Note that we start with the value of 72 g AgCl given in the question and multiply it by three fractions
that have in common the property that the numerator is equivalent to (or “corresponds to”) the
quantity in the denominator: 1 mol of AgCl is equivalent to 144 g of AgCl; 1 mol CaCl 2 gives us 2 mol
AgCl in this reaction; 1 mol CaCl 2 is equivalent to 110 g CaCl 2 . Because of the equivalence between
numerator and denominator, we can switch the two and not affect the fraction as a whole. The way
we have decided which to use as numerator and which to use as denominator is dictated by which
units we want to cancel. For example, we want to get rid of the weight of AgCl and obtain the
number of moles instead, and so we have chosen to put 144 g AgCl in the denominator to cancel the
unit of (g AgCl) in the starting value:
You can verify that all the units do cancel to yield at the end “g CaCl2,” which is what we want. The
way the equation has to be set up to give the right unit tells us how to manipulate the numbers,
without having to spend too much time trying to recall, “Should I divide by the molar mass or
multiply?”