Concentration Terms and Calculations
Concentration can be expressed in various ways depending on which form is the most useful for
the given solution. In chemical laboratory, molarity (M) is the most common way to express
concentration. Molarity is defined as moles of solute per liter of solution. Note that a solution is
composed of the solvent and solute. Molarity may be expressed in equation form as
Moles of solute may be found by dividing solute mass by its molar mass.
In the medical field, intravenous solutions are sometimes labeled using m/v% concentration.
This concentration term expresses the mass of solute in grams per 100 mL of solution. The equation
form may be written
The alcoholic beverage industry uses v/v% to express the amount of ethanol in their products.
This concentration term is similar to m/v% except the volume of solute in mL replaces the mass. In
equation form this becomes
In the case of environmental pollution, much smaller concentration terms are needed. Two of
the most common ones are parts per million (ppm) and parts per billion (ppb). Both terms express the
amount of solute per million or billion parts of solution. The solute units and solvent units must be
identical when doing the calculation.
If the solvent is water and parts are grams, ppm may be redefined in mass/volume units by using 1.00
g/mL for the density of a dilute aqueous solution.
Similarly, ppb becomes µg solute/L solution by substituting 109 for 106 and 106 µg for 1000 mg in the
above equation.
How to calculate the molarity of a solution from m/v%
The normal level of glucose in the blood usually averages 100. mg/100. mL. Calculate the average
molarity of glucose in blood.
Solution:
Write down the given values with their
units and what they represent.
In this problem, mg/v% is given for the solute glucose and
the solvent blood. The chemical formula for glucose is
C6H12O6. The molar masses of C, H, and O can be obtained
from the Periodic Table.
mg/v% = 100. mg glucose/100. mL blood
1 mol C = 12.01 g/mol
1 mol O = 16.00 g/mol
1 mol H – 1.008 g/mol
molar mass glucose = 180.16 g/mol
Write down the unknown and its unit.
The molarity of the glucose in blood is unknown.
M glucose = ?
Develop a strategy to solve the problem.
The definition of molarity is
A dimensional analysis problem is set up to give mol
glucose/L blood.
Round off the answer to the correct
number of significant digits and give the
appropriate units.
The mg/v% has 3 significant digits while the molar mass has
5. The molarity is rounded off to 3 significant digits. The
conversion factors for mg and mL are metric definitions and
do not affect the number of significant digits in the answer.
The answer is
M = 0.00555 M glucose or 5.55 x 10-3 M glucose
Check whether the answer is reasonable.
The mass of C6H12O6 is 0.0005 the molar mass of glucose
in 0.1 L of blood; the answer is reasonable.
How to calculate the m/v% of a solution from molarity
A certain pre-mixed antifreeze solution is 8.45 M ethylene glycol. What is the m/v% for ethylene glycol
in this mixture?
Solution:
Write down the given values with their
units and what they represent.
In this problem, molarity is given for the solute ethylene
glycol and the solvent water. The chemical formula for
ethylene glycol is C2H6O2. The molar masses of C, H, and O
can be obtained from the Periodic Table.
mg/v% = 100. mg glucose/100. mL blood
1 mol C = 12.01 g/mol
1 mol O = 16.00 g/mol
1 mol H = 1.008 g/mol
molar mass ethylene glycol = 62.07 g/mol
Write down the unknown and its unit.
The m/v% of ethylene glycol in antifreeze is unknown.
m/v% ethylene glycol = ?
Develop a strategy to solve the problem.
The definition of m/v% is
A dimensional analysis problem is set up to give g
C2H6O2/100 mL antifreeze.
Round off the answer to the correct
number of significant digits and give the
appropriate units.
The molarity has 3 significant digits while the molar mass
has 4. The m/v% is rounded off to 3 significant digits. The
answer is
m/v% = 52.4 m/v% ethylene glycol
Check whether the answer is reasonable.
The moles of C2H6O2 in 100 mL is  0.1 that in a liter. The
mass of C2H6O2 in 100 mL antifreeze is  0.8 its molar mass;
the answer is reasonable.
How to calculate grams of solute needed to make a solution
The average molarity of sucrose in (C12H22O11) in a soft drink can (355 mL) is 0.37 M. Calculate the
average grams of sucrose needed per can of soft drink.
Solution:
Write down the given values with
their units and what they represent.
In this problem, molarity is given for the solute sucrose and the
solvent water. The chemical formula for sucrose is C12H22O11. The
molar masses of C, H, and O can be obtained from the Periodic
Table.
M sucrose = 0.37 mol sucrose/L soda
V soda = 355 mL
1 mol C = 12.01 g/mol;
1 mol O = 16.00 g/mol
1 mol H – 1.008 g/mol
molar mass sucrose = 342.30 g/mol
Write down the unknown and its
unit.
The grams of sucrose in a soda can is unknown.
g sucrose = ?
Develop a strategy to solve the
problem.
The definition of molarity is
A dimensional analysis problem is set up to give g sucrose in 355
mL soda.
Round off the answer to the correct The molarity has 2 significant digits while the volume has 3 and
number of significant digits and give molar mass 5. The mass is rounded off to 2 significant digits. The
the appropriate units.
answer is
mass = 45 g sucrose
Check whether the answer is
reasonable.
The volume of a soda can is 1/3 L.  0.1 moles of sucrose is
needed; the answer is reasonable.
How to calculate grams of solute in a given volume of solution
A physiological saline solution is 0.154 M NaCl. How many grams of sodium chloride are in 500. mL of
the solution?
Solution:
Write down the given values with their
units and what they represent.
In this problem, molarity is given for the solute NaCl and
the solvent water. The molar masses of Na and Cl can be
obtained from the Periodic Table.
M NaCl = 0..154 mol NaCl/L saline
V saline = 500. mL
1 mol Na = 22.99 g/mol
1 mol Cl = 35.45g/mol
molar mass NaCl = 58.44 g/mol
Write down the unknown and its unit.
The grams of NaCl in 500. mL of saline is unknown.
g NaCl = ?
Develop a strategy to solve the problem.
The definition of molarity is
A dimensional analysis problem is set up to give g NaCl in
500. mL saline.
Round off the answer to the correct
number of significant digits and give the
appropriate units.
The molarity and volume saline have 3 significant digits
while the molar mass has 4. The mass is rounded off to 3
significant digits. The answer is
mass = 4.50 g NaCl
Check whether the answer is reasonable.
The volume of saline is 0.5 L.  0.075 moles of NaCl would
be present; the answer is reasonable.