Long Island University Department of Chemistry Chem 3 (Introduction to Chemistry) Dr. Nikita Matsunaga

## Dimensional Analysis

Dimensional analysis is a very powerful method of calculating quantities one needs as a chemist. One can calculate, for example, an amount of substance needed to make a certain concentration of aqueous solution. The main idea of the method is to cancel out the units appearing in both the numerator and the denominator. Therefore, the units of the particular numerator-denominator pair has to be exactly the same. At the end, the unit that you are looking for on the left-hand side of the equation, is matched by the unit that is left over by canceling all other units on the right-hand side. The details of the method are shown below.

Step-by-step Dimensional Analysis:
The analysis is illustrated by the following example:

How many grams of NaCl is needed to make 150 mL solution of 2.00 M NaCl, given that the molar mass of NaCl is 58.4 g/mol?

• What is the question?
The first thing one has to know, obviously, is what units you are trying to get at the end. Hence, this becomes your question. Put unit next to the question mark on the left-hand side of the equation. At the end of a calculation, the unit is compared with the uncanceled unit on the right-hand side of the equation. In our example, we start our problem by writing:

It is important to write down not only the unit itself, but substance to which the unit belongs. So, we write g(NaCl), and not just g. You will see in later examples why it is important to include the substance.

Place the cursor over the equation to see what I mean!

• How do you start?
We have to equate the grams of NaCl on the left-hand side of the equation to something that is appropriate for us to start our calculation. What is appropriate to put on the right-hand side of the equation? Generally, in this kind of problem, there are many givens available for you. One group is conversion factors that are made of compounded units, such as g/mol, g/mL, or mol/L. Another group has one unit alone, and this is the one you want to start your calculation with. In our case, the only simple unit that is given is the volume of NaCl solution needed. Hence,

• The first conversion factor:
As it was mentioned in the introduction, the name of the game here is to cancel out the unit that you start with. This unit should appear in the denominator of the conversion factor. In our example, we would like to cancel the unit, mL(solution), by extending our equation with a conversion factor in parentheses. There are two givens with compounded units. They are 58.4 g/mol (or 58.4 g(NaCl)/mol(NaCl)) and 2.00 mol(NaCl)/1.00 L(solution) (or 2.00 mol(NaCl)/1000 mL(solution)) as a definition of 2.00 M NaCl. In order to cancel the mL(solution), we shall use the definition of 2.00 M NaCl solution. Therefore, we get

Check to see if the unit that was given, mL(solution), matches with the denominator of the 2.00 M definition. Again, place the cursor over the equation above.

• The second conversion factor:
We use the other conversion factor, 58.4 g/mol, to finish off this problem. By placing 58.4 g(NaCl) in the numerator and 1 mol(NaCl) in the denominator, we accomplished our mission.

When the product of the numerators is divided by the product of the denominators, you get an answer, 17.533 g(NaCl). The significant figure in this problem is three, therefore you should report your answer as 17.5 g(NaCl). So, you will need 17.5 g(NaCl) to make 150 mL solution of 2.00 M NaCl solution. Pretty cool, eh!

Additionally, there is no reason for us to keep 58.4 g(NaCl) in the numerator in all calculations. It can be placed in the denominator as well, in which case your numerator is 1 mol(NaCl). The 58.4 g/mol really means 58.4 g(NaCl) = 1 mol(NaCl)!