Microanalytical data give C,H,N:65.73%;15.06%;19.21%. What is the molecular formula if the vapour density=37?

1 Answer
anor277
Dec 21, 2016

We assume that we have been given C;H;N microanalysis = 65.73%;15.06%;19.21%.
We get (eventually) molecular formula=C4H11N

Explanation:

We find the empirical formula in the usual way; that is assume that there are 100⋅g of unknown compound, and divide thru by the atomic masses of each constituent:

C: 65.73⋅g12.011⋅g⋅mol−1=5.47⋅mol

H: 15.06⋅g1.00794⋅g⋅mol−1=14.94⋅mol

N: 19.21⋅g14.01⋅g⋅mol−1=1.37⋅mol

And now we divide thru by the smallest molar quantity, that of N, to give the empirical formula:

C4H11N

Now, vapour density is the density of a vapour in relation to that of dihydrogen.

And thus here, Molar mass of gasMolar mass of dihydrogen=37.

And thus Molar mass of gas=74⋅g⋅mol−1.

As is typical, the molecular formula is a whole number multiple of the empirical formula, and thus,

74⋅g⋅mol−1≅n×(4×12.011+11×1.00794+14.01)⋅g⋅mol−1

The vapour density was a bit out, but in fact this persuades me that these data came from an actual experiment, and not a problem someone pulled out of their posterior. Typically, vapour density measurements are not perfect.

Clearly, n=1, and the empirical formula = molecular formula = C4H11N

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