The Law of Conservation of Mass:
The total mass of the substances does not change during a chemical reaction. The
number of substances might change (e.g. in the reaction below we have 9 total molecules
on the left, and 11 total molecules on the right), their properties might change, but the
total amount of matter must remain constant. In the reaction below: there are 5 carbons
on both sides of the reaction. There are 12 hydrogens on both sides of the reaction, and
there are 16 oxygens total on both sides of the reaction. The moral is: matter can be
neither created nor destroyed. The law of conservation of mass can be explained as:
atoms are neither created nor destroyed during a chemical reaction, they just rearrange to
form a new substance.
C5H12
(l) +
8O2 (g)
5 CO2 (g) +
6 H2O (l)
We took a hydrocarbon (C5H12) known as pentane, added some oxygen to it and with an
input of energy we generate CO2 gas and water.
The Law of Definite Proportion/Constant Composition:
Constant composition refers to the elemental composition by mass of a given compound
being the same for all samples of that compound. This means that every, every, every
time water is decomposed, there are 88.8 grams of oxygen present for every 11.2 g of
hydrogen. And a 100-gram sample of NaCl always contains 39.3 grams of Na and 60.7 g
of Cl.
How does this work?
NaCl is made up of 1 Na and 1 Cl.
The molar mass of Na = 22.9 grams/mole
The molar mass of Cl = 35.4 grams/mole
Thus for one mole of NaCl, the amount of
Na =
22.9
22.9  35.4
*
100 = 39.3 %
If you had a 100 gram sample, 39.3% of that 100 grams = 39.3 grams
Constant composition/definite proportion is explained by assuming that atoms combine
in fixed ratios when they form a compound. Thus, if one oxygen atom combines with 2
hydrogen atoms to form water, then all samples of water must have the same
composition.
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The Law of Multiple Proportions:
In different compounds containing the same elements, the masses of one element
combined with a fixed mass of the other element are in the ratio of the smallest whole
numbers.
Consider SO2 and SO3. Both compounds contain only sulfur and oxygen atoms. The
formulas suggest that for a given mass of sulfur, the masses of oxygen in SO3 compared
to SO2 will be in a 3/2 ratio. Other examples include CO and CO2, FeCl2 and FeCl3, UF3
UF4 and UF6.
How does this work?
SO2
molar mass of S = 32.06 grams/mole
molar mass of O = 16.00 grams/mole
grams of S in a 100 g sample = 50.05 grams
grams of O in a 100 g sample = 49.95 grams
gramsO
= 0.998 ≈ 1
gramsS
SO3
molar mass of S = 32.06 grams/mole
molar mass of O = 16.00 grams/mole
grams of S in a 100 g sample = 40.04 grams
grams of O in a 100 gram sample = 59.96 grams
gramsO
= 1.498 ≈ 1.5
gramsS
Since we do not want decimals (remember we are looking for the ratio of the smallest
whole numbers!!) for ratios we must get rid of the 0.5 and turn it into a whole number. If
we multiply a 0.5 by 2 we get 1 (for 0.33 multiply by 3 to get the new whole number, for
0.25 multiply by 4, for 0.75 multiply by 4 etc . . .)
So – comparing the O in SO3 to the O in SO2 for a given 1 atom of sulfur results in 3 O
atoms (1.5*2) to 2 O atoms (1*2). REMEMBER!! Whatever we do to one ratio number we
must mathematically do to the other. Therefore a 1.5:1 ratio is equal to 3:2 ratio if we
want the answer in only whole numbers (which we DO!!)
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It is very important to begin thinking in terms of moles when examining chemical reactions as it
clarifies the amount of each substance present. Comparing molar masses does not clarify the
ratio of each substance present. Thinking in terms of masses does nothing but show us that the
mass during the chemical reaction will be conserved (the grams at the beginning will be the
same number of grams at the end of the chemical reaction). For example, consider the following
relationships:
2.016 g of H2 + 38.00 g F2  40.02 g HF
1 mole of H2 + 1 mole F2  2 mole HF
This information shows that equal amounts of H2 and F2 molecules combine to form twice as a
large a population of HF molecules.
This chemical equation is made up of chemical formulas that express the identity and quantity
of substance that undergo a chemical or physical change. These chemical equations are like
sentences, which means you need to KNOW the names of the chemical species that are coming
together and undergoing a reaction. The left side of the equation shows the chemical species
that are coming together and doing the reacting, thus they are called the reactants, or reagents.
The chemical species that we end up with are called the products. The reactant side shows the
amount of substance(s) present before the change and the right side shows the amount of
material(s) produced. For the equation to accurately depict the amounts, it must be balanced.
That means the same atoms must appear on both sides of the equation and there must be the
same number of moles of each atom on both sides of the equation.
Rules for Balancing Chemical Reactions
1.) Write a skeletal equation: this will include the reactants and the products
2.) Balance the number of atoms: the number of atoms on the left side of the
arrow should be the same as the number of atoms on the right side of the arrow.
3.) Make the coefficients whole numbers: even though we can accurately
balance a chemical equation using fractions, at this time, we will want all the
numbers as whole numbers. This means multiplying through the entire equation.
4.) Indicate the state of matter: are the reactants solids, liquids or gases? Wh
What about the products? Identify the state of matter!
