CONCENTRATION UNITS
This handout will deal with units of concentration and how to convert from one
concentration unit to another. It will be important to understand a few terms dealing with
solutions, so let's define them:
Solution -- a mixture consisting of a solute and a solvent
Solute -- component of a solution present in the lesser amount
Solvent -- component of a solution present in the greater amount
Concentration -- amount of a solute present in a solution per standard amount of solvent
There are numerous ways of expressing concentrations. It will be important to
know the units used to express each concentration, as these units essentially define the
concentration. Let's look at some ways to express concentration.
Weight/Weight Percent (w/w%): This unit of concentration is often used for
concentrated solutions, typically acids and bases. If you were to look on a bottle of a
concentrated acid or base solution the concentration is often expressed as a weigh/weight
percent. A weight/weight percent is defined as:
grams ofsolute
x 100
grams of solution
Note: Mass percent is also another term used for weight/weight percent.
w/w % 
Molarity (M): This unit of concentration relates the moles of solute per liter of solution.
Molarity 
moles of solute
L solution
Molarity is the most common concentration unit involved in calculations dealing with
volumetric stoichiometry.
Molality (m): This unit of concentration relates the moles of solute per kilogram of
solvent.
moles of solute
Molality 
kg solvent
Molality is often used as the concentration unit involved in calculations dealing with
colligative properties, such as freezing point depression, boiling point elevation and
osmotic pressure.
Parts per million (ppm): This unit of concentration may be expressed in a number of
ways. It is often used to express the concentration of very dilute solutions. The
"technical" definition of parts per million is:
grams of solute
x 106
grams of solution
ppm 
Since the amount of solute relative to the amount of solvent is typically very small, the
density of the solution is to a first approximation the same as the density of the solvent.
For this reason, parts per million may also be expressed in the following two ways:
ppm 
mg solute
L solution
ppm 
mg solute
kg solution
Parts per billion (ppb): This concentration unit is also used for very dilute solutions.
The "technical" definition is as follows:
ppb 
grams solute
x 109
grams of solution
Owing to the dilute nature of the solution, once again, the density of the solution will be
about the same as the density of the solvent. Thus, we may also express parts per billion
as:
ppb 
g solute
L solution
ppb 
g solute
kg solution
Mole fraction (X) and mole percent (%X): A fraction is defined as a part over a whole.
Multiplying this fraction by 100 would give the percent. Thus, a mole fraction involves
knowing the moles of solute or component of interest over the total moles of all
components in the solution mixture:
Xsolute 
mol solute
total moles of all components
Xsolute% 
mol solute
x 100
total moles of all components
You may find it necessary to be able to convert from one concentration unit to
another. The key to solving this type of problem is to realize that you may make an
assumption to get started. You may need to know the density of the solution, which
would be given in the problem. Then, by using dimensional analysis, you try to get to the
units of the concentration unit you are seeking to find. To get started, assume the quantity
of solution found in the denominator unit of the concentration unit you are trying to
convert. For example, if you are trying to convert weight/weight percent to molarity,
assume 100 grams of solution. If you are trying to convert molarity to weight/weight
percent, assume 1 liter of solution.
Let's look at a typical example. Suppose you are given a concentrated solution of
HCl which is known to be 37.0% HCl and has a solution density of 1.19 g/mL. What is
the molarity, molality and mole fraction of HCl?
Begin with the assumption of 100 g of solution. With this assumption, you now
know a few other facts. In 100 g of solution, 37.0 g is due to HCl (grams of solute) and
63.0 g is due to water (grams of solvent).
To find molarity, we need to determine the moles of HCl (solute) per liter of solution.
First, convert the known amount of HCl (37.0 g) to moles:
mol HCl  37.0 g HCl x
1 mol HCl
 1.01 mol HCl
36.5 g HCl
Next, convert the known mass of solution, 100 g solution, to liters of solution, using the
density of the solution:
L soln  100 g solution x
1 mL solution
1 L solution
x
 0.0840 L solution
1.19 g solution 1000 mL solution
Since the moles of solute (HCl) and volume of solution in liters is now know, calculate
the molarity (M) as the moles of solute per liter of solution:
M 
1.01 mol HCl
 12.0 mol HCl/ Lsolution  12.0 M HCl
0.0849 L solution
From the information above, let's find the molality of the HCl solution. The moles of
solute are already known (1.01 mol HCl). We need to find the kilograms of solvent:
63.0 g H2O x
1 kg H2O
 0.0630 kg H2O
1000 g H2O
Since molality (m) is defined as the moles of solute per kilogram solvent, it becomes easy
to find the molality:
m 
1.01 mol HCl
 16.0 mol HCl/kg H2O  16.0 m H2O
0.0630 kg H2O
Finally, let's tackle the mole fraction of HCl. The moles of HCl is known to be 1.01 mole.
