Unit 2  Quantities in Chemistry
Are You Ready?  Page 76    Answers

Atomic Mass

The sum of individual atoms can be used to find the mass of a molecule.

The mass of hydrogen peroxide, H2O2 would be calculated like this:

H2O2 has 2 hydrogen atoms and 2 oxygen atoms in it.

Therefore the mass is
2 X H   =  2 X 1.01 u   = 2.02 u

2 X O  = 2 X 16.00 u = 32.00 u
So one molecule of hydrogen peroxide weighs in at 34.02 u.

The periodic table provides you with individual atomic masses.  If you know the number and type of elements in a molecule you can add up the individual masses to find the molecular mass or molecular weight.

Find the molecular mass of calcium phosphate, Ca3(PO4)2

The molecule has 3 calcium atoms, 2 phosphate atoms and 8 O atoms in it. Stop and verify this for yourself. The Ca has a subscript 3 with it. The P has an assumed 1 and the O has a 4. However the PO4  group has a set of brackets around it with a subscript 2. The 2 means multiply everything inside the brackets by 2. So we end up with the 2 P and 8 O atoms.

Calculation:  3 X Ca = 3 X 40.08 u = 120.24 u

      2 X P   = 2 X 30.97 u =   61.94 u

      8 X O  = 8 X 16.00 u = 128.00 u
      310.18 u

One molecule of calcium phosphate weighs 310.18 u 

Stop here and do the  Atomic Masses Exercise

Percent Composition

Percent composition is a simple little calculation that has a large impact on chemistry. When a brand new substance is discovered one of the first things that is determined is its chemical make-up. That means how much of each type of atom is in the molecule. What is the percent composition of strontium oxide? SrO

There is 1 atom of Sr and 1 atom of O.

The mass of 1 Sr is 87.62 u and the mass of an O is 16.00 u.

The percent composition is the fraction of the SrO that is just Sr.  This fraction is:

Percentage of Sr = mass of Sr in the molecule / molecular mass X 100%

            = 87.62 u / 103.62 u X 100%

            = 0.8456 X 100%

            =  84.56%

Strontium makes up 84.56% of the molecule. The oxygen makes up 100% - 84.56% = 15.44%. The u's are units that cancel out. The % sign means out of 100.

The general equation for finding the percentage composition of an element is:

percentage of an element = the total mass of just that element
                                                   molecular mass X 100%

Stop here and do the  Percent Composition Exercise

Homework  Section 2.2  Questions  Page 106    Answers

Molecular Amounts versus Labratory Amounts
The Mole: The Start of Chemical Calculations

The creation of formulas and the finding of their molecular masses is only partially complete.
What exactly does that number you have found for the molecular mass mean?  Is it the mass of one molecule, or a million, or ten billlion?  What unit is it measured in?  

The Carbon-12 Based Atomic Mass Scale
In the years following Dalton's presentation of his atomic theory, chemists worked very hard at determining a complete set of relative masses for all the elements that were known.  By knowing these relative masses the chemists were able to select amounts of elements in grams needed for any desired atom ratio.    In establishing a table of atomic amsses, it is necessary to have a reference point against which to compare the relative masses. Currently, the agreed-upon reference is the most abundant isotope of carbon, which is carbon-12. By definition, an atom of this isotope is defined as having the mass of exactly 12.000 u. (atomic mass units)  In other words, an amu is defined as 1/12th of the mass of one atom of carbon-12.

The definition of the size of the atomic mass unit was quite arbitrary.  It could just as easily have been selected to be 1/24th of the mass of one atom of a carbon atom, or 1/10th the mass of a calcium atom, or any other value. Why 1/12th the mass of carbon-12?  Carbon is a very common element, available to any scientist  and by choosing the amu to be of this size, the atomic masses of nearly all the other elements are almost whole numbers, with the lightest atom having a mass of approximately 1. (hydrogen-1 has a mass of 1.007825 amu when carbon-12 is assigned a mass of exactly 12 amu.)

The Mole
Suppose we want to make molecules of carbon dioxide, CO2, in such as way that there would be no extra carbon and oxygen atoms left. If we took ten atoms of carbon and twenty atoms of oxygen we would make 10 molecules of carbon dioxide.  Suppose we wanted to make more, lets say 40,000 molecules of carbon dioxide.  Once again, we could count out 40,000 atoms of carbon and 80,000 atoms of oxygen, let them react and we'd have 40,000 molecules of carbon dioxide.  This sounds all very nice and neat unless you realize the trap.  Have you ever seen an atom?  Atoms are too tiny to count individually.  Saying they are tiny is even wrong because tiny can be seen.   Even with the best scanning tunneling electron microscope ever invented the largest atoms known, look just like fuzzy cloud tops.  We have never seen individual atoms.  There are no lenses with the resolving power or balances fine enough to measure an individual atom.

