5 Steps to a 5 AP Chemistry

The theoretical yield is 49.5 g.

Note: All the units except % must cancel. This includes canceling g CaCl 2 with

g CaCl 2 , not simply g.

Molarity and Solution Calculations

We discuss solutions further in the chapter on solutions and colligative properties, but solu-

tion stoichiometry is so common on the AP exam that we will discuss it here briefly also.

Solutionsare homogeneous mixtures composed of a solute(substance present in smaller

amount) and a solvent(substance present in larger amount). If sodium chloride is dissolved

in water, the NaCl is the solute and the water the solvent.

One important aspect of solutions is their concentration, the amount of solute dissolved

in the solvent. In the chapter on solutions and colligative properties we will cover several con-

centration units, but for the purpose of stoichiometry, the only concentration unit we will use

at this time is molarity. Molarity (M) is defined as the moles of solute per liter of solution:

M=mol solute/L solution

Let’s start with a simple example of calculating molarity. A solution of NaCl contains

39.12 g of this compound in 100.0 mL of solution. Calculate the molarity of NaCl.

Answer:

Knowing the volume of the solution and the molarity allows you to calculate the moles or

grams of solute present.

Next, let’s see how we can use molarity to calculate moles. How many moles of ammo-

nium ions are in 0.100 L of a 0.20 M ammonium sulfate solution?

Answer:

Stoichiometry problems (including limiting-reactant problems) involving solutions can

be worked in the same fashion as before, except that the volume and molarity of the

solution must first be converted to moles.

If 35.00 mL of a 0.1500 M KOH solution is required to titrate 40.00 mL of

a phosphoric acid solution, what is the concentration of the acid? The reaction is:

2KOH (aq) +H 3 PO 4 (aq) →K 2 HPO 4 (aq) +2H 2 O (l)

Answer:

()

(. )( )

(

35.00 mL

1500 mol KOH 1 mol H PO

1000 mL

0 34

))( )

()

()

()

2molKOH

40.00 mL

1L

1000 mL

=0.06562 MHPO 34

0.20 mol (NH ) SO

L

2molNH

1mol(NH )SO

42 4 +

42

⎡ 4

⎣

⎢

⎤

⎦

⎥

44

(0.100 L) = 0.040 mol NH 4 +

⎡

⎣

⎢

⎤

⎦

⎥

(39.12 g NaCl)

1molNaCl

58.45 g NaCl

(1

⎡⎣ ⎤⎦

⎡⎣ ⎤⎦

000.0 mL)

1L

1000 mL

6.693 M NaCl

⎡⎣ ⎤⎦

⎡⎣ ⎤⎦

=

37.5 g CaCl

49.478 g CaCl

2

2

×= 100 %.%75 8

94  STEP 4. Review the Knowledge You Need to Score High