We need to find the moles of H2O:
mol H2O  63.0 g H2O x
1 mol H2O
 3.50 mol H2O
18.0 g H2O
Since the moles of solute (HCl) and moles of solvent (H2O) are known, the mole fraction
of HCl may be calculated:
XHCl 
1.01 mol HCl
 0.244
1.01 mol HCl  3.50 mol H2O
Problems
1. If 15.0 grams of sodium sulfate is dissolved in water to form 400.0 grams of water. The
density of the solution is 1.056 g/mL at 25 °C.
a.
b.
c.
d.
e.
f.
mass percent (w/w%) of sodium sulfate
molarity of sodium sulfate
molality of sodium sulfate
parts per thousand sodium sulfate
parts per million sodium ion
mole fraction sulfate ion
2. A 12.0 M solution of HCl has a density of 1.107 g/mL at 25° C. Calculate:
a. mass percent (w/w%) of HCl
b. molality
c. mole fraction HCl
3. Given a 0.300 m solution of calcium nitrate with a density of 1.009 g/mL at 25°C, calculate:
a.
b.
c.
d.
e.
f.
g.
mass percent (w/w%) of calcium nitrate
mass percent (w/w%) of nitrate ion
mass percent (w/w%) of calcium ion
molarity of calcium nitrate
molarity of nitrate ion
mole fraction of calcium ion
mole fraction of nitrate ion
4. Given a solution containing magnesium chloride with a concentration of 25.0 ppm magnesium ion and
a solution density of 1.00 g/mL at 25 °C, calculate:
a.
b.
c.
d.
e.
f.
g.
mass percent (w/w%) of magnesium ion
mass percent (w/w%) of chloride ion
mass percent (w/w%) of magnesium chloride
molarity of magnesium chloride
molarity of magnesium ion
molarity of chloride ion
molality of magnesium chloride
Answers:
1 a. 3.61% Na2SO4
3. a 4.69% Ca(NO3)2
4. a. 0.00250 %Mg2+
b.
c.
d.
e.
f.
0.270 M Na2SO4
0.265 m Na2SO4
36.1 ppt Na2SO4
1.24 x 104 ppm Na+
0.0117 = mole fraction of
sulfate ion
2 a. 39.5% HCl
b. 17.4 m HCl
c. 4.18 = mole fraction of HCl
b. 3.55% NO3¯
c. 1.14 %Ca2+
d. 0.288 M Ca2+
e. 0.576 M NO3¯
f. 0.00532 = mole fraction of
calcium ion
g. 0.0106 = mole fraction of
nitrate ion
b.
c.
d.
e.
f.
g.
0.00729 % Cl¯
0.00929 %MgCl2
0.00103 M MgCl2
0.00103 M Mg2+
0.00206 M Cl¯
0.00103 m MgCl2
Answers
1 a. The formula for sodium sulfate is Na2SO4.
grams Na2 SO4
w/w% Na2 SO4 =
x 100
grams solution
The mass of solute is 15.0 g Na2SO4. The mass of solvent (water) is 400.0 g, so the mass of solution is:
15.0 g Na2SO4 + 400.0 g H2O = 415.0 g solution
w/w% Na2 SO4 =
15. 0 g Na2 SO4
x 100 = 3.61% Na2 SO4
415.0 g solution
1 b. The molar mass of Na2SO4. must be calculated to find the moles of Na2SO4.
Molar mass of Na2SO4. = 2 AW of Na + AW of S + 4 AW of O (where AW = atomic weight)
= 2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.05 g Na2SO4./mol Na2SO4.
Find the moles of Na2SO4.:
15.0 g Na2 SO4 x
1 mole Na2 SO4
= 0.106 mol Na2 SO4
142.05 g Na2 SO4
Find the volume of solution in liters. Recall from part a, the mass of solution is 415.0g.
415.0 g solution x
1 mL solution
1 L solution
x
= 0.3930 L solution
1.056 g solution 1000 mL solution
Calculate the molarity:
Molarity of Na2 SO4 =
1 c.