We get around this problem because each element has its own characteristic atomic mass and each formula has its own unique molecular mass.    We know, that oxygen weighs in at 1.33 times that of carbon, because of this we get a ratio of their masses:

                   16.0 amu (for one atom of O) = 1.33
                   12.0 amu (for one atom of C)       1

If we take a sample of oxygen and carbon in a ratio of 1.33 to 1, we must obtain equal numbers of their atoms.  That is, if we actually had a balance that could measure amu directly we could mass out  32 amu of oxygen and 12 amu's of carbon - a mass ratio of 2.66 to 1 - we would have exactly 2 atoms of oxygen for every atom of carbon and a 2 to 1 ratio by atoms.

When we mass out a sample of an element such that its mass in grams is numerically equal to the element's atomic weight, we always obtain the same number of atoms no matter what element we choose.  Thus 12.0 g of carbon has the same number of atoms as 16.0 grams of oxygen, or 32.1 g of sulphur, or 55.8 g or iron.

This relationship also extends to compounds. The formula mass of water, H2O, is 18.0 amu. If we take 18.0 grams of water then it should have the same number of molecules as there were atoms in 12.0 grams of carbon.  The carbon-12 isotope, which makes up 98.89% of all naturally occuring carbon, is the reference used by SI Metric in its definition of the base unit for a chemical substance, the mole, abbreviated mol. 

This mole concept is the most important in all of chemistry. Once this concept is grasped all the rest of chemistry will appear easy.

Avogadro's Number
The mole is defined as 6.02 x 1023 units.  It is called Avogadro's number in honour of Italian scientist, Amadeo Avogadro (1776-1856). It is a pure number with a special name, just like so many others.  For example:
                                                 2 = pair
                                               12 = dozen
                                             144 = gross
                                             500 = ream
                                6.02 X 1023 = Avogadro's number
This number is not an odd number at all. It became inevitable once the amu was defined.   The relationships needed are:

                                          1 mole = 6.02 X 1023 particles     (General expression)
                                         12 amu = 1 atom of  C      (Specific example)
                                    1 mol of C = 6.02 X 1023 atoms of C
                                    1 mol of C = 12 g of  C

Another useful relationship is that 1 amu = 1.66 X 10-24 g

SO                            1  amu               =  6.0 X 1023 particles
                      1.66 X 10-24  grams

The mass in grams of a substance, that equals one mole is often called its molar mass, and the units are grams/mole or g/mol.   For example, aspirin has a molecular mass of 180 grams. Therefore if we massed out exactly 180 grams of aspirin we would have Avogadro's number of aspirin molecules.
Stop here and do the Atoms and Moles Exercise Sheet
Using the Mole Concept
One mole of any substance can be calculated from its formula mass. Since this is true it is absolutely essential that when you are using the mole concept that the correct formula be used. It is not enough to say "use 1 mole of nitrogen".  Do we mean atomic elemental nitrogen or nitrogen gas?  There is a difference!  One mole of  N consists of Avogadro's number of nitrogen atoms (and has a mass of 14.01 g), whereas 1 mole of N2 consists of Avogadro's number of molecules, each molecule having two nitrogen atoms. One mole of N2 molecules would have a mass of 2 X 14.01 g = 28.02 g.

One of the advantages of the mole concept is that it lets us think about formulas on two levels at the same time.  One level is that of atoms or molecules or ions, and the other level is that of lab-sized practical quantitites, such as moles and grams.  Look at the equation below:

    H2O   consists of                   2 H                               +    O
1 molecule of H2O                    2 atoms of H                      1 atom of O
1 dozen H2O molecules            2 dozen atoms of H            1 dozen O atoms
6.02 X 1023 H2O molecules     12.04 X 1023 H atoms        6.02 X 1023 O atoms
1 mole of H2O molecules         2 moles of H atoms            1 moles of O atoms
18.0 g of H2O                            2.0 g of H                           16.0 g O atoms

When we think about H2O at the first level, we can easily see that a dozen of its molecules are made from two dozen atoms of H and one dozen atoms of O.  However, if we switch to the more practical lab-sized level, it is just as easy to think about one mole of H2O and to view this quantity as consisting of two moles of H and one mole of O.

The numbers in all but the last row are in the same ratio regardless of the scale, whether we deal with single particles or with moles of them. After planning an experiment at the mole level, it is easy to convert numbers of moles into corresponding masses of chemicals to meet any desired needs.

