A gaseous hydrocarbon of vapor dwnsity 14 contains 85.2â„…

1.	A gaseous hydrocarbon of vapor dwnsity 14 contains 85.2â„…
Answer» Answer: Carbon=2 and hydrogen = 5 Thus Empirical formula of the compound is GIVEN by, C2H5 Empirical formula mass of C2H5 = 212 + 51 = 29 g Now we can DETERMINE molecular formula mass from vapor density = 2 * 28 = 58 g Therefore n = 58/29 = 2 Thus Molecular formula of the compound = 2 * C2H5 = C4H10

Answer»

Answer:

Carbon=2 and hydrogen = 5

Thus Empirical formula of the compound is GIVEN by, C2H5

Empirical formula mass of C2H5 = 2*12 + 5*1 = 29 g

Now we can DETERMINE molecular formula mass from vapor density = 2 * 28 = 58 g

Therefore n = 58/29 = 2

Thus Molecular formula of the compound = 2 * C2H5 = C4H10

Discussion