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`        an orgnic copound contain 69.4% corbon 5.8%hydrogen sample of 0.303g of this compound was analijsed for nitrogen by kjeldonl's method .the ammonia evolved was avorbed in 50ml of 0.05M H2so4 .the exess acid required 25ml of 0.1m Naoh for neutralisation determine the molecule formula of the compound if its malecular mass is 121`
2 years ago

```							Ammonia evolved on Kjeldhal method neutralizes some milli equivalents of H2SO4 while rest will be neutralised by NaOH.Mill equivalents of NaOH = N.V = 0.1 x 25 = 2.5 [The charge on Na and OH is 1, therefore N = M]Initial milli equivalents of H2SO4 = N.V  = 0.05 x 2 x 50 = 5 [The charge on H and SO4 is 2, therefore N = 2 x M]This means milliequivalents of H2SO4 neutralised by NH3 = 5 – 2.5 = 2.5Since ammonia is formed by nitrogen present in the organic compound, therefore milli equivalents of N = 2.5Hence mass of Nitrogen in the compound, w = NVE/1000 = 2.5 x 14/1000 = 0.035 gNow percentage of N in the organic compound = 0.035 x 100/0.303 = 11.55 %Since total percentage of C, H and N is not 100 therefore rest is O (oxyegen).Therefore C = 69.40%, H = 5.80%, N = 11.55% and O = 13.25%.Dividing these values by respective atomic masses we get, C = 5.783, H = 5.80, N = 0.825, O = 0.828.This on simplification gives C = 7, H = 7, N = 1 and O = 1.Hence empirical formula is C7H7NO and empirical formula mass = 121.Therefore molecular formula is C7H7NO and it is C6H5CONH2 that is benzamide.
```
2 years ago
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