Extension Exercise 2.1 Ratios in Chemical Equations
Extension Exercise 2.1 Amounts in Chemistry

Converting Moles to Grams
This is probably the single most used equation in chemistry.  It is the one that allows the conversion of moles into grams and grams back into moles.  Moles are a "theoretical value" which looks good on paper.  Grams is the "practical value" that you take out of a stock bottle and place on the balance.  Lab balances read in grams, not moles, and so in practical work all mole values have to first be converted.

Moles to grams formula:     grams(g)  = moles(n)  X  molecular mass(M)

Example:  Sodium bicarbonate, NaHCO3, is one ingredient of baking powder.  How many grams of sodium bicarbonate are in 0.673 moles?
Step 1.  Find the molecular mass:
          NaHCO3 = 1 Na = 1 X 22.99 g/mol =  22.99 g/mol
                             1 H   = 1 X   1.01 g/mol =    1.01 g/mol
                               1 C = 1 X 12.01 g/mol =  12.01 g/mol
                               3 O = 3 X 16.00 g/mol =  48.00 g/mol
                                                                        84.01 g/mol
Step 2. Use the equation:   g = n  X  M
                                            = 0.673 mol X 84.01   g
                                            = 56.54 g

Step 3: Provide a written answer:  There are 56.54 grams of sodium bicarbonate in 0.673 moles.
Grams to moles formula:

   n =     g    

Example:  Potassium permanganate, KMnO4, at one time was used for an anti-fungal agent. You
                could always tell someone who had just been treated because their feet were purple. The
                pharmacy gives you 250 grams of the stuff in a bottle.  How many moles of it do you
Step 1:  Find the molecular mass of KMnO4:
              1  K  = 1 X 39.10 g/mol  =  39.10 g/mol
              1 Mn = 1 X 54.94 g/mol  =  54.94 g/mol
              4 O   =  4 X 16.00 g/mol  = 64.00 g/mol
                                                       158.04 g/mol
Step 2:  Use the equation:
 n =           g            
       molecular mass

     250 g         
   158.04 g/mol

=  1.58 mol
Step 3:  Write a written answer.
A sample of 250 grams is equivalent to 1.58 moles of potassium permanganate.

Finding the molecular mass
In the above two examples you have had a molecular formula to work with.  As long as you have a molecular formula, or name, from which you can make a formula up, you will always have a molecular mass.   In order to find the molecular mass, without a name to go on, you need two pieces of information. The number of grams and the number of moles that it represents.
   molecular mass(M) =    grams     =        g    
                                     moles               n
Example:   A fellow student comes to you with a sample of an unknown chemical. They tell you that it has a mass of 34.91 grams and that it is exactly 0.20 moles.   What is the molecular mass of the substance?

Step 1:  Use the equation:     molecular mass =   34.91  g   =  174.55     g
                                                                                0.20 mol                    mol

Step 2: Write a sentence.  The unknown substance has a molecular mass of 174.55 g/mol.
Stop here and go to the Grams, Moles and Molar Mass Worksheet

Converting Moles into Molecules
The formula needed is:     molecules = n * NA

Multiply the number of moles by Avogadro's number to find the number of molecules.
This is similar to multiplying 4 cartons of eggs by 12 eggs/carton to find the total number of eggs present.

Ex.   How many molecules are there in 3 moles of CO2?
          molecules = n * NA
                           = 3 moles * 6.0 X 1023 molecules/mole
                           = 18.06 X 1023 molecules
                           = 1.806 X 1024 molecules

There are 1.806 X 1024 molecules in 3 moles of CO2.

Converting Molecules into Moles
The necessary formula is:     moles (n)  = molecules

This concept is similar to counting 144 eggs and dividing by 12 to get the number of cartons.

Example: You have a sample of 5.6 X 1024 molecules of XeF6.  How many moles of this particle sample of XeF6 do you have?

moles =    molecules
           =        5.6 X 1024 molecules      
                6.02 X 1023 molecules/mole

           =   9.3 moles

There are 9.3 moles of XeF6 in this sample.

 Converting Molecules into Atoms
You have to be able to look inside a molecule and determine the total number of atoms in its makeup.
This is the atoms subscript total.  For example H2O has a subscript total of 3 atoms.  H2SO4 has a subscript total of 7 atoms.  MgSO4•7 H2O has a subscript total of 27 atoms.

The formula needed is:    atoms = molecules X subscript total

Example: You have 1.2 X 1023 molecules of hydrogen peroxide, H2O2.  How many atoms in total do you have?

atoms = molecules X subscript total
           = 1.2 X 1023 molecules X 4 atoms/molecule
           = 4.8 X 1023 atoms

There are 4.8 X 1023 atoms in total in this sample of H2O2.

Converting Atoms into Molecules
The formula needed is:   molecules =    total atoms  
                                                               subscript total

Exmaple:  You have a sample of calcium hydroxide, Ca(OH)2.   You have a total of 5 million atoms. How many molceuls are presnet?

molecules =    total atoms  
                     subscript total

                 =    5,000,000 atoms  
                      5 atoms/molecule

                 =   1,000,000 molecules

There are 1 million molecules in this sample of Ca(OH)2.

