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

Mar 6, 2016

$C {H}_{3} C l$

#### 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 $100 k g$ 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 $M {M}_{C} l = 35.5 g \cdot m o {l}^{-} 1$, $M {M}_{C} = 12 g \cdot m o {l}^{-} 1$, $M {M}_{H} = 1 g \cdot m o {l}^{-} 1$

So we have

${n}_{C} l = {m}_{C} \frac{l}{M {M}_{C} l} = \frac{70}{35.5} = 1.97 \cong 2$
${n}_{C} = {m}_{C} / \left(M {M}_{C}\right) = \frac{24}{12} = 2$
${n}_{H} = {m}_{H} / \left(M {M}_{H}\right) = \frac{6}{1} = 6$

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 ${C}_{2} {H}_{6} C l$ to $C {H}_{3} C l$, 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.)