(^94) › STEP 4. Review the Knowledge You Need to Score High
The theoretical yield is 49.5 g.
×=
37.5 gCaCl
49.478gCaCl
(^2) 100% 75.8%
2
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 Chapter 13, Solutions and Colligative Properties, but solu-
tion stoichiometry is so common on the AP exam that we will discuss it here briefly also.
Solutions are 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 Chapter 13 we will cover several concentration 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:
=
(39.12gNaCl) [
1mol NaCl]
[58.45gNaCl]
(100.0mL) [1L]
[ 1000 mL]6.693MNaClKnowing 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:0.20 mol(NH)SO
L2mol NH
1mol (NH)SO(^4244) (0.100L) 0.040molNH
42 4
4
=
+
+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:=
(35.00mL)(0.1500molKOH)(1mol HPO)
(1,000mL)(2mol KOH)
(40.00mL)(1 L)
(1,000mL)0.06562 HPO
34
M 34