Extension Exercise 2.2 Calculations Involving the Mole
Stop here and do the
Marvin Da Mole Worksheet

Stop here and do the Marvin Da Mole Strikes Again Worksheet

Determining Chemical Formulas (Empirical Formulas)  
When the quantitative analysis of a compound gives us the masses of each element in the same-size sample, we can convert their masses into proportions by moles by a grams to moles calculation.  With atomic particles, proportions by moles are numerically the same as proportions by atoms, which are just the numbers we need to construct a chemical formula.  The formula obtained in this way is called an empirical formula.  An empirical formula is the one that uses as subscripts the smallest whole numbers that describe the ratios of atoms in a compound.

Calculating an Empirical Formula from data obtained by quantitative analysis.
Example #1
A sample of an unknown compound with a mass of 2.571 grams was found to contain 1.102 grams of C and 1.469 grams of oxygen.  What is its empirical formula?

1.102 g of C X 1 mole of C  = 0.09176 moles of C
                        12.01 g of C

1.469 g of O X 1 mole of O = 0.09181 moles of O
                         16.00 g of O

Therefore the formula is C0.09176O0.09181   Pick the smallest of the subscript numbers as the divisor and divide it into all the other numbers.

                        C1.0000O1.0001 = C1O1

As a general rule, if the calculated subscript differs from a whole number by only several units in the last place, we can safely round to the whole number.

Example #2
When a sample with a mass of 2.448 grams of compound present in liquified petroleum gas was analyzed, it was found to contain 2.003 grams of carbon and 0.4448 grams of hydrogen.   What is its empirical formula?

2.003 g of C X 1 mole of C  = 0.16678 moles of C
                        12.01 g of C

0.4448 g of H X 1 mole of H = 0.4404 moles of H
                           1.1 g of H

Therefore the formula is C0.16678H0.4404 = C1.000H2.667
The ratios don't come out to the expected nice ratios.  The solution to this is to multiply to see if we can get a better whole number ratio.
      C0.16678 X 2H0.4404 X 2 = C2H5.334             Still not good enough

      C0.16678 X 3H0.4404 X 3 = C3H8.001   =  C3H8

Example #3
A 5.438 gram sample of ludlamite, a greyish-blue mineral sometimes used in ceramics, was found to contain 2.549 grams of iron, 1.947 grams of oxygen, and 0.9424 grams of phosphorus.  What is its empirical formula?

2.549 grams of Fe X 1 mole of Fe = 0.0456 moles of Fe
                                   55.85 g Fe

1.947 grams of O X 1 mole of O = 0.1217 moles of O
                                   16.00 g O

0.9424 grams of P X 1 mole of P = 0.0304 moles of P
                                    30.97 g P

Therefore the formula is Fe0.0456O0.1217P0.0304 = Fe1.5O4.0P1 =Fe3O8P2

Calculating Empirical Formulas from Percentage Compositions
Example #1
Barium carbonate, a white powder used in paints, enamels, and ceramics, has the following composition: Ba, 69.58%; C, 6.090%; O, 24.32%.  What is its empirical formula?
Assume we have a 100 gram sample since it is out of 100 percent.  Therefore all the percentages change to grams and solve it like the above three examples.
69.58  grams of Ba X 1 mole of Ba = 0.5067 moles of Ba
                                    137.33 g Ba

6.090 grams of C X 1 mole of C = 0.5071 moles of C
                                   12.01 g C

24.33 grams of O X 1 mole of O = 1.520 moles of O
                                   16.00 g O

Therefore the formula is Ba0.5067C0.5071O1.520 = Ba1C1.001O2.999 = BaCO3

Example #2
Calomel is the common name of a white powder once used in the treatment of syphilis.  Its composition is 84.98% mercury and 15.02% chlorine.  What is its empirical formula?
84.98 grams of Hg X 1 mole of Hg = 0.4237 moles of Hg
                                    200.59 g Hg

15.02 grams of Cl X 1 mole of Cl =  0.4237 moles of Cl
                                   35.45 g Cl

Therefore the formula is Hg0.4237Cl0.4237 = HgCl

Combustion Analysis
In combustion analysis a sample of the chemical being analysed is burned with oxygen.  The resulting products are used with a balanced combustion equation to stoichiometrically determine the moles of the starting elements.  The mass spectrometer will give you the peercentages of the atoms as they are burned.  You can then use these in a percentage calculation to get your empirical formula.
Read Page 108-110  for a more complete description of the mass spectrometer and combustion analayzer.

