Saved Bookmarks
| 1. |
Butyric acid contains only C, H and O. A 4.24 mg sample of butyric acid is completely burned. It gives 8.45 mg of CO_(2) and 3.46 mg og H_(2)O. The molecular mass of butyric acid was determined by experiment to be 88 amu. What is molecular formula? |
|
Answer» `"i.e.,44 g "CO_(2)" contains C= 12 g"` `THEREFORE 8.45 MG CO_(2)" will contain C"=(12)/(44)xx8.45mg` This is present in 4.24 mg of the compound. `therefore %" of C in the compound "=(12)/(44)xx(8.45)/(4.24)xx100=54.4%` `"or use the formula directly"%" of C"=(12)/(44)xx("Mass of CO"_(2))/("Mass of compound")xx100` `"Similarly, "%" of H"=(2)/(18)xx("Mass of "H_(2)O)/("Mass of compound")xx100=(2)/(18)xx(3.46)/(4.24)xx100=9.1%` `therefore""%" or O"=100-(54.4+9.1)=36.5%` CALCULATE E.F. Its comes out to be `C_(2)H_(4)O`. E.F. mass = 44 u, Mol mass = 88 u. Hence, n = Mol. mass/E.F mass = 2 `therefore"Mol. formula"=2xxE.F.=C_(4)H_(8)O_(2)` |
|