Introduction to Solution Stoichiometry
Whenever possible, reactions are carried out with all of the reactants in the same fluid phase.

This means that it is preferable to have liquids reacting with liquids, or gases reacting with gases. This is because no matter what type of particle we talk about, in order for them to react they must collide with each other.  Solids have little or no movement and therefore offer few opportunities for collision.  Liquids and gases are both fluids and offer much more opportunity for frequent collision.  Liquids are used in the lab because they are fairly easy to create and use.  Gases are used but specialized gas handing equipment is required.

A solution is a uniform mixture of particles of atomic, ionic, or molecular size.  A minimum of two substances are present.  One is called the solvent and all the others are called the solutes.  The solvent is the fluid medium in which all of the solutes are dissolved.  A solvent can be a solid, liquid or a gas but the most common solvent is water and therefore we deal almost exclusively with aqueous solutions.   Solutes are any substance dissolved in the solvent.

A dilute solution is one in which the ratio of solute to solvent is very small, for example, a few crystals of sugar in a glass of water.  In a concentrated solution the ratio of solute to solvent is large.  Maple syrup is a concentrated solution of sugar in water.  A saturated solution is one in which no more solute can be dissolved at a particular temperature.  An unsaturated solution is one in which the ratio of solute to solvent is lower than that of the corresponding saturated solution.  If more solute is added to an unsaturated solution, at least some of it should dissolve.   A supersaturated solution is an unstable system in which the ratio of dissolved solute to solvent is higher than that of a saturated solution.  A supersaturated solution can be made by gently cooling a hot saturated solution.  At a lower temperature the dissolved solute can be made to precipitate out when a seed crystal is added.  The process is called precipitation and the substance that forms is the precipitate.

The amount of solute needed to make a saturated solution in a given quantity of solvent at a specific temperature is called the solubility of the solution

Units of Concentration (v/v, w/v, w/w and ppm)
These units of concentration are most often seen and used with commercial products.  Expect for 'ppm' they are not used often in the lab.
 
Percent Concentration Volume/Volume (v/v):  used with 2 liquids.
      % Concentration = Vsolute    X 100%
                                       Vsolution

    e.g..   5 mL of vinegar are dissolved in 100 mL of vinegar solution.  What is its v/v concentration.

        % concentration =    5 mL of vinegar    X 100%  = 5%
                                        100 mL of solution
 

Example #1   A photographic stop bath contains 140 mL of pure acetic acid in a 500 mL bottle of solution.  What is the v/v concentration?
 
        % concentration = 140 mL X 100% = 28%
                                        500 mL
 

Percent Concentration Weight/Volume (w/v): used with one solid and one liquid
This means there is a certain mass, in grams, in every 100 mL of solution.
 
e.g..   a 3% H2O2 topical antibiotic solution means that there is 3 grams of H2O2 in every 100 mL of solution.
 

Percent Concentration Weight/Weight (w/w): used with two solids
Useful when dealing with alloys of precious metals.
 
e.g..  A ring with a mass of 12.0 grams contains 11.1 grams of pure silver.  What is the w/w%?

    % concentration = 11.1 grams X 100% = 92.5% w/w of silver
                                    12.0  grams
 


Parts per Million Concentration (ppm)
Environmental solution are often very low in concentration.  We often use terms like:
1 part per million (ppm): 1 part out of 1 X 106 parts
1 part per billion (ppb): 1 part out of 1 X 109 parts
1 part per trillion (ppt): 1 part out of 1 X 1012 parts

1 ppm = 1 drop in a full bathtub
1 ppb = 1 drop in a full swimming pool
1 ppt = 1 drop in 1000 full swimming pools
 

We express ppm concentration in a variety of units depending on what we need to use.  But they are all interrelated.
 
ppm =   1 g        =       1 g      =        1 mg       =    1 mg     =   1 microgram
            106 mL        1000 L             1 L               1 kg                    1 g
 
Example #2   Dissolved O2 in water shows a concentration of 250 mL of water At SATP and 2.2 mg of O2.  What is the concentration in ppm?