Determining a Molecular Formula from an Empiricial Formula and a Formula Weight
Example #1
Styrene, the raw material for polystyrene plastics, has the empirical formula CH. Its formula wieght is 104.08 g/mole.  What is its molecular formula?
Find the molecular mass of the empirical formula: the empirical moelcular mass
CH = 13.1 g/mole
Divide the actual molecular mass by the empirical molecular mass:  104.08 g/mole / 13.1 g/mole = 8
Multiply the empirical formula by this number:  C1H1 X 8 = C8H8  

Example #2
Calomel has a formula weight of 472.08.  The empirical formula is HgCl.  What is its actual molecular formula?
HgCl = 200.59 + 35.45 = 236.04 g/mole
472.08 g/mole / 236.04 g/mole = 2
Therefore Hg1Cl1 X 2 = Hg2Cl2

Stop here and go to the Empirical Calculations Worksheet
Homework Section 2.3 Questions Page 120
Extension Exercise 2.3 Determining Chemical Formulas
Investigation 2.4  Percentage Composition by Mass of Magnesium Oxide
Alternative Exercise 2.4  Percentage Composition of a Copper Compound
Self Quiz 2.1 - 2.4

Introduction to Solutions and Concentration
Whenever possible, reactions are carried out with all of the reactants in the same fluid phase.
This means that it is preferrable 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 solvnet 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 concnetrated 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 tempeature is called the solubility of the solution.
Units of Concentration (v/v, w/v and ppm)
These units of concentration are most often seen and used with commerical 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%

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

        % concentration =   5 mL    X 100%  = 5%
                                        100 mL

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% = 38.9%
                                        360 mL

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.

eg.   a 3% H2O2 topical antibiotic solution means that there is 3 grams of H2O2 in every 100 mL of solution.

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 concenrtaion in ppm?
ppm concentration = 1 mg   =   2.2 mg   =  8.8 mg/L = 8.8 ppm
                                   1 L          0.25 L
Stop here and go to the
Concentration Unit Calculations Other than Molarity Workshsheet

Molar Concentrations and Molarity
Molarity is a way of specifing the amount of solute in one litre 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 litre 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 litre 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 millilitres.
Use the information on the bottle.  There is 0.400 moles of solute per litre.  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.

Practise 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?
Extension Exercise 2.5  Concentrations of Solutions
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.
Practise problem:  How can we prepare 250 mL of 0.200 M NaHCO3?
Stop here and go to the
Molarity and Solution Creation Worksheet
Homework Section 2.5 Page 137
Alternate Exercise 2.5 - House Hold product Safety Testing
Tec Extension 2.6 The Breathalyzer
Activity 2.7  Determining the Concentration of a Solution
Alternate Activity 2.7 Concentration Using a Spectrophotometer

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 numebr of moes per unit volume decreases.  The solution is said to be diluted.  Such dilutiosn are a natural part of chemistry in the lab.  If two solutiosn of diffreent solutes are moixed, 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 requir for a much less conencrtated 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?
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.

Practise Problem:  Describe how to make 500 mL of 0.20 M NaOH from 0.50 M NaOH.
How many millilitres 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.

Practise Problem: How many millilitres of water would have to be added to 300 mL of 0.5 M NaOH to give a solution with a conentration of 0.2 M?
Stop here and go to the Solution Dilution Worksheet  
Homework Section 2.5 Questions Page 137
What is Stoichiometry?
Stoichiometry is the quantitative description of the proportions by moles of the substances in a chemical reaction.

When doing experiments, it is unscientific and dangerous to mix chemicals in a haphazard manner.  Before any practical laboratory work is done, chemists almost always start with a balanced reaction equation. This is true for a reaction that we know works, or for one that we predict works.  A balanced equation helps tell the reaction's stoichiometry.

The stoichiomentry of a reaction is the description of the relative quantitites by moles of the reactants and products as they appear by the coefficients of the balanced equation.  Stoichiometry is the molar bookkeeping of chemistry, and in nature the books must balance.  Quantitative information is only available through its stoichiometry.  This applies to the pure study of chemistry, but also to the uses of chemistry in a wide variety of fields such as agriculture, clinical analysis, pharmaceuticals, food chemistry, inhalation therapy, nutrition, forensic science, geochemistry and the list can go on and on.  The work begins with the writing and balancing of an equation, because this is the work that gives the coefficients - the numbers that disclose the proportions by mole.
Stoichiometric Calculations: Mole to Mole Calculations 
When we balance an equation it is important to think if it in terms of atoms of each element. For example, in a simple reaction between hydrogen and oxygen to make water, the equation we get is

                           2 H2   +   O2   ------->  2 H2O

which can mean

 2 molecules of H2  +  1 molecules of O2  --->  2 molecules of H2O

However, when we use a balanced equation to plan how much of each reactant to use in an actual experiment, we have to shift our thinking to huge collections of molecules - to moles. The shift from molecules to moles is done by taking advantage of a simple rule from mathematics. Multiplying a set of numbers, such as the coefficients, by any constant number does not alter the ratios among the members of the set.  If we select Avogadro's number as the multiplier then we get lab-sized units of each chemical.

