CHM 235 – Dr. Skrabal
Brief review of important concepts
for quantitative analysis
 Some important units of quantification
 Units for expressing concentrations in solids and liquids
 Concentration-dilution formula
Fundamental SI units
Remember the correct abbreviations!
Mass
kilogram (kg)
Volume
liter (L)
Distance
meter (m)
Temperature
kelvin (K)
Time
second (s)
Current
ampere (A)
Amount of substance
mole (mol)
Luminous intensity
candela (cd)
Some other SI and non-SI units
Length
angstrom (Å)
Force
newton (N)
Pressure
pascal (Pa), atmosphere (atm)
Energy, work, heat
joule (J)
Power
watt (W)
Electric charge
coulomb (C)
Electric potential
volt (V)
Electric resistance
ohm ()
Electric capacitance
farad (F)
Temperature
degree Celsius (°C)
degree Fahrenheit (°F)
Some common prefixes for
exponential notation
Remember the correct
abbreviations!
1012
tera (T)
109
giga (G)
106
mega (M)
103
kilo (k)
10-1
deci (d)
10-2
centi (c)
10-3
milli (m)
10-6
micro (μ)
10-9
nano (n)
10-12
pico (p)
10-15
femto (f)
10-18
atto (a)
Commonly used equalities
103 mg = 1 g
1 mg = 10-3 g
milli = thousandth
106 μg = 1 g
1 μg = 10-6 g
micro = millionth
109 ng = 1 g
1 ng = 10-9 g
nano = billionth
1012 pg = 1 g
1 pg = 10-12 g
pico = trillionth
Concentration scales
 Molarity (M) =
Moles of solute
Liter of solution
 Molality (m) =
Moles of solute
kg solvent
• Molarity is a temperature-dependent scale because
volume (and density) change with temperature.
• Molality is a temperature-independent scale because
the mass of a kilogram does not vary with temperature.
Concentration scales (cont.)
 Formality (F) =
Moles of solute (regardless of chemical form)
Liter of solution
Formality is sometimes used to express the
concentration of substances, such as electrolytes, acids,
and bases, that turn into different species in solution.
For example:
• 0.1 M NaCl gives 0.1 M Na+ and 0.1 M Cl- in solution
• 0.5 M HCl gives 0.5 M H+ and 0.5 M Cl- in solution
Based on D.Harvey (2000) Modern Analytical Chemistry
Concentration scales (cont.)
Weight / weight (w/w) basis
% (w/w) =
 mass solute ( g )  2

10
mass
sample
(
g
)


 percent
ppt (w/w) =
 mass solute ( g )  3

10
 mass sample ( g ) 
 ppt = parts per thousand
ppm (w/w) =
 mass solute ( g )  6

10  ppt = parts per million
mass
sample
(
g
)


ppb (w/w) =
 mass solute ( g )  9

10  ppt = parts per billion
 mass sample ( g ) 
ppt (w/w) =
 mass solute ( g )  12

10  ppt = parts per trillion
 mass sample ( g ) 
This scale is useful for solids or solutions.
Concentration scales (cont.)
Weight / volume (w/v) basis
% (w/v) =
 mass solute ( g )  2

10  percent
 vol. sample (mL) 
ppt (w/v) =
 mass solute ( g )  3

10
 vol. sample (mL) 
ppm (w/v) =
 mass solute ( g )  6

10  ppt = parts per million
vol
.
sample
(
mL
)


ppb (w/v) =
 mass solute ( g )  9

10  ppt = parts per billion
 vol. sample (mL) 
ppt (w/v) =
 mass solute ( g )  12

10  ppt = parts per trillion
 vol. sample (mL) 
 ppt = parts per thousand
Concentration scales (cont.)
Volume / volume (v/v) basis
% (v/v) =
 vol. solute (mL)  2

10
vol
.
sample
(
mL
)


 percent
ppt (v/v) =
 vol. solute (mL)  3

10
 vol. sample (mL) 
 ppt = parts per thousand
ppm (v/v) =
 vol. solute (mL)  6

10
vol
.
sample
(
mL
)


 ppt = parts per million
ppb (v/v) =
 vol. solute (mL)  9

10
 vol. sample (mL) 
 ppt = parts per billion
ppt (v/v) =
 vol. solute (mL)  12

10
 vol. sample (mL) 
 ppt = parts per trillion
Concentration examples
 Concentrated HCl
 Alcoholic beverage
 Color indicator for
titrations
 37.0 g HCl  2