0.106 mol Na2 SO4
= 0.270 M Na2 SO4
0.3930 L solution
The moles of Na2SO4 is already known from part b. Find the kilograms of solvent (water):
400.0 g H2 O x
Calculate the molality of Na2SO4 :
1 kg H2 O
= 0.4000 kg H2 O
1000 g H2 O
molality of Na2 SO4 =
0.106 mol Na2 SO4
= 0.265 m Na2 SO4
0.4000 kg water
1 d. Calculate the parts per thousand of Na2SO4 from the mass of solute and mass of solution:
ppt Na2 SO4 =
15.0 g Na2 SO4
x 1000 = 36.1 ppt Na2 SO4
415 g solution
1 e. Parts per million may also be expressed as mg solute/L solution. In this case, converting the
molarity to ppm will be the simplest route:
0.270 mol Na2 SO4
2 mol Na+
22.99 g Na+ 1000 mg Na+
x
x
x
= 1.24 x 104 ppm Na+
L solution
1 mol Na2 SO4
1 mol Na+
1 g Na+
1 e. For the mole fraction of sulfate, we need to have:
mole fraction SO2−
=
4
Find the moles of Na+ :
0.106mol Na2 SO4 x
mol SO2−
4
mol SO2−
4
+ mol Na+ + mol H2 O
2 mol Na+
= 0.212 mol Na+
1 mol Na2 SO4
Find the moles of SO42- :
0.106 mol Na2 SO4 x
1 mol SO2−
4
= 0.106 mol SO2−
4
1 mol Na2 SO4
Find the moles of H2O:
400.0 g H2 O x
1 mol H2 O
= 22.20 mol H2 O
18.02 g H2 O
Calculate the mole fraction of sulfate:
X of SO2−
4 =
0.212 mol Na+
0.106 mol SO2−
4
= 0.00471
+ 0.106 mol SO2−
+ 22.20 mol H2 O
4
2 a. Assume exactly 1 liter (1000 mL) of solution, which is the quantity present in the denominator of the
known concentration unit (molarity). This also means you have 12.0 mol HCl present. Convert the
moles of HCl to a mass of HCl:
36.46 g HCl
12.0 mol HCl x
= 437.5 g HCl
1 mol HCl
Find the mass of solution using the density of the solution:
1000 mL solution x
Calculate the mass percent of HCl:
1.107 g solution
= 1107 g solution
1 mL solution
2 b.
437.5 g HCl
x 100 = 39.5 %HCl
1107 g solution
Find the mass of solvent:
1107 g solution – 437.5 g HCl = 670 g H2O
Convert the mass of solvent to kilograms:
1 kg H2 O
= 0.670 kg H2 O
1000 g H2 O
670g H2 O x
Calculate the molality:
12.0 mol HCl
= 17.9 m HCl
0.670 kg H2 O
2 c.
Find the moles of H2O:
670 g H2 O x
1 mol H2 O
= 37.2 mol H2 O
18.02 g H2 O
Calculate the mole fraction of HCl:
X of HCl =
12.0 mol HCl
= 0.244
12.0 mol HCl + 37.2 mol H2 O
3 a. Assume exactly 1 kg (1000 g) of water (solvent), which is the quantity present in the denominator of
the known concentration unit (molality). This also means you have 0.300 mol Ca(NO3)2.
To find the mass of calcium nitrate, you need the molar mass of Ca(NO3)2:
Molar mass of Ca(NO3)2 = AW of Ca + 2 AW of N + 6 AW of O (where AW = atomic weight)
= 40.08 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol) = 164.10 g Ca(NO3)2/mol Ca(NO3)2
Calculate the mass of Ca(NO3)2 :
0.300 mol Ca(NO3 )2 x
164.10 g Ca(NO3 )2
= 49.2 g Ca(NO3 )2
1 mol Ca(NO3 )2
Find the mass of solution:
mass of solution = 49.2 g Ca(NO3)2 + 1000.0 g H2O = 1049.2 g solution
Calculate the mass percent of Ca(NO3)2 :
49.2 g Ca(NO3 )2
x 100 = 4.69 %Ca(NO3 )2
1049.2 g solution
3 b.
To find the mass of nitrate ion, you need the molar mass of nitrate ion (NO3¯ ):
Molar mass of NO3¯ = AW of N + 3 AW of O
(where AW = atomic weight)
= 14.01 g/mol + 3(16.00 g/mol) = 62.01 g NO3¯/mol NO3¯
Calculate the mass of nitrate ion:
0.300 mol Ca(NO3 )2 x
2 mol NO−
62.01 g NO−
3
3
x
= 37.2 g NO−
3
1 mol Ca(NO3 )2
1 mol NO−
3
Calculate the mass percent of nitrate ion:
37.2 g NO−
3
x 100 = 3.55 %NO−
3
1049.2 g Ca(NO3 )2
3 c. Find the mass of calcium ion:
0.300 mol Ca(NO3 )2 x
1 mol Ca2+
40.08 g Ca2+
x
= 12.0 g Ca2+
1 mol Ca(NO3 )2
1 mol Ca2+
Calculate the mass percent of calcium ion:
12. 0 g Ca2+
x 100 = 1.14 %Ca2+
1049.2 g Ca(NO3 )2
Notice how the sum of %Ca2+ and %NO3¯ sum to the %Ca(NO3)2.