2 X (6.02 X 1023 molecules) of  H2  +   1 X (6.02 X 1023 molecules) of O2 
  2 X (6.02 X 1023 molecules) of H2O

The essential 2:1:2 ratio has not been changed by this multiplication. But the scale of the reaction has shifted to the mole level.

          2 moles of H2   + 1 moles of O2   --------> 2 moles of H2O

The ratio of moles of molecules is identical to the ratio of molecules - it has to be, since equal numbers of moles have equal numbers of molecules.

The ratio of the coefficients for any given chemical reaction is set by nature. You cannot change this ratio.  It is set when you write the formulae correctly and then balance the equation properly.  Once this is done the coefficient numbers can be used as the basis for chemical calculations.  The decision that is left for us is the scale of the reaction - how much do we want to use or make?  The number of options is infinite.  We could have
         0.02 moles of H2   +   0.01  moles of O2   -------->   0.02 moles of H2O
        1.36 moles of H2   +   0.68 moles of O2   -------->   1.36 moles of H2O
        88 moles of H2   +   44 moles of O2   -------->   88 moles of H2O

In every case, the relative mole quantities of H2 to O2 to H2O are 2:1:2.  We could say that 2 moles of H2, 1 mole of O2, and 2 moles of H2O are equivalent to each other in this reaction.  This does not mean that one chemical can actually substitute for any other chemical.  It does mean that a specific mole quantity of one substance requires the presence of a specific mole quantity of each of the other substance in accordance with the ratio of coefficients.

Below shows five different scales for the reaction of iron with sulphur to make iron sulphide, FeS.  Notice that the mole ratios are the same regardless of the scale.

                 1 atom  of Fe   +   1 atom  of S  ---->   1 molecule of FeS

                10 atoms of Fe  + 10 atoms of S ----> 10 molecules of FeS

                 55.8 mg of Fe  +      32.1 mg S  ---->   87.9 mg  FeS

                 5.58 g of Fe     +      3.21 g of S ---->  8.79 g of FeS

                 55.8 g of Fe     +      32.1 g of S  ----> 87.9 g of FeS

Mole to mole calculations:
This is an example of how to do mole to mole type problems:
Two atoms of sulphur react with three molecules of oxygen to form two molecules of sulphur trioxide, which is an air pollutant.

                            2 S  +  3 O2   ------->  2 SO3

How many moles of sulphur react in this way with 9 moles of O2?
Solution: From the balanced equation you can see that  2 S react with 3 O2
Set up your ratio like this:    2 S  =   3 O2
                                            x       9 moles

Cross multiply to get    2 * 9 moles = 3 * x

                                   x = (2 * 9 moles) / 3 = 6 moles

Therefore if 9 moles of oxygen are reacted then 6 moles of S must also be present.
Note that the unit "moles" was carried through the calculation.
Stop here and go to the Mole to Mole Calculations Worksheet
Extension 2.9 The Mole and Chemical Equations
Alterate Extension 2.9 Combustion and Carbom Monoxide
Self Quiz 2.5 - 2.9

The following files are recommended reading.  Not mandatory, but highly recommended.
Eggsamples of Concrete Stoichiometry
Bridging the Stoichiometry Gap

Stoichiometric Calculations: Gram to Gram Calculations
The type of problems involved here can be solved easily with 2-3 simple steps.  Use the Bridge Method for solving only after the equation is balanced. 

Never, ever assume that an equation is balanced.
Example Solution #1
During its combustion, ethane C2H6, combines with oxygen O2 to give carbon dioxide and water.  A sample of ethane was burned completely and the water that formed has a mass of 1.61 grams.  How much ethane, in moles and in grams, was in the sample?

1.  Set up the equation based on the words in the problem.  Then balance it correctly.

                              2  C2H6    +  7 O2      -------->  4  CO2   +  6  H2O

2.  Compute the needed formula molecular masses.   We need the molecular mass for water and for      ethane.    H2O = 18.02 g.mol;   C2H6 = 30.08 g/mol
3.   Draw a roadmap outline the math steps that need to be taken:

                  moles of water  --------->  moles of ethane
                           ^                                        |
                           |                                        V
                 grams of water                   grams of ethane

Therefore this type of question requires a three step solution.
     Step #1  Convert 1.61 grams of water to moles of water.
     Step #2  Using the equation, compare the moles of water made from moles of ethane.
     Step #3  Convert the moles of ethane back into grams of ethane.
Draw a roadmap outlining the mathematical steps that need to be taken:

                  moles of water  --------->  moles of ethane
                            ^                                        |
                            |                                        V
                  grams of water                  grams of ethane

Math Step #1  Convert 1.61 grams of water into moles of water

Moles = g/mm = 1.61 g/ 18.02 g/mol = 0.09 mol of water were used.


