBSEENCH11007641

Discuss a suitable method for the estimation of halogens.

Solution

Carius method for the estimation of halogens: A known mass of the organic compound containing halogens is heated with an excess of fuming nitric acid and silver nitrate in a sealed tube called Carius tube. Cabron, hydrogen and sulphur (if present) are oxidised to CO2, H2O and H2SOrespectively whereas halogen forms a precipitate of silver halide. The precipitate is separated, washed with distilled water, dried and weighed. The percentage of halogens is then calculated. Calculations:
Let the mass of the compound taken = Wg Let the mass of halide (AgX) formed = ag

Atomic mass of halogen = x
space space therefore space space
Molecular space mass space of space silver space halide space left parenthesis AgX right parenthesis space equals space 108 space plus straight x
space space space AgX space equals space straight X
space space left parenthesis 108 plus straight x right parenthesis space equals space straight x
space Now space left parenthesis 108 plus straight x right parenthesis space straight g space of space silver space halide space contains space halogen space equals space xg
space space space therefore space space space
space straight a space straight g space of space silver space halide space contains space hydrogen space
space space space space space equals space fraction numerator straight x over denominator left parenthesis 108 plus straight x right parenthesis end fraction cross times space ag
therefore space space
Percentage space of space halogen space equals space fraction numerator straight x space cross times space straight a over denominator left parenthesis 108 plus straight x right parenthesis space cross times space straight W end fraction cross times 100
therefore space space Percentage space of space chlorine
space space space space equals space fraction numerator 35.5 space cross times space Mass space of space AgCl space cross times space 100 over denominator 143.5 space cross times space Mass space of space organic space compound end fraction
Percentage space of space bromine
space space space space space equals space fraction numerator 80 space cross times space Mass space of space AgBr space cross times space 100 over denominator 188 space cross times space Mass space of space organic space compound end fraction
Percentage space of space iodine
space space space space space space space space space space space space space equals space fraction numerator 127 space cross times space Mass space of space AgI space cross times space 100 over denominator 235 space cross times space Mass space of space organic space compound end fraction

Question
CBSEENCH11007642

0.3780 of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. 

Solution

Pecentage space of space Cl equals

fraction numerator Atomic space mass space of space Cl space straight x space mass space of space AgCl space form space space straight x space 100 over denominator Molecular space mass space of space AgCl space space straight x space straight m end fraction
Molecular mass of AgCl = 108 + 35.5 = 143.5 g mol-1.
Now 143.5 g AgCl contains 35.5 g chlorine
  0.5740 g AgCl would contain
                 equals fraction numerator 35.5 over denominator 143.5 end fraction cross times 0.5740 space straight g space chlorine
Mass of organic compound is =0.3780
therefore space space Percentage space of space chlorine
space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 35.5 over denominator 143.5 end fraction cross times fraction numerator 0.5740 over denominator 0.3780 end fraction cross times 100
space space space space space space space space space space space space space space space space space space space space space equals space 37.57 percent sign

Sponsor Area

Question
CBSEENCH11007643

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound. 

Solution

Mass of organic compound 0.15g
Mass of AgBr = 0.12g
Molecular mass of AgBr = 108+80 = 188 g mol-1
Now 188g AgBr contains 80 g bromine
space space therefore space 0.12 straight g space AgBr space would space contain space 80 over 188 cross times 0.12 space bromine
space space space space equals space 0.05 space bromine
space space therefore space space space space Percentage space of space bromine space equals space fraction numerator 0.05 over denominator 0.15 end fraction cross times 100
space space space space space space space equals space 33.3 percent sign

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Question
CBSEENCH11007644

0.15 g of iodoform gave 0.2682 g of silver iodide. Calculate the percentage of iodine.

Solution

Mass of the compound  = 0.15 g,   Mass of silver iodide = 0.2682 g
The molecular mass of silver iodide. 
              (Ag I) = 108 + 127 = 235
              AgI space identical to space straight I
235g of AgI contains  = 127 g of iodine
 therefore space space 0.2682 space straight g space of space AgI space contains space equals space 127 over 235 cross times 0.2682
therefore space space Percentage space of space iodine space equals space fraction numerator 127 space cross times space 0.2682 over denominator 235 space cross times space 0.15 end fraction cross times 100 space equals space 96.6