Saved Bookmarks
| 1. |
The analysis of an organic compound gave the following data: (a) 0.4020g gave 0.6098g and 0.2080g of CO_(2) and H_(2)O respectively. (b) 1.01g by Kjeldahl method produced amonia whise was neutralised by 23.2mL of N/2HCl (c ) 0.1033g of the compound gave 0.2772g of BaSO_(4) (d) 0.1015g when vaporised in Victor Meyer's apparatus displaced 27.96mL of air at 15^(@)C and 766 mm pressure Calculate the molecular formula. (Aqueous tension at 15^(@)C=16mm) |
|
Answer» Solution :Moles of `C=1 xx` moles of `CO_(2)=1 xx (0.6098)/(44)= 0.0138` Moles of `H= 2xx` moles of `H_(2)O = 2 xx (0.2080)/(18)= 0.0231` Calculation of moles of N: m.e. of `HCl= 23.2 xx (1)/(2)=11.6` `therefore` m.e. of `NH_(3)=11.6` m mol of `NH_(3)=11.6` Moles of `NH_(3)= (11.6)/(1000)= 0.0116` `therefore` moles of `N =1 xx` moles of `NH_(3)=0.0116` Moles of N in 0.4020g of compound `=(0.0116)/(1.01)xx 0.4020` =0.0046 Moles of `S=1 xx` moles of `BaSO_(4)= (0.2772)/(233.3)= 0.00119 (BaSO_(4)=233.3)` Moles of S in 0.4020g of compound `=(0.00119)/(0.1033) xx 0.4020` = 0.0046 `therefore` mole of `C: H: N: S= 0.0138: 0.0138: 0.0046: 0.0046` `=3:5:1:1` Hence the empirical formula is `C_(3)H_(5)NS` Calculation of molecular weight: PRESSURE due to dry AIR only `=766-16=750mm` Volume of vapour of 0.1015g of compound at NTP `=(750 xx 27.96 xx 273)/(288 xx 760)=26.155mL` `therefore` moles of vapour `=(26.155)/(22400)=0.00116` Molecular weight `=("weight in grams")/("no. of moles") = (0.1015)/(0.00116)=86.97 ~~87` Since empirical formula weight is also equal to 87, the molecular formula is `C_(3)H_(5)NS` |
|