ppm concentration = 1 mg   =   2.2 mg   =  8.8 mg/L = 8.8 ppm
                                   1 L          0.25 L

Go to the Concentration Unit Calculations Other than Molarity Worksheet
 
Mole Fractions and Mole Percents
One way to describe the relative numbers of molecules of all components of a solution is by their mole fractions.  The mole fraction of any component in a mixture is the ratio of the number of moles of it to the total number of moles of all the components present.  Expressed mathematically,
Xa =                      na                 
               na + nb + nc + ... + nx
Where Xa is the mole fraction of component 'a', and na, nb, nc ... nx, are the numbers of moles of each component in the mixture, 'a', 'b', 'c', ....'x', respectively.  The sum of all the mole fractions for a mixture must always be equal to 1.

Even though mole fractions have no formal units, always remember that they stand for the ratio of moles of a component to total moles in the mixture.

Sometimes the mole fraction is multiplied by 100% and the result is referred to as the mole percent.

Example of how to calculate mole fractions and mole percent.
What are the mole fractions of each component in a solution that consists of 1.00 mole of C2H5OH,   0.500 mole of CH3OH, and 6.00 mole of H2O?
Component

Mole
Fraction
Mole
Percent
C2H5OH XC2H5OH       ___ 1.00               =   0.133
  1.00 + 0.500 + 6.00
0.133 13.3%
CH3OH
XCH3OH       ___ 0.500               =   0.0667
  1.00 + 0.500 + 6.00
0.0667
6.70%
H2O
XH2O =           ___ 6.00                =   0.800
  1.00 + 0.500 + 6.00
0.800
80.0%

*Note the Sums --->
1
100

Molar Concentrations and Molarity  (M)
Molarity is a way of specifying the amount of solute in one litter of solvent.   Molarity is also known as the concentration of a solution.  The concentration of a solution is the ratio of solute to a given quantity of solvent. We can use whatever units we wish but the most commonly used units are moles of solute per litter of solution.  The special name for this ratio is the molar concentration or molarity which is abbreviated 'M'.

       M =    mol solute    =        mol solute
                     L soln             1000 mL soln
 

Suppose we had a bottle with the label "0.5 M KBr".  This means that it contains a solution of potassium bromide with a concentration of 0.10 mol of potassium bromide per litter of solution (or per 1000 mL of solution).    It does not tell us the amount of solution in the bottle.
 
Solutions are considered to be homogenous substances once the solute has completely dissolved.
 
Sample Problem
A student requires 0.250 moles of NaCl for an experiment.  The only thing available to them is a bottle with a solution labeled "0.400 M NaCl."  What volume of the solution should be used?  Give the answer in milliliters.
 
Solution:
Use the information on the bottle.  There is 0.400 moles of solute per litter.  We need 0.250 moles.

                0.400 moles = 0.400 moles = 0.250 moles
                         1 L             1000 mL              x

                   x = 625 mL.  Thus 625 mL of 0.400 M NaCl contains 0.250 mol of NaCl.
 

Practice Problem:
A glucose solution with a molar concentration of 0.200 M is available.  What volume of this solution must be measured to obtain 0.001 mol of glucose?
 
 
 
 
 
 
 
Preparation of Solutions
Another very common problem involving molar concentration is the calculation of the number of grams of solute needed to make a given volume of solution having a specific molarity.  The best way to see this is by example:
 
Problem:  How can 500 mL of 0.150 M Na2CO3 solution be prepared?
 
Solution: This is stated in a way that normally arises in the lab?  What we really want to know is how many grams of Na2CO3 that are going to be in 500 mL of 0.150 M Na2CO3 solution.
Although the label reads in M, the balance reads in grams.  Before we can calculate the number of grams we need to know the number of moles.

                   0.150 M  =  0.150 mol  = 0.150 mol  =         x
                                              L            1000 mL        500 mL

          x = 0.075 mol of solute.  Therefore we need to weigh out 0.075 mol of Na2CO3.

