Easy way to determine the type of hydrocarbon from its vapour density.

Easy way to determine the type of hydrocarbon from its vapour density. 

The best best, for sure. :)

Students often encounter questions like "The vapour density of an aliphatic hydrocarbon is X. Determine whether it is an alkane, alkene or alkyne". Quite a few calculations follow, and one single misstep can ruin the correctness of the answer. However, during my year as a student of class IX, I devised a new, extremely way to immediately determine the type of aliphatic hydrocarbon, with minimal calculation. And in this post, I intend to share it with you all.

This method I am going to mention works only for basic aliphatic ones, and most questions regarding determination of type from V.D are regarding alkanes, alkenes and alkynes anyway ones anyway. So, there are three types of basic hydrocarbons, as you all know, namely:

1. Alkanes, the hydrocarbons with only single carbon-carbon bonds, represented by the common formula C(n)H(2n+2). 
2. Alkenes, the hydrocarbons with at least one double C-C bond, given by the common formula C(n)H(2n). 
3. Alkynes, the ones with at least one triple C-C bond, given by C(n)H(2n-2). 

So, usually the question is bout vapour densities. Say, the vapour density of a hydrocarbon is provided, and you need to determine whether it is an alkyne, alkene or alkane. For those of you who'd just like to know the trick, it is basically dividing the V.D value by 7 and finding out the remainder. So, if the remainder is 1, the hydrocarbon is an alkane. If the remainder is 0, i.e the vapour density an integral multiple of 7, it is an alkene. And if the remainder is 6, the hydrocarbon is an alkyne. 

Ok, so these are the three guys. 

So, the process is very simple. Divide X by 7, and the hydrocarbon is: 

• Alkane, if remainder is 1
• Alkene, if remainder is 0
• Alkyne, if remainder is 6

Additionally, the method can also tell you whether the value of X is a possible value for the vapour density of an aliphatic hydrocarbon. In that case, if the remainder is anything other than 1,0 or 6, it is not a possible value. 

Thus, this is basically the process I figured out. I am not sure if anyone else has done it as well, at least in our books it wasn't mentioned, and the teachers weren't familiar with it either. 

Now, let us explore why this trick works. We know that vapour density of any hydrocarbon is mathematically half the value of its molecular weight. Now, for an alkane, the general formula is C(n)H(2n+2). The molecular weight of carbon is 12, while that of hydrogen is 1. So, the total molecular weight of the alkane is given as: 12*n+(2n+2) = 14n+2 . Now, the vapour density of alkanes is given by the general formula 1/2*(14n+2) = 7n+1. And undoutedly, (7n+1) yields a remainder of 1, when divided by 7 (n being a natural number). That's why the remainder of always 1 for alkanes. 

For alkenes, similarly the general formula is C(n)H(2n). So, the molecular weight will be 12n+2n = 14n, and the V.D will be 14n/2 = 7n, which is an integral multiple of 7, yielding remainder 0 when divided by 7. 

And lastly, for alkynes, the formula is C(n)H(2n-2), and as such the molecular weight will be 12n+(2n-2) = 14n - 2, and V.D = 1/2*(14n - 2) = 7n - 1. Since n is a natural number, 7n-1 when divided by 7 will always yield remainder 6. 

Thus, the process actually works.