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

1 Answer
Mar 6, 2016

Answer:

#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.)