 The number of grams of Na2CO3 are therefore:

                                           g = n * mm
                                              = 0.075 mol * 105.99 g/mol
                                              =  7.95 grams
 

To answer the question:   Weigh out 7.95 grams of sodium carbonate.  Add it to 100 mL of water in a 500 mL volumetric flask.    Swirl to dissolve.  Top the flask up with 400 mL more water up to the 500 mL meniscus mark.
 
Practice problem:  How can we prepare 250 mL of 0.200 M NaHCO3?
 
 
 
 
 
 
 
Go to the Molarity and Solution Creation Worksheet
Go to the Stoichiometry of Solution Molarity Worksheet

Molality or Molal Concentration (m)
The number of moles of solute per kilogram of solvent is called the molal concentration or the molality of a solution.  The usual symbol for molality is m.

molality = m = moles of solute
                          kg of solvent

If for example, we dissolve 0.5000 moles of sugar in 1000 g of water (1.000 kg), we have made a 0.500 molal solution of sugar.  The need to use a volumetric flask is gone since the water was weighted.

The molality of a solution is an indirect expression of the ratio of moles of solute to moles of solvent, because the mass of the solvent is known.  If needed, we could convert the 1 kg of solvent into moles of solvent.  

To contrast molality with molarity, suppose we put the same quantity of sugar 0.500 moles, into a one liter volumetric flask and add water to the 1-L mark, making sure that everything is dissolved first.  The calibration mark on the volumetric flask correlates with the flask's specified volume only at a given temperature, which is etched on the flask.   If our solution were at this temperature, then we would have a 0.5000 molar solution.  But we would not now the mass of water used, and it would very unlikely that this mass of water would be 1.00 kg.  Unless we actually measured the mass of water used to make the 0.500 molar solution, or knew the density of the solution, we would have no way to calculate its molality.

It is very important that you learn the distinction between molarity and molality.  Molarity is useful in determining the moles of solute obtained when a specified volume of solution is poured.   Molality is useful is describing the ratio of moles of solute to moles of solvent.  The value of a concentration given in units of molality does not change with temperature, but that of molarity does. 
Ex. 1    An experiment calls for an aqueous 0.150 m solution of sodium chloride, To prepare a solution with this concentration, how many grams of NaCl would have to be dissolved in 500 g of water?

Step 1: Prepare two conversions factors:

  0.150 m  is the same as  0.150 g  NaCl         or           1 000 g H2
                                            1000 g H2O                         0.150 g NaCl

1 kg of water was changed to 1000 g!

Step 2: Multiple by the mass needed using the first of the conversion formulas above so that the units cancel out.

          500 g H2  X     0.150 g NaCl    =   0.0750 mol of NaCl
                                       1000 g  H2O

Step 3:  Convert moles into grams.
               g = n X mm
                  = 0.0750 moles X 58.5 g/mol
                  = 4.39 g of NaCl

By placing 4.39 g of NaCl into 500 g of water we obtain a 0.150 m solution.

Converting from Weight Percent to Molality
What is the molality of a 10.0% w/w NaCl solution?
Solution:
10% w/w means we have a ratio of 10.0 g of NaCl to 100 grams of solution.   We need to convert w/w concentration units into mol/kg of solvent units.

Step 1:  Calculate the moles of NaCl in 10.0 g of NaCl.

                       n = g/mm
                          = 10.0g / 58.5 g/mol
                          = 0.171 mol of NaCl.

Step 2:  Calculate the kilograms of H2O in 100 grams of this solution.  (Remember it is grams of solution with percent concentration, but kilograms (or 1000 g) of solvent with molality.

There is 10.0 grams of NaCl in 100.0 grams of solution.  So there must be 100.0 g - 10.0 g  = 90.0 grams of water present.  90.0 grams  = 0.090 kg of water.

Step 3:  Calculate molality, m,  the ratio of moles to kg of solvent.
 
m = moles of solute / kg of solvent
    =  0.171 mol of NaCl / 0.090 kg of H2O
    =  1.90 m

Therefore a 10.0% w/w NaCl solution is also 1.90 m NaCl

Converting From Weight Percent to Mole Fractions
What are the mole fractions and mole percents of the components in 10.0% w/w NaCl?
 