Math Step #2   The chemical equation shows us that 2 moles of C2H6 is needed to make 6 moles of water.  Setup a ratio:

                                          6 mol H2O = 2 mol C2H6
                                              0.09 mol            x                

x = 0.03 moles of ethane are needed.


Math Step #3   Convert the moles of ethane back into grams of ethane.
g = n * mm = 0.03 mol * 30.08 g/mol = 0.90 grams of ethane

Finish off the question with a statement.  1.61 grams of water can be made from 0.03 moles of ethane or 0.90 grams of ethane.

Example Question #2
Calculate how many grams of K2Cr2O7 are needed to make 35.8 grams of I2 according to the following equation.

            K2Cr2O7  +  6 NaI  +  7 H2SO4   ------->
                                       Cr2(SO4)3 + 3 I2 + 7 H2O + 3 Na2SO4 + K2SO4

The equation is already balanced but you should check it over just to be sure.

Next calculate the molecular masses that you need. The question deals with iodine, I2, and potassium dichromate, K2Cr2O7.  Calculate the molecular masses for these only.  The other parts of the equation can be forgotten about.

                    I2 =  253.82 g/mol      K2Cr2O7 = 294.20 g/mol

Draw a roadmap outlining the mathematical steps that need to be taken:

                  moles of iodine  --------->  moles of potassium dichromate
                           ^                                        |
                           |                                        V
                 grams of iodine                  grams of potassium dichromate

Math Step #1  Convert grams of iodine into moles of iodine.

               moles = g/mm = 35.8 g / 253.82 g/mol = 0.14 moles of iodine

Math Step #2  Compare using the equation coefficients, the number of moles of potassium dichromate is need to make how many moles of iodine.

                 3 I2    =   1 K2Cr2O7
              0.14 mol          x                             

x = 0.05 moles of potassium dichromate

Math Step #3 Convert the moles of potassium dichromate back into grams of potassium dichromate.

       g = n * mm = 0.05 mol *  294.20 g/mol = 14.71 grams

To make 35.8 grams of iodine you must start with 14.71 grams of potassium dichromate.
Stop here and go to the  Stoichiometry Gram to Gram Calculations Worksheet
Stop here and go to the More Gram to Gram Calculations Worksheet

Limiting Reagents and Percentage Yield
"If one reactant is entirely used up before any of the other reactants, then that reactant limits the maximum yield of the product."

Problems of this type are done in exactly the same way as the previous examples, except that a decision is made before the ratio comparison is done. The decision that is made is "What reactant is there the least of?"

Example Problem #1
Methane, CH4, burns in oxygen to give carbon dioxide and water according to the following equation:
                          CH4  + 2 O2 ------>  CO2 + 2 H2O
In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel vessel.  Find the limiting reactant, if any, and calculate the theoretical yield, (in moles) of water.
In any limiting reactant question, the decision can be stated in two ways.  Do it once to get an answer, then do it again the second way to get a confirmation.
According to the equation:    1 mol CH4 = 2 mol O2
If we use up all the methane then:
                              1 mol CH4 = 2 mol O2
                                 0.25 mol          x                    

 x = 0.50 mol of O2 would be needed.

We have 1.25 mol of O2 on hand. Therefore we have 0.75 mol of O2 in excess of what we need.
If the oxygen in is excess, then the methane is the limiting reactant.
Confirmation:  If we use up all the oxygen then

                            1 mol CH4 = 2 mol O2
                                    x             1.25 mol         

x = 0.625 mol of methane.

We don't have 0.625 moles of methane. We have only 0.25 moles.  Therefore the methane will be used up before all the oxygen is.  Again the methane is the limiting reactant.
We now use the limiting reactant to make the mole comparison across the bridge to find the amount of water produced.
                           1 mol CH4 = 2 H2O
                              0.25 mol         x            

x = 0.50 mol of H2O  would be produced.

Finish off with a statement:  When 0.25 mole of methane and 1.25 mole of oxygen are mixed and reacted according to the equation, the methane is the limiting reactant and the maximum yield of water will be 0.50 moles.

