Unit
2 Quantities in Chemistry 
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76 Answers
Atomic Mass
The sum of individual atoms can be used to find the mass of a molecule. The mass of hydrogen peroxide, H_{2}O_{2} would be calculated like this: H_{2}O_{2} has 2 hydrogen atoms and 2 oxygen atoms in it. Therefore the mass is 2 X O = 2 X 16.00 u = 32.00 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, Ca_{3}(PO_{4})_{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 PO_{4} 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
8 X O = 8 X 16.00 u = 128.00 u One molecule of calcium phosphate weighs 310.18 u Stop here and do the Atomic Masses Exercise 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 makeup. 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%
= 0.8456 X 100% = 84.56% 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

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 Carbon12 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 agreedupon reference is the most abundant isotope of carbon, which is carbon12. 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 carbon12. 
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 carbon12? 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. (hydrogen1 has a mass of 1.007825 amu when carbon12 is assigned a mass of exactly 12 amu.) 
The Mole 
Suppose we want to make molecules of carbon dioxide, CO_{2}, 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 
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, H_{2}O, 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 carbon12 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. 

Avogadro's Number 
The mole is defined as 6.02 x 10^{23} units.
It is called Avogadro's number in honour of Italian scientist, Amadeo Avogadro
(17761856). 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 10^{23} = 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 10^{23} particles (General
expression) 
Another useful relationship is that 1 amu = 1.66 X 10^{24}
g
SO
1 amu
= 6.0 X 10^{23} particles 
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 
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, NaHCO_{3}, is
one ingredient of baking powder. How many grams of sodium bicarbonate
are in 0.673 moles? 
Step 1. Find the molecular mass: NaHCO_{3} = 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 mol = 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, KMnO_{4},
at one time was used for an antifungal 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 have? 
Step 1: Find the molecular mass of KMnO_{4}:
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 * N_{A} 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 CO_{2}? molecules = n * N_{A} = 3 moles * 6.0 X 10^{23} molecules/mole = 18.06 X 10^{23} molecules = 1.806 X 10^{24} molecules There are 1.806 X 10^{24} molecules in 3 moles of CO_{2}. 
Converting
Molecules into Moles The necessary formula is: moles (n) = molecules N_{A} 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 10^{24} molecules of XeF_{6}. How many moles of this particle sample of XeF_{6} do you have? moles = molecules N_{A} = 5.6 X 10^{24} molecules 6.02 X 10^{23} molecules/mole = 9.3 moles There are 9.3 moles of XeF_{6} 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 H_{2}O has a subscript total of 3 atoms. H_{2}SO_{4} has a subscript total of 7 atoms. MgSO_{4}•7 H_{2}O has a subscript total of 27 atoms. The formula needed is: atoms = molecules X subscript total Example: You have 1.2 X 10^{23} molecules of hydrogen peroxide, H_{2}O_{2}. How many atoms in total do you have? atoms = molecules X subscript total = 1.2 X 10^{23} molecules X 4 atoms/molecule = 4.8 X 10^{23} atoms There are 4.8 X 10^{23} atoms in total in this sample of H_{2}O_{2}. 
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 samesize 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

Therefore the formula is C_{0.09176}O_{0.09181}
Pick the smallest of the subscript numbers as the divisor and divide it into
all the other numbers.
C_{1.0000}O_{1.0001} = C_{1}O_{1} 
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

Therefore the formula is C_{0.16678}H_{0.4404}
= C_{1.000}H_{2.667} 
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. 
C_{0.16678 X 2}H_{0.4404
X 2} = C_{2}H_{5.334
}Still not good enough
C_{0.16678 X 3}H_{0.4404
X 3} = C_{3}H_{8.001} = C_{3}H_{8}

Example #3 
A 5.438 gram sample of ludlamite, a greyishblue 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
0.9424 grams of P X 1 mole of P = 0.0304 moles of P

Therefore the formula is Fe_{0.0456}O_{0.1217}P_{0.0304}
= Fe_{1.5}O_{4.0}P_{1} =Fe_{3}O_{8}P_{2}

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

Therefore the formula is Ba_{0.5067}C_{0.5071}O_{1.520} = Ba_{1}C_{1.001}O_{2.999} = BaCO_{3} 
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 
Therefore the formula is Hg_{0.4237}Cl_{0.4237}
= 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 108110 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: C_{1}H_{1} X 8 = C_{8}H_{8} 
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 Hg_{1}Cl_{1} X 2 = Hg_{2}Cl_{2 } 
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 
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 = V_{solute}
X 100% V_{solvent} 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% 
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% H_{2}O_{2} topical antibiotic solution means that there is 3 grams of H_{2}O_{2} 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 10^{6} parts 1 part per billion (ppb): 1 part out of 1 X 10^{9} parts 1 part per trillion (ppt): 1 part out of 1 X 10^{12} parts 1 ppm = 1 drop in a full bathtub 
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 10^{6} mL 1000 L 1 L 1 kg 1 g 
Example #2 Dissolved O_{2} in water shows a concentration of 250 mL of water At SATP and 2.2 mg of O_{2}. 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 
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. 
Solution: 
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
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 Na_{2}CO_{3}
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 Na_{2}CO_{3} that are going to be in 500 mL of 0.150 M
Na_{2}CO_{3} 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 x = 0.075 mol of solute. Therefore we need to weigh out 0.075 mol of Na_{2}CO_{3}. The number of grams of Na_{2}CO_{3} are therefore:
g = n * mm 
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 NaHCO_{3}? 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 V_{c} and M_{c} are the concentrated solutions
volume and molarity. V_{d} and M_{d} are then the diluted
solutions volume and molarity.
V_{c }M_{c} = V_{d }M_{d}