The amount of solute from the example above is 0.171 mole of NaCl.  We also found from the above example that there is 90.0 g of water in 100 grams of the solution.  Moles of water is therefore n= g/mm = 90.0 g/18.02 g/mol = 5.00 mol of water.

So the total number of moles is 0.171 moles + 5.00 moles  = 5.171 moles total
Calculate the mole fractions and mole percent.



Mole Fractions
Mole Percent
NaCl
XNaCl  0.171  =   0.033
       5.171       
0.033
3.3%
H2O
XH2O  5.000  =   0.967
       5.171       
0.967
96.7%

Calculating Percents by Weight from Mole Fractions
So called "100-proof" alcohol has the following mole fractions of components: for water, 0.765, and fr ethyl alcohol, 0.235.  Calculate the percents (w/w) of water and ethyl alcohol in 100 proof ethyl alcohol.  The formula weights are: H2O: 18.0 and ethyl alcohol, C2H5OH, 46.1

AS in all these kinds of calculations, we work from the meanings of the given concentration units toward the final units.  To work toward percents by weight, we have to work toward a ratio calculated from grams of components.  Let's start with water.   When the mole fraction of water is 0.765, we know that there are 0.765 mole of H2O in a total of 1 mole of all components.  So, in terms of water we have:

g = n X mm
   =  0.765 mol  X 180.8 g/mol
   =  13.8 g of H2O

Next, we look at the ethyl alcohol.

 g = n X mm
    =  0.235 mol  X 46.1 g/mol
    =  10.8 g of C2H5OH

Thus,  in "1.0 mole" of this ethyl alcohol solution there are 13.8 g of H2O and 10.8 g of C2H5OH.  To convert to percents by weight, we need the total mass of the solution, which is 13.8 g + 10.8 g = 24.6 g.       For the percents:

H2O
mass of water   =     13.8 g    X  100%  =   56.1% (w/w)
     total mass                 24.6 g
C2H5OH
mass of alcohol   =     10.8 g    X  100%  =   43.9% (w/w)
     total mass                 24.6 g
Thus, 100 proof alcohol is 56.1% water and 43.9% alcohol!

Converting Molality into Mole Fractions
Ammonium nitrate is sometimes a component in nitrogen fertilizers.  Suppose that an aqueous solution of NH4NO3 has a concentration of 2.48 m.  What are the mole fractions of NH4NO3 and water in this solution?

Solution:
The given molal concentration means that we can create two conversion units.

2.48 m  means   2.48 mol NH4NO3          and                1 kg water       
                                 1 kg water                                 2.48 mol NH4NO3

We need to work our way toward a ratio of moles of solute to total number of moles, so we need to find the moles of water in 1 kg.  Remember to treat the 1 kg as exact, therefore 1 kg of water  = 1000 g of water and water has a molecular mass of 18.02.   To find moles of water :    n  =   g  /  mm
                   =  1000 g  / 18.0 g/mole
                   =   55.6 mol of H2O

So the total moles, water + NH4NO3, is 

2.48 moles  +  55.6 moles  =  58.1 mol (rounded correctly)

Now we can calculate the mole fraction of NH4NO3 and H2O

          XNH4NO3 =       2.48 moles    =  0.0427
                                   58.1 moles

The mole fraction of water is XH2O = 1.000 - 0.0427 = 0.9573

Converting a Percent by Weight into a Molar Concentration
A certain supply of concentration hydrochloric acid is 36.0% (w/w) in HCl.  The density of the solution is 1.19 g/mL.  Calculate the molar concentration of HCl in this solution.

Solution:  We are converting from w/w to mol/L.  We need the density to get from g solution into mL solution.  The density is given as 1.19 g/ mL and from this we get two possible conversion units.

either  1.19 g solution   or     1 mL solution
             1 mL solution           1.19 g solution

Since we have to deal with 1000 mL of solution as we work from grams of solution toward the molar concentration, we now calculate how many grams of this solution are in 1000 mL.