10
 100.0 g solution 
 4.00 mL CH 3CH 2OH

 38.5 mL beverage
 37.0 % ( w / w)
 2
10

 0.050 g phenolphthalein  2

10
50.0 mL solution


 10.4 % (v / v)
 0.10 % ( w / v)
Concentration example: %(w:v)
What is the concentration, on a %(w:v) basis, of V in a
solution that contains 281.5 mg/L of V?
 mass solute ( g )  2
10
% ( w : v)  
 vol. sample (mL) 
 281.5 mg V
% ( w : v)  
L

 1 g V

 1000 mg V
 1L  2
 
10
  1000 mL 
 0.02815 % ( w : v) or 2.815 x 10  2 % ( w : v)
Concentration scales (cont.)
• Parts per million, billion, trillion are very often used to
denote concentrations of aqueous solutions:
 1 g solute
1 ppm   6
 10 g solution

 103 mg  1 g solution  1000 mL solution 




 1 g  1 mL solution  1 L solution 

 1
 1 g solute
1 ppb   9
 10 g solution

 106 g  1 g solution  1000 mL solution 




 1 g  1 mL solution  1 L solution 

 1
 1 g solute
  9
 10 g solution

 109 ng  1 g solution  1000 mL solution 




 1 g  1 mL solution  1 L solution 

 1
1 ppt
Note: ppt = parts per trillion
mg
L
g
L
ng
L
Concentration scales (cont.)
It will be very useful to memorize:
 1 part per million (ppm) = 1 mg / L
 1 part per billion (ppb) = 1 μg / L
 1 part per trillion (ppt) = 1 ng / L
Concentration examples
Conversion of molarity to ppm
Solution of 0.02500 M K2SO4
mg K 2 SO4
 0.02500 mol K 2 SO4  174.26 g K 2 SO4  1000 mg 





4356




L
g
L

 mol K 2 SO4 

 4356 ppm
Concentration examples
What is concentration (in ppm) of K+ in this solution?
Solution of 0.02500 M K2SO4

mg K 
 0.02500 mol K 2 SO4  2 mol K  39.10 g  1000 mg 


  1955


 
L
g
L

 1 mol K 2 SO4  mol K 

 1955 ppm
Concentration-dilution formula
A very versatile formula that you
absolutely must know how to use
• C1 V1 = C2 V2
where C = conc.; V = volume
• M1 V1 = M2 V2
where M = molarity
• Cconc Vconc = Cdil Vdil
where “conc” refers to the more concentrated solution
and “dil” refers to the more dilute solution
Concentration-dilution formula example
Problem: You have available 12.0 M HCl (conc. HCl) and
wish to prepare 0.500 L of 0.750 M HCl for use in an
experiment. How do you prepare such a solution?
Cconc Vconc = Cdil Vdil
Write down what you know and what you don’t know:
Concentration-dilution formula example
Problem: You have available 12.0 M HCl (conc. HCl) and
wish to prepare 0.500 L of 0.750 M HCl for use in an
experiment. How do you prepare such a solution?
Cconc Vconc = Cdil Vdil
Cconc = 12.0 mol L-1
Vconc = ?
Cdil = 0.750 mol L-1
Vdil = 0.500 L
Vconc = (Cdil)(Vdil) / Cconc
Vconc = (0.750 mol L-1) (0.500 L) / 12.0 mol L-1
Vconc = 3.12 x 10-2 L = 31.2 mL
Concentration-dilution formula example
Great! So how do you prepare this solution of 0.750 M HCl?
1.
2.
3.
4.
5.
Use a pipet or graduated cylinder to measure exactly
31.2 mL of 12.0 M
Transfer the 31.2 mL of 12.0 M HCl to a 500.0 mL
volumetric flask
Gradually add deionized water to the volumetric flask
and swirl to mix the solution
As the solution gets close to the 500.0 mL graduation on
the flask, use a dropper or squeeze bottle to add water
to the mark
Put the stopper on the flask and invert ~20 times to mix