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Explain principle of estimation of oxygen in organic compound |
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Answer» Solution :(a) The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition 100 and the sum of the percentage of all the other elements. % O= 100- (sum of % of all the other elements) (b) A definite mass of an organic compound is decomposed by heating in a steam if nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoixde. (CO). This mixture is passed through VORM iodine pentoxide `(I_(2)O_(5))` when carbon monoxide is oxidised to carbon dioxide producing iodine `(I_(2))` compound `= underset(Delta)overset("heat")rarr underset(1373K darr +"cock" 2C)(O_(2)) +` other GASES (A) `2C + O_(2) overset(1373K)rarr 2CO` ...(A) (B) `I_(2)O_(5) + 5CO rarr I_(2) + 5 CO_(2)` ...(B) Now eq. (A) `xx 5: 10C + 5O_(2) rarr 10CO`....(C ) eq. (B) `xx 2 : 2 I_(2)O_(5) + 10CO rarr 2I_(2) + 10 CO_(2)` ....(D) `(C + D) : 10 C + 5O_(2) + 2I_(2) O_(5) = 2I_(2) + 10 CO_(2)` `THEREFORE` 5 mole `O_(2) rarr 100 "mole " CO_(2)` `therefore` 1 mole `O_(2) rarr 2 "mole " CO_(2)` So, 32gm `O_(2) rarr 88gm CO_(2)` In oxygen estimation method ..32 gm oxygen is converted into 88gm carbon dioxide... (c ) Calculation: Initially take compound = m gm and weight of produced `CO_(2)= m_(1)gm` `therefore` weight of O in `m_(1) gm CO_(2) = (32 xx m_(1))/(88) gm O_(2)` `therefore % (O) =(32)/(88) xx (m_(1))/(m) xx 100` (d) Presently the estimation of elements in an organic compound is carried out by using micro quantities of substance and AUTOMATIC experimented techniques. The elements carbon (C) , HYDROGEN (H) and nitrogen (N) present in a compound are determined by an apparatus known as CHN elemental analyzer. The analyzer requires only a very small amount of the substance (1-3gm) and displays the values on a screen within a short time. |
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