3 d. Convert the mass of solution to volume of solution in liters:
1049.2 g solution x
1 mL solution
1 L solution
x
= 1.040 L solution
1.009 g solution 1000 mL solution
Calculate the molarity:
0.300 mol Ca(NO3 )2
= 0.288 M Ca(NO3 )2
1.040 L solution
3 e. Calculate the molarity of nitrate ions:
0.288 mol Ca(NO3 )2
2 mol NO−
3
x
= 0.576 M NO−
3
1 L solution
1 mol Ca(NO3 )2
3 f. Find the moles of calcium ion:
0.300 mol Ca(NO3 )2 x
1 mol Ca2+
= 0.300 mol Ca2+
1 mol Ca(NO3 )2
Find the moles of nitrate ion:
2 mol NO−
3
0.300 mol Ca(NO3 )2 x
= 0.600 mol NO−
3
1 mol Ca(NO3 )2
Find the moles of water:
1 mol H2 O
= 55.49 mol H2 O
18.02 g H2 O
Calculate the mole fraction of calcium ion:
1000 g H2 O x
0.300 mol Ca2+
0.300 mol Ca2+
= 0.00532
+ 0.600 mol NO−
3 + 55.49 mol H2 O
3 g. Calculate the mole fraction of nitrate ion:
0.300 mol Ca2+
0.600 mol NO−
3
= 0.0106
+ 0.600 mol NO−
3 + 55.49 mol H2 O
4. Recall that parts per million may also be expressed as mg solute/liter of solution or mg solute per
kilograms solution. Either case will work here because the density of the solution is given as 1.00 g/mL.
Assume exactly 1 liter of solution (1000 g solution) so that the mass of solute (25.0 mg of magnesium
ion) is known.
4 a. Convert milligrams of magnesium ion to grams magnesium ion:
1 x 10−3 g Mg 2+
= 0.0250 g Mg 2+
1 mg Mg 2+
25.0 mg Mg 2+ x
Calculate the w/w% magnesium ion:
0.0250 g Mg 2+
x 100 = 0.00250 %Mg 2+
1000 g solution
4 b. Find the number of grams of chloride ion keeping in mind the chemical formula of magnesium
chloride is MgCl2:
0.0250 g Mg 2+ x
1 mol Mg 2+
2 mol Cl−
35.45 g Cl−
x
x
= 0.0729 g Cl−
24.30 g Mg 2+ 1 mol Mg 2+
1 mol Cl−
Calculate the w/w% of Cl¯:
0.0729 g Cl−
x 100 = 0.00729 %Cl−
1000 g solution
4 c. Find the number of grams of magnesium chloride:
0.0250 g Mg 2+ x
1 mol Mg2+ 1 mol MgCl2 95.20 g MgCl2
x
x
= 0.0979 g MgCl2
24.30 g Mg 2+ 1 mol Mg2+
1 mol MgCl2
Calculate the w/w % MgCl2:
0.0929 g MgCl2
x 100 = 0.00929 %MgCl2
1000 g solution
4 d. Find the moles of magnesium chloride:
0.0979 g MgCl2 x
1 mol MgCl2
= 0.00103 mol MgCl2
95.20 g MgCl2
Calculate the molarity of MgCl2 :
0.00103 mol MgCl2
= 0.00103 M MgCl2
1 L solution
4 e. Calculate the molarity of magnesium ions:
0.00103 mol MgCl2 1 mol Mg 2+
x
= 0.00103 M Mg2+
1 L solution
1 mol MgCl2
4 f. Calculate the molarity of chloride ions:
0.00103mol MgCl2
2mol Cl−
x
= 0.00206 mol Cl−
1 L solution
1 mol MgCl2
4 g. Find the mass of water:
1000 g solution – 0.0979 g MgCl2 = 999.9021 g H2O
Convert this mass to kilograms:
999.9021 g H2 O x
1 kg H2 O
= 0.9999021 kg H2 O
1000 g H2 O
Calculate the molality of MgCl2:
0.00103 mol MgCl2
= 0.00103 m MgCl2
0.9999021 kg H2 O