Example Problem #2
Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl.  In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed.  Which is the limiting reactant?  What is the maximum yield of CCl4 in moles and in grams?
Start with the equation:        CHCl3  +  Cl2 ------->  CCl4  +  HCl
Did you check to see if it was balanced?
Calculate the molecular masses of the species needed in the problem.
CHCl3 = 1 C = 1(12.01)  =    12.01                   Cl2 = 2 (35.45) = 70.90 g/mol
               1 H =  1(1.01)  =       1.01       H2O = 2 H = 2  (1.01)  =   2.02
               3 Cl =  3(35.45) = 106.35                    1 O = 1 (16.00) = 16.00
                                            119.37 g/mol                                        18.02 g/mol
Then calculate the moles of each of the reactants to be used.
moles of CHCl3 =    =    25.00 g       
                             mm   119.37 g/mol

=  0.21 moles of CHCl3 are present.

moles of Cl2   g   =     25.00 g      
                        mm        70.90 g/mol 

=  0.35 moles of chlorine are present.  

Decision time.  Which of the two reactants do you have the least of?
From the balanced equation you can see that the chloroform and chlorine reactant in a one to one ratio.  If we use all the chloroform then we get the following equation.

                      1 CHCl3  1 Cl2
                      0.21 mol           x                    

x = 0.21 moles of chlorine are needed.

We need 0.21 moles of chlorine.  We have 0.35 moles of chlorine.  Therefore chlorine is in excess.  The chloroform must be the limiting reactant.
Confirmation:  IF we use all the chlorine then:

                       1 CHCl3  1 Cl2
                           x             0.35 mol             

x = 0.35 moles of chloroform are needed.

If we use all the chlorine then we need 0.35 moles of chloroform.  We have only 0.21 moles of chloroform.  It is the reactant that we will run out of first.  Therefore it is the limiting reactant.
Use the limiting reactant to cross the ratio bridge and find the number of moles of water made.

                             1 CHCl3  =  2 H2O
                             0.21 mol         x           

x = 0.42 moles of H2O will be made.

Calculate the grams of water produced.   

grams = moles * molecular mass

           = 0.42 mol * 18.02 g/mol
           = 7.57 grams of water
Finish off with a statement:   When 25 grams of each reactant are mixed according to the equation, the chloroform is the limiting reagent and the maximum yield of water will be 0.42 moles or 7.57 grams.

Example Problem #3
Aluminum chloride, AlCl3, can be made by the reaction of aluminum with chlorine according to the following equation:
                                      2 Al  + 3 Cl2  ------> 2  AlCl3
What is the limiting reactant if 20.0 grams of Al and 30.0 grams of Cl2 are used, and how much AlCl3 can theoretically form?
Have you checked to make sure the equation is balanced correctly?
Find the molecular masses of all species involved.
Al = 26.98 g/mol      Cl2 =  70.90 g/mol      AlCl3 =  133.33 g/mol
Convert the grams into moles.
moles of Al = g/mm = 20.00 g/26.98 g/mol = 0.74 moles of aluminum on hand.

moles of Cl2 = g/mm = 30.00 g/70.90 g/mol = 0.42 moles of chlorine on hand.

Decision time:  Which is the limiting reagent?
IF we use all aluminum then:

                               2 Al       =   3  Cl2
                            0.74 mol          x         

x = 1.11 moles of chlorine are needed.

We don't have 1.11 moles of chlorine.  We have 0.42 moles of chlorine. Therefore we will run out of chlorine first. It is the limiting reactant.
If we use all the chlorine then:

                                2 Al       =   3  Cl2
                                   x            0.42 mol       

x = 0.28 moles of aluminum are needed.

We have 0.74 moles of aluminum, therefore it is in excess. If it is in excess then the chlorine is the limiting reactant.
Use the limiting reactant to cross the ratio bridge and find the moles of AlCl3 that will be produced.
                                      3 Cl2   =        2 AlCl3
                                  0.42 mol             x                  

x = 0.28 moles of AlCl3 are produced

Grams of aluminum chloride are found with
g = n * mm = 0.28 mol * 133.33 g/mol  = 37.33 g

Finishing statement:  When 20.0 grams of aluminum and 30.0 grams of chlorine are reacted according to the above equation, the chlorine is the limiting reactant and the maximum yield of aluminum chloride is 0.28 moles or 37.33 grams.

Stop here and go to the Limiting Reagents Worksheet
Homework Section 2.10 Questions Page 153

Stop here and go to the Percentage Yield Problems Worksheet
Homework Section 2.12 Questions Page 159

Extension 2.10  Limiting and Excess reagents
Investigation 2.11 The Limiting Reagent in a Chemical Reaction
Alternative Exercise Activity 2.11 The Limiting reagent
Extension 2.12 Percentage Yield
Investigation 2.13 Percentage Yield
Self Quiz 2.10 - 2.14
Unit 2 Summary
Homework Unit 2 Review  Page 169 through 173

Unit 2 Self Quiz