Section 6.2—Concentration
How do we indicate how much of the electrolytes are in the drink?
Concentrated versus Dilute
solute
Lower
concentration
Not as many solute (what’s
being dissolved) particles
solvent
Higher
concentration
More solute (what’s being
dissolved) particles
Concentration
Concentration gives the ratio of amount
dissolved to total amount
There are several ways to show
concentration
Percent Weight/Volume
 This is a method of showing concentration
that is not used as often in chemistry
 However, it’s used often in the food and
drink industry
For example, your diet drink can might say you
have less than 0.035 g of salt in 240 mL.
That would give you a concentration of
0.035 g / 240 mL, which is 0.015% solution
grams solute
%(W / V ) 
100
mL solvent
%(W/V) Example
grams solute
%(W / V ) 
100
mL solvent
Example:
If you dissolve 12 g
of sugar in 150 mL
of water, what
percent
weight/volume is
the solution?
%(W/V) Example
grams solute
%(W / V ) 
100
mL solvent
Example:
If you dissolve 12 g
of sugar in 150 mL
of water, what
percent
weight/volume is
the solution?
12 g
%(W / V ) 
100
150mL
8.0% (W/V)
Concentration using # of molecules
 When working with chemistry and molecules,
it’s more convenient to have a concentration
that represents the number of molecules of
solute rather than the mass (since they all
have different masses)
 Remember, we use moles as a way of
counting molecules in large numbers
Quick Mole Review
1 mole = 6.02 × 1023 molecules
The molecular mass of a molecule is
found by adding up all the atomic masses
in the atom
Molecular mass in grams = 1 mole of that
molecule
Quick Mole Example
Example:
How many
moles are
in 25.5 g
NaCl?
Quick Mole Example
Na 1  22.99 g/mole = 22.99 g/mole
Cl 1  35.45 g/mole = + 35.45 g/mole
58.44 g/mole
Example:
How many
moles are
in 25.5 g
NaCl?
25.5 g NaCl
1 mole NaCl molecules = 58.44 g
1
mole NaCl
58.44 g NaCl
0.44
= _______
mole NaCl
Molarity
 Molarity (M) is a concentration unit that uses
moles of the solute instead of the mass of the
solute
moles solute
M
L solvent
Molarity Example
Example:
If you dissolve 12 g
of NaCl in 150 mL
of water, what is
the molarity?
Molarity Example
Na 1  22.99 g/mole = 22.99 g/mole
Cl 1  35.45 g/mole = + 35.45 g/mole
58.44 g/mole
Example:
If you dissolve 12 g
of NaCl in 150 mL
of water, what is
the molarity?
12 g NaCl
1 mole NaCl molecules = 58.44 g
1
mole NaCl
0.21
= _______
mole NaCl
58.44 g NaCl
Remember to change mL to L! 150 mL of water = 0.150 L
moles solute
M
L solvent
M
0.21moles
0.150 L
1.4 M NaCl
Converting between the two
If you know the %(W/V), you know the
mass of the solute
You can convert that mass into moles
using molecular mass
You can then use the moles solute to find
molarity
Converting from % to M Example
Example:
What molarity is a
250 mL sample of
7.0 %(W/V) NaCl?
Converting from % to M Example
7.0%(W / V ) 
Example:
What molarity is a
250 mL sample of
7.0 %(W/V) NaCl?
?g
100
250mL
? = 17.5 g NaCl
Na 1  22.99 g/mole = 22.99 g/mole
Cl 1  35.45 g/mole = + 35.45 g/mole
58.44 g/mole
1 mole NaCl molecules = 58.44 g
17.5 g NaCl
1
mole NaCl
0.30
= _______
mole NaCl
58.44 g NaCl
Remember to change mL to L! 250 mL of water = 0.250 L
M
0.30moles
0.250 L
1.2 M NaCl
Concentration of Electrolytes
An electrolyte breaks up into ions when
dissolved in water
You have to take into account how the
compound breaks up to determine the
concentration of the ions
CaCl2  Ca+2 + 2 Cl-1
For every 1 CaCl2 unit that dissolves, you will produce 1 Ca+2 ion and
2 Cl-1 ions
If the concentration of CaCl2 is 0.25 M, the concentration of Ca+2 is 0.25
M and Cl-1 is 0.50 M
Let’s Practice #1
Example:
You want to make
200 mL of a 15%
(W/V) solution of
sugar. What mass
of sugar do you
need to add to the
water?
grams solute
%(W / V ) 
100
mL solvent
Let’s Practice #1
Example:
You want to make
200 mL of a 15%
(W/V) solution of
sugar. What mass
of sugar do you
need to add to the
water?
grams solute
%(W / V ) 
100
mL solvent
?g
15%(W / V ) 
100
200mL
30 g of sugar
Let’s Practice #2
Example:
What is the %(W/V)
of a 500. mL
sample of a 0.25 M
CaCl2 solution?
Let’s Practice #2
Example:
What is the %(W/V)
of a 500. mL
sample of a 0.25 M
CaCl2 solution?
0.25M 
? moles
0.500 L
? = 0.125 moles CaCl2
Ca 1  40.08 g/mole = 40.08 g/mole
Cl 2  35.45 g/mole = + 70.90 g/mole
110.98 g/mole
1 mole CaCl2 molecules = 110.98 g
0.125 moles CaCl2 110.98 g CaCl2
13.9
= _______
g CaCl2
1
mole CaCl2
13.9 g
%(W / V ) 
100
500mL
2.8 %(W/V) CaCl2
Let’s Practice #3
Example:
What are the
molarities of the ions
made in a 0.75 M
solution of Ca(NO3)2
Let’s Practice #3
Example:
What are the
molarities of the ions
made in a 0.75 M
solution of Ca(NO3)2
Ca(NO3)2  Ca+2 + 2 NO3-1
For every 1 Ca(NO3)2, there will be 1 Ca+2 and 2 NO3-1 ions
Ca+2 = 0.75 M
NO3-1 = 1.5 M