1000 mL solution  X   1.19 g solution  =  1190 g solution
                                     1 mL solution

But this is the mass of a 36.0% w/w in HCl; 36.0% of this mass is pure HCl.  So the quantity of pure HCl in 1190 g of solution is:
 
                    1190 g  X   36.0   =  428 g of HCl
                                       100

Thus, there is 428 grams of HCl in 1190 grams of solution.  But, remember that this mass of solution has a volume of 1000 mL or 1 L.  To get the molarity we need to convert 428 grams of HCl into moles of HCl.  The formulas weight os HCl is 36.5 so:

  n = g / mm
     =  428 g / 36.5 g/mol
     =  11.7 mol of HCl

Therefore the molarity of the HCl solution is 11.7 mol/L or 11.7 M

Converting From Molarity to Percent by Weight
Perchloric acid, HClO4, (formula weight, 100.5), can be purchased at a concentration of 9.20 M.  The density of this solution is 1.54 g/mL. What is the percent by weight of perchloric acid in this solution?

When the molarity is 9.20 M, we have the following ratio.

9.20 mol HClO4 
1000 mL solution

What we want is the ratio of grams of HClO4 to grams of solution, taken as a percent.  this is what a percent by weight means.  So, in 9.2 mol of HClO4, we have:

9.20 mol HClO4   X     100.5 g HClO4   =  925 g of HClO4
                                      1 mol HClO4

And in 1000 mL solution we have the following mass of solution.

1000 mL solution   X    1.54 g solution  =    1540 g of solution
                                      1 mL solution

So the ratio of grams of solute to grams of solution, taken as a percent,  is

 925 g of HClO4     X  100 %   =  60.0% (w/w)
1540 g (solution)

Thus, 9.20 M HClO4 is 60.0% (w/w) HClO4.

Go to the Molality, Mole Fraction and Mole Percent Worksheet
 
Stoichiometry of Reactions in Solution
Up until now we have used pure substances and added them to water in order to make solutions.  The point of using solutions is that they are often easier to use in the lab than pure substances.  We will now find out how to do stoichiometry with these solutions.
 
Problem:  What volume of 0.556 M HCl has enough hydrochloric acid to combine exactly with 24.5 mL of aqueous sodium hydroxide with a concentration of 0.458 M?  The equation for the reaction is:
 
                     HCl(aq)  +  NaOH(aq) ----->  NaCl(aq)  +   H2O
 
Solution:  This is a simple mL --> mol --> mol ---> mL solution.
 
Step #1
Start with what you know:  You have 24.5 mL of 0.458 M NaOH.  How many moles is this?

            0.458 mol  =       x               
              1000 mL       24.5 mL
 
x = 0.011 mol of NaOH are being used.

Step #2    The HCl and NaOH react on a 1:1 basis.  Therefore

                          1 HCl   =     1 NaOH
                              x            0.011 mol

                               x = 0.011 mol of HCl are needed.
 

Step #3   We need 0.011 moles of HCl and the solution we have available is 0.556 M. Therefore

                          0.556 mol  =   0.011 mol
                           1000 mL              x

                            x =  19.78 mL are needed.

If  we measure out 19.78 mL of the HCl solution it will exactly neutralize the NaOH.
 
 

Example Problem #2:    How many milliliters of 0.114 M H2SO4 solution provide the sulfuric acid required to react with the sodium hydroxide in 32.2 mL of 0.122 M NaOH according to the following equation?

                       H2SO4(aq)  +  2 NaOH(aq) ------>  Na2SO4(aq)  + 2 H2O
 

Solution:  Again we need the mL --> mol --> mol --> mL solution.
 
Step #1
    Start with what you know.    You have 32.3 mL of 0.122 NaOH.

                             0.122 mol  =       x
                               1000 mL       32.3 mL

                               x = 0.004 mol of NaOH are present.
 

