Saved Bookmarks
| 1. |
Explain about the estimation of carbon and hydrogen. |
|
Answer» Solution :(i) Principle: A known weight of organic substance is brunt in excess of oxygen and the carbon and hydrogen present in it are oxidised to `CO_(2)` and `H_(2)O` respectively. `C_(x)H_(y) + underset("excess")(O_(2))rarr xCO_(2) +(y)/(2)H_(2)O` The weight of carbon dioxide and water thus FORMED are determined and the amount of carbon and hydrogen in the organic substance are calculated. (ii) Description of the apparatus  (a) The oxygen supply (b) combustion tube (c) Absorption tube Oxygen supply: To remove the moisture from oxygen, it is allowed to bubble through sulphuric acid and then passed through a U-tube containing sodalime to remove `CO_(2)` . The oxygen gas free from moisture and `CO_(2)` enters the combustion tube. Combustion tube: A hard glass tube open at both ends USED for the combustion. It contains (i) an oxidized copper gauze to prevent the hackward diffusion of the products of combustion (ii) a porcelain boat containing a known weight of the organic substance (iii) coarse copper oxide on either side and (iv) an oxidised copper gauze placed towards the END of the combustion tube. The combustion tube is heated by a gas burner. Absorption apparatus: The combustion products containing moisture and `CO_(2)` are then passed through the absorption apparatus which consists of (i) a weighed U-tube packed with pumice soaked in conc. `H_(2)SO_(4)` to ABSORB Water (ii) a set of bulbs containing a strong solution of KOH to absorb `CO_(2)` and finally (iii) a guard tube filled with anhydrous `CaCl_(2)` to prevent the entry of moisture from atmosphere. (iii) Procedure: The combustion tube is heated strongly to dry its content. It is then cooled and connected to absorption apparatus. The other end of the combustion tube is open for a while and the boat containing weighed organic substance is introduced. The tube is again heated strongly till all the substance in the boat is burnt AWAY. This takes about 2 hours. Finally a strong current of oxygen is passed. Then the U-tube and potash bulbs are then detached and increase in weight of each of them is determined. (iv) Calculation: Weight of organic substance = Wg Increase in weight of `H_(2)O` = xg Increase in weight of `CO_(2)` = yg 18g of `H_(2)O` contains 2g of hydrogen `:.` xg of `H_(2)O` contain `(2)/(18) xx x g` of hydrogen `:. "Percentage of hydrogen"= ((2)/(18) xx (x)/(w) xx 100)%` 44g of `CO_(2)` contains 12g of carbon `:.` y g of `CO_(2)` contain `(12)/(44) xx y` g of carbon `:. "Percentage of carbon"=((12)/(44) xx (y)/(w) xx 100)%` |
|