Most chemical reactions are performed in solution. Therefore we need a method of stating concentrations and there are several ways to do this.

**I. Molarity**

One way would be to give the number of moles of compound per liter of solution and to do this you would divide the number of moles by the number of liters so that you would have the units of moles/liter. This is called molarity and given the symbol M where:

M = (moles of compound)/(liters of solution)

Note that this liters of solution, not liters of solvent. The solute also occupies volume and we can't ignore it. So let's work an example. What is the molar concentration (or what is the molarity) of a 200. ml solution of 26.3 g of NaHCO_{3} and
H_{2}O? The volume of the solution is 0.200 L so we need to know the moles of the compound. The solute has a molecular mass of 84.0 g/mol so the number of moles is:

26.3 g NaHCO_{3} * (1 mol/84.0 g) = 0.313 mol NaHCO_{3}

The molarity of the solution is then

M = 0.313 mol NaHCO_{3}/(0.200 L) = 1.57 M or we say that the solution is 1.57 M.

We can also state the ion concentrations. Since NaHCO_{3} is an ionic compound, it dissociates in water into it's constituent ions:

NaHCO_{3} Na^{+} + HCO_{3}^{-}

The solution is then 1.57 M in Na^{+} and it is 1.57 M in HCO_{3}^{-}.

What are the ion concentrations in a 0.50 M solution of (NH_{4})_{2}SO_{4}?

Well,

(NH_{4})_{2}SO_{4} 2 NH_{4}^{+} + SO_{4}>^{-2}

So the solution is 1.0 M in NH_{4}^{+} and it is 0.50 M in SO_{4}^{-2}.

Now let's work a problem from the opposite direction. How many grams of Na_{2}CO_{3} are needed to make 2.0 L of a 1.5 M solution of Na_{2}CO_{3}? The first thing to do is to determine the number of moles of
Na_{2}CO_{3} which would be in such a solution:

2.0 L of solution * (1.5 mol Na_{2}CO_{3})/(1.0 L of solution)

= 3.0 mol Na_{2}CO_{3}

So now we need to know how many grams is needed to make 3.0 moles of Na_{2}CO_{3}:

3.0 mol * (106 g/mol) = 3.2 x 10^{2} Na_{2}CO_{3}

Fini!

Work another example. Your boss asks you to prepare 500. ml of a 0.200 M solution of KMnO_{4}. How do you do it and keep your job (do it correctly)?

First realize from the definition of molarity (M = # moles/L solution) we can solve for the number of moles of the compound needed: # moles = M * L solution so for this problem

# moles = .200 moles/(liter of solution) * 0.500 liters of solution = 0.100 mols of compound

So now we know the moles of KMnO_{4} needed so we can calculate the number of grams needed:

0.100 mol KMnO_{4} * (158 g/mol) = 15.8 g of KMnO_{4}

**Dilution**

Laboratories will often store large quantities of common solutions of a stated concentration and you need to make up a small volume of the solution to a different concentration. So it is important for you to know how to do this. The principle used is that the number of moles of the solute will be the same in both containers. And from our definition of molarity, how do we obtain the number of moles? # moles = M * V where V is the volume of the solution and M is the molarity. So for two solutions containing the same number of moles of solute:

# moles of solute in solution 1 = # moles of solute in solution 2 so

M_{1} * V_{1} = M_{2} * V_{2}

So the above is our working formula for dilution problems.

Let's work an example. You need 300. ml of a 1.00 M NaOH solution and your lab has a large bottle of 3.00 M NaOH. How do you obtain what you need? Both solutions must contain the same number of moles of NaOH so

M_{1} * V_{1} = M_{2} * V_{2} is our working formula where M_{1} = 1.00, V_{1} = 0.300 L, M_{2} = 3.00, and V_{2} is our unknown (what volume of the 0.3 M solution has the same number of
moles of NaOH as 300 ml of a 1.00 M solution). Solving for V_{2} gives

V_{2} = M_{1} * V_{1}/M_{2}

= 1.00 * 0.300/3.00 = 0.100 liters

So we need to take 100 ml of the 3.00 M solution(this will give us the number of moles of NaOH needed) and add 200 ml of water and you will have 300 mol of a 1.00 M NaOH solution. You keep your job!

Practice, practice, practice.

Here is an excellent practice site.

This is also an excellent place to practice.

**II. Molality**

Molality is another way of expressing solution concentrations. The molality (m) is defined as

Molality = m = moles of solute/kilogram of solvent

The numerator is the same as in molarity, but the denominator is in weight instead of volume and it is in terms of solvent rather than solution.

**III. Mass percent**

Mass percent (sometimes called weight percent) is the percent by mass of the solute in solution:

Mass percent = (mass of solute/mass of solution) * 100

**IV. Mole fraction **

The mole fraction is the number of moles of one of the solution components divided by the total number of moles of the solution.

Mole fraction of A = moles of A/(total number of moles of solution) or for a binary solution (A and B):

X_{A} = n_{A}/(n_{A} + n_{B})

Practice Problem:

You make a solution of 50.0 g of CsCl in 50.0 g of water. The total volume of the solution is 63.3 ml. Calculate the mass percent, molarity, molality, and mole fraction of CsCl. The answers are 50.0 %, 4.69 M, 5.94 m, and 0.0968.

Here is a place to go practice and after working there for awhile go to this site for further practice.

Now take a practice quiz to help you understand if you understand the basic concepts. |

You must use your real name when it asks for a name. |

The test will only submit when you have answers all of the questions correctly. |

If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests. |