Step #2               1 H2SO42 NaOH
                                  x            0.004 mol

                                 x = 0.002 mol of H2SO4 are required.
 

 Step #3     Therefore we need:

                        0.114 mol  =  0.002  mol
                         1000 mL             x                       x = 17.54 mL are needed.

Therefore we need 17.54 mL of 0.114 M H2SO4 to exactly neutralize 32.3 mL of a 0.122 M solution of NaOH.
 

Practice Problems
1.  What volume of 0.337 M KOH provides enough solute to combine with the sulfuric acid in 18.6 mL of 0.156 M H2SO4?   The reaction is:

                2 KOH(aq)  +  H2SO4(aq) ---->  K2SO4(aq)  +  2 H2O
 
 
 
 
 
 
 
 
 

2.  Hydrochloric acid reacts with sodium carbonate as follows:

                2 HCl(aq)  + Na2CO3(aq)  ------->  2 NaCl(aq)  + CO2(g)  +  H2O

    What volume of 0.224 M HCl is neutralized by the Na2CO3 in 24.2 mL of 0.284 M Na2CO3?
 
 
 
 
 
 
 
 
Go to the Stoichiometry Involving Solutions Worksheet


Dilutions from Concentrated Stock Solutions
When a dissolved solvent is added to a solution, the solute is spread out through the large volume and the number of moles per unit volume decreases.  The solution is said to be diluted.  Such dilutions are a natural part of chemistry in the lab.  If two solutions of different solutes are mixed, the total volume increases and becomes occupied by both solutes.  As a result the concentration of both solutes is less.  At other times dilution is deliberate and essential.  Stock solutions of common chemicals are often prepared in large volumes in preset concentrations.  The chief supply of HCl in the lab may be 1.00 M HCl.  In your experiment, you may require for a much less concentrated solution, so the concentrated solution must be diluted first.
 
  When carrying out a dilution with an acid.  Always add acid to water.  Never add water to the acid.   When acid and water mix a great deal of heat is released.  Acid added to water gives a large heat sink for the heat to dissipate in.  If you add water to acid, the heat generated will cause the water to boil, and may splatter you with the hot acid solution.
 
Dilution Calculation
All the calculations that you need to do are based on a simple fact.  As a solution is diluted, the number of moles of solute doesn't change; the solute simply spreads out through a larger volume.
 
We will use the following symbols: V for volume and M for molarity. Therefore Vc and Mc are the concentrated solutions volume and molarity. Vd and Md are then the diluted solutions volume and molarity.

                                            Vc Mc  =  Vd Md
 

Problem:  How can we prepare 100 of 0.040 K2Cr2O7 from 0.200 M K2Cr2O7?
Solution:
First assemble the data into a table:
                 Vc =  ?               Vd = 100 mL
                Mc = 0.200 M    Md = 0.040 M

Next, use the formula from above.  Vc X 0.20 M = 100 mL X 0.040 M
                                                      Vc = 100 mL X 0.040 M
                                                                      0.20 M
                                                      Vc = 20.0 mL

Place 20.0 mL of 0.200 M K2Cr2O7 into a 100 mL volumetric flask and then add water until the final volume is 100 mL.
 
 

Practice Problem:  Describe how to make 500 mL of 0.20 M NaOH from 0.50 M NaOH.
 
 
 
 
 
 
 
Problem:
How many milliliters of water would have to be added to 100 mL of 0.40 M HCl to give a solution with a concentration of 0.10 M?
 
First assemble the data.         Vc = 100 mL         Vd = ?
                                                Mc = 0.40 M         Md = 0.10 M

Use the equation:   100 mL X 0.40 M = Vd  X 0.10 M
                                                     Vd  = 400 mL

The final volume must be 400 mL.  But since we started with 100 mL we will add 300 mL of water until the volume reaches 400 mL.
 

Practice Problem: How many milliliters of water would have to be added to 300 mL of 0.5 M NaOH to give a solution with a concentration of 0.2 M?
 
 
 
 
 
Go to the Solution Dilution Worksheet