Question

What is the empirical formula of a molecule containing 24% carbon, 6% hydrogen, and 70% chlorine?

Answer 1

`CH_3Cl`

Explanation:

The first step we need to do is to pick our base, i.e.: a random value of our choosing that will let us change this from percentages to actual mass. For convenience's sake we can choose something like `100 g` or `100 kg` or something of the like, using `100 g`:

In `100 g`, since any percentage of `100` is that number, we have

`70% 100g = m_Cl = 70g`
`24%100g = m_C = 24g`
`6%100g = m_H = 6g`

Now we need to divide these values by their molar masses, because then we'll know how much atoms we can find in a molecule. For practical purposes like this we say `MM_Cl = 35.5 g*mol^-1`, `MM_C = 12 g* mol^-1`, `MM_H = 1g*mol^-1`

So we have

`n_Cl = m_Cl/(MM_Cl) = 70/35.5 = 1.97 ~= 2`
`n_C = m_C/(MM_C) = 24/12 = 2`
`n_H = m_H/(MM_H) = 6/1 = 6`

The empirical formula wants us to have all the values as indivisible integers, and since all of these values can divide by 2, we can reduce them from `C_2H_6Cl` to `CH_3Cl`, which conveniently enough gave us a molecule that could exist from one that couldn't. (Why the first molecule can't exist and the last can is a matter for another time though, and if you have some grounding in organical chemistry, should be easy enough to see why, if not and you're curious just send me a message and I'll happy to explain why.)