# Brief review of important concepts for quantitative analysis

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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
1012    tera (T)
109     giga (G)
Remember the correct   106     mega (M)
abbreviations!         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       milli = thousandth
1 mg = 10-3 g

106 μg = 1 g       micro = millionth
1 μg = 10-6 g

109 ng = 1 g       nano = billionth
1 ng = 10-9 g

1012 pg = 1 g      pico = trillionth
1 pg = 10-12 g
Concentration scales
Moles of solute
 Molarity (M) =
Liter of solution

Moles of solute
 Molality (m) =
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 (= 0.1 F NaCl) gives 0.1 M Na+ and 0.1 M
Cl- in solution
• 0.5 M HCl (= 0.5 F HCl) gives 0.5 M H+ and 0.5 M Cl- in
solution
Concentration scales (cont.)
Weight / weight (w/w) basis

 mass solute ( g )  2
 percent
% (w/w) =           
 mass sample ( g ) 10

                   

 mass solute ( g )  3
ppt (w/w) =          mass sample ( g ) 10
                         ppt = parts per thousand
                   

 mass solute ( g )  6
ppm (w/w) =          mass sample ( g ) 10  ppt = parts per million
                   
                   

 mass solute ( g )  9
ppb (w/w) =          mass sample ( g ) 10  ppt = parts per billion
                   
                   
 mass solute ( g )  12
ppt (w/w) =         
 mass sample ( g ) 10  ppt = parts per trillion

                   

This scale is useful for solids or solutions.
Concentration scales (cont.)
Weight / volume (w/v) basis

 mass solute ( g )  2
% (w/v) =      vol. sample (mL) 10  percent
                   
                   
 mass solute ( g )  3
ppt (w/v) =   
 vol. sample (mL) 10
      ppt = parts per thousand
                   

 mass solute ( g )  6
ppm (w/v) =    vol. sample (mL) 10  ppt = parts per million
                   
                   

 mass solute ( g )  9
ppb (w/v) =    vol. sample (mL) 10  ppt = parts per billion
                   
                   
 mass solute ( g )  12
ppt (w/v) =   
 vol. sample (mL) 10  ppt = parts per trillion

                   
Concentration scales (cont.)
Volume / volume (v/v) basis

 vol. solute (mL)  2
% (v/v) =     
 vol. sample (mL) 10
       percent
                  

 vol. solute (mL)  3
ppt (v/v) =    vol. sample (mL) 10
                         ppt = parts per thousand
                  

 vol. solute (mL)  6
ppm (v/v) =   
 vol. sample (mL) 10
       ppt = parts per million
                  

 vol. solute (mL)  9
ppb (v/v) =   
 vol. sample (mL) 10
       ppt = parts per billion
                  
 vol. solute (mL)  12
ppt (v/v) =   
 vol. sample (mL) 10
       ppt = parts per trillion
                  
Concentration examples

 Concentrated HCl         37.0 g HCl  2

 100.0 g solution 10
      37.0 % ( w / w)
                  

 4.00 mL CH 3CH 2OH  2
 Alcoholic beverage      
 38.5 mL beverage 10
         10.4 % (v / v)
                    

 Color indicator for    0.050 g phenolphthalein  2

                         10
       0.10 % ( w / v)
titrations                 50.0 mL solution    
Concentration example: %(w:v)

What is the concentration, on a %(w:v) basis, of vanadium
in a solution that contains 281.5 mg/L of vanadium?

 mass solute ( g )  2
 vol. sample (mL) 10
% ( w : v)                     
                   

 281.5 mg V    1 g V         1L  2
% ( w : v)                
 1000 mg V     1000 mL 10
          
     L                               

 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       103 mg  1 g solution  1000 mL solution          mg
1 ppm   6                                                           1
 10 g solution    1 g  1 mL solution  1 L solution 
                                
                                                                        L

 1 g solute       106 g  1 g solution  1000 mL solution          g
1 ppb   9                                                           1
 10 g solution    1 g  1 mL solution  1 L solution 
                                
                                                                       L

 1 g solute       109 ng  1 g solution  1000 mL solution          ng
1 ppt   9                                                           1
 10 g solution    1 g  1 mL solution  1 L solution 
                                
                                                                        L

Note: ppt = parts per trillion
Concentration scales (cont.)

It is important 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

 0.02500 mol K 2 SO4  174 .26 g K 2 SO4  1000 mg         mg K 2 SO4
                      
                   
           4356

         L            mol K 2 SO4           g               L

 4356 ppm
Concentration examples

What is concentration (in ppm) of K+ in this solution?

Solution of 0.02500 M K2SO4

 0.02500 mol K 2 SO4  2 mol K  39 .10 g  1000 mg 

mg K 
                       1 mol K SO  mol K  
                              1955

         L                   2   4          g              L

 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. Note that
you can use any units you want for conc. and vol. as
long as they are the same on both sides of the
equation.
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         Cdil = 0.750 mol L-1
Vconc = ?                    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.   Use a pipet or graduated cylinder to measure exactly
31.2 mL of 12.0 M
2.   Transfer the 31.2 mL of 12.0 M HCl to a 500.0 mL
and swirl to mix the solution
4.   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
5.   Put the stopper on the flask and invert ~20 times to mix

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