Problem: How can we prepare 100 of 0.040 K_{2}Cr_{2}O_{7}
from 0.200 M K_{2}Cr_{2}O_{7}? Solution: 
First assemble the data into a table: V_{c} = ? V_{d} = 100 mL M_{c} = 0.200 M M_{d} = 0.040 M Next, use the formula from above. V_{c} X 0.20
M = 100 mL X 0.040 M Place 20.0 mL of 0.200 M K_{2}Cr_{2}O_{7}
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. 
Problem: 
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.
V_{c} = 100 mL V_{d}
= ? M_{c} = 0.40 M M_{d} = 0.10 M Use the equation: 100 mL X 0.40 M = V_{d}
X 0.10 M 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 
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 H_{2} + O_{2} > 2 H_{2}O 
which can mean
2 molecules of H_{2}
+ 1 molecules of O_{2}
> 2 molecules of H_{2}O

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 labsized units of each chemical. 
2 X (6.02 X 10^{23}
molecules) of H_{2} + 1 X (6.02 X 10^{23} molecules) of O_{2}
> 2 X (6.02 X 10^{23} molecules) of H_{2}O 
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 H_{2} + 1 moles of O_{2} > 2 moles of H_{2}O 
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 H_{2} +
0.01 moles of O_{2}
> 0.02 moles of H_{2}O
or 1.36 moles of H_{2} + 0.68 moles of O_{2} > 1.36 moles of H_{2}O or 88 moles of H_{2} + 44 moles of O_{2} > 88 moles of H_{2}O 
In every case, the relative mole quantities of H_{2} to O_{2} to H_{2}O are 2:1:2. We could say that 2 moles of H_{2}, 1 mole of O_{2}, and 2 moles of H_{2}O 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 O_{2} > 2 SO_{3} 
How many moles of sulphur react in this way with 9 moles
of O_{2}? Solution: From the balanced equation you can see that 2 S react with 3 O_{2} Set up your ratio like this: 2 S = 3 O_{2} 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 23 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 C_{2}H_{6}, combines with oxygen O_{2} 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? 
Solution: 
1. Set up the equation based on the words in the problem.
Then balance it correctly.
2 C_{2}H_{6} + 7 O_{2}
> 4 CO_{2} + 6 H_{2}O

2. Compute the needed formula molecular masses.
We need the molecular mass for water and for
ethane. H_{2}O = 18.02 g.mol; C_{2}H_{6}
= 30.08 g/mol 
3. Draw a roadmap outline the math steps that
need to be taken:
moles of water > moles 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 
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 C_{2}H_{6} is needed to make 6 moles of water. Setup a ratio: 
6 mol H_{2}O = 2 mol C_{2}H_{6} 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 K_{2}Cr_{2}O_{7} are needed to make 35.8 grams of I_{2} according to the following equation. 
K_{2}Cr_{2}O_{7} + 6 NaI +
7 H_{2}SO_{4} > Cr_{2}(SO_{4})_{3} + 3 I_{2} + 7 H_{2}O + 3 Na_{2}SO_{4} + K_{2}SO_{4} 
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, I_{2}, and potassium dichromate, K_{2}Cr_{2}O_{7}.
Calculate the molecular masses for these only. The other parts of the
equation can be forgotten about.
I_{2} = 253.82 g/mol K_{2}Cr_{2}O_{7} = 294.20 g/mol 
Draw a roadmap outlining the mathematical steps that need
to be taken:
moles of iodine > moles 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 I_{2} = 1 K_{2}Cr_{2}O_{7}
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. 
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, CH_{4}, burns in oxygen to give carbon dioxide
and water according to the following equation: CH_{4} + 2 O_{2} > CO_{2} + 2 H_{2}O 
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.

Solution: 
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 CH_{4}
= 2 mol O_{2} 
If we use up all the methane then: 1 mol CH_{4} = 2 mol O_{2} 0.25 mol x x = 0.50 mol of O_{2} would be needed. 
We have 1.25 mol of O_{2} on hand. Therefore we have
0.75 mol of O_{2} 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 CH_{4} = 2 mol O_{2} 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 CH_{4} = 2 H_{2}O
0.25 mol x x = 0.50 mol of H_{2}O 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, CHCl_{3}, reacts with chlorine, Cl_{2},
to form carbon tetrachloride, CCl_{4}, 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 CCl_{4}
in moles and in grams? 
Solution: Start with the equation: CHCl_{3} + Cl_{2} > CCl_{4} + HCl Did you check to see if it was balanced? 
Calculate the molecular masses of the species needed in the
problem. CHCl_{3} = 1 C = 1(12.01) = 12.01 Cl_{2} = 2 (35.45) = 70.90 g/mol 1 H = 1(1.01) = 1.01 H_{2}O = 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 CHCl_{3} = g = 25.00 g mm 119.37 g/mol = 0.21 moles of CHCl_{3} are present. moles of Cl_{2} = g
= 25.00 g
= 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 CHCl_{3} = 1 Cl_{2} 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 CHCl_{3} = 1 Cl_{2} 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 CHCl_{3} = 2 H_{2}O
x = 0.42 moles of H_{2}O 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, AlCl_{3}, can be made by the reaction
of aluminum with chlorine according to the following equation:
2 Al + 3 Cl_{2} > 2 AlCl_{3} 
What is the limiting reactant if 20.0 grams of Al and 30.0
grams of Cl_{2} are used, and how much AlCl_{3} 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 Cl_{2} = 70.90 g/mol AlCl_{3} = 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 Cl_{2} = 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 Cl_{2}
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. 
Confirmation: If we use all the chlorine then:
2 Al = 3 Cl_{2}
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 AlCl_{3} that will be produced. 3 Cl_{2 } = 2 AlCl_{3} 0.42 mol x x = 0.28 moles of AlCl_{3} 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 