Concept Test
Balance the following equations:
______Na
______H3PO4
______C8H18
+
+
______Cl2 
______NaCN
+

______O2 
3
______NaCl
______HCN
______CO2 +
+
______Na3PO4
______H2O
Keep in mind that when you are balancing the equations that the coefficient operates on all the
atoms (or ions) in the formula. Just like the parentheses when we write a chemical formula, the
subscript applies to all atoms inside the parentheses, the coefficient operates on all atoms in the
formula. You cannot change the chemical formulas to balance the number of atoms. The
formulas are set by the type of compound it is. For example, in the reactions above, you cannot
change Na3PO4 to NaPO4 in order to “balance” the number of sodium ions! You also cannot
add new elements to chemical reactions in order to balance the reaction. You do not balance the
reaction by adding 2 Na+1 ions to the reactant side so that you have 3 sodium ions on each side.
Use the reactants and products given and balance by using coefficients. A balanced chemical
equation will stay balanced no matter if you multiply or divide each coefficient by the same
value. You can multiply the system by 2 or 10 or 1000, but the atoms/ions will still be balanced.
However, generally speaking, a balanced equation should contain the lowest whole number
ratio of coefficients!
A balanced chemical equation contains a lot of information relating the individual chemical
entities, amounts of chemical entities, and the masses of reactants and products. A BALANCED
chemical equation is essential for all calculations involving reactants and products. When the
equation is balanced, we can derive stoichiometric relationships between reactants and other
reactants, reactants and products, or even products and other products using ratios called mole:
mole ratios (mole to mole ratios).
Let’s consider the following balanced equation:
1C3H8 + 5O2

3CO2
+ 4H2O
From the balanced equation, the following stoichiometric relationships can be derived:
1 mole C 3 H 8
1 mole C 3 H 8
1 mole C 3 H 8 3 mole CO 2
5 mole O 2
4 mole H 2 O
;
;
;
;
;
5 mole O 2
1 mole C 3 H 8 3 mole CO 2 1 mole C 3 H 8 4 mole H 2 O
1 mole C 3 H 8
5 mole O 2 3 mole CO 2 5 mole O 2
4 mole H 2 O 3 mole CO 2 4 mole H 2 O
;
;
;
;
;
3 mole CO 2 5 mole O 2 4 mole H 2 O 5 mole O 2 4 mole H 2 O 3 mole CO 2
The mole : mole ratios can be used as conversion factors to calculate values for any species in
the chemical reaction.
Concept Test
If we use (consume) 15 moles of C3H8 how many moles of oxygen will be used?
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This problem could not have been solved without first balancing the chemical equation. The
general approach for solving any problem that involved the chemical equation is:
1. Balance the chemical equation
2. Convert the given mass or volume or some other measureable value of the first substance
into moles
3. Use the correct mole : mole ratio from the balanced equation to calculate the moles of the
other substance
4. Convert the moles of the substance into mass or volume or some other measureable
value of the asked about species.
General Template for Chemical Reaction Conversions
given mass in grams of A
1 mol A
Y mol B molar mass in grams of B
 mass in grams of B
molar mass in grams of A X mol A
1 mol B
Concept Test
Solid copper (I) sulfide reacts in the presence of oxygen gas to form solid copper (I) oxide and
sulfur dioxide gas.
1. write a balanced chemical reaction with states of matter for each species
2. how many moles of oxygen gas will react with 10.0 moles of copper (I) sulfide?
3. how many grams of sulfur dioxide are formed when 10.0 moles of copper (I) sulfide react
with oxygen?
4. how many kilograms of oxygen are required to form 2.86 kg of copper (I) oxide?
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Sometimes chemical reactions occur and we run out of one of the reagents. Thus, the formation
of the product is limited. Imagine if you wanted to make 4 batches of chocolate chip cookies,
and each batch requires 1 bag of chocolate chips. If you only have 3 bags of chocolate chips, you
cannot possible hope to make 4 batches of cookies! Thus, our product (the cookies) would be
limited to 3 batches as that is the number of bag of chips that we have. It is the same with
chemical reactions. Once a reagent had been used up, no more product(s) can be made.
In choosing the limiting reactant, choose the reactant that will make (yield) the LOWEST
amount of product. The limiting reagent is NOT the reactant present in the fewest amount of
moles, it is the reactant that yields (forms) the fewest amount of moles of product. The limiting
reactant is NOT the reactant with the lowest mass. It is the reactant that forms the lower mass
of product.
Step 1: Balance the chemical reaction
Step 2: Pick a product (circle it) or if one is chosen for you (asked about in the question) circle it
Step 3: Using the starting amounts of each reactant given, perform a separate calculation
starting with each reactant and calculating the amount of product (the one that you circled!)
formed based on THAT starting amount. Please note that you must report all final answers of
product created in measurable units (e.g. grams or mL, you cannot stop at moles as a final
answer unless specifically asked to do so. Why? Because we can’t measure moles in the lab!)
Step 4: Look at the amounts of products formed based on the reactants. Compare them in your
mind. Circle the SMALLEST amount of product formed. The circled amount of product
formed is called your THEORETICAL YIELD.
Step 5: Draw an arrow from the theoretical yield that you circled BACK to the reactant that you
used to calculate that amount of product. The arrow is pointing at the LIMITING REACTANT.
Step 6: If asked, determine the amount of excess reagent (coming soon)
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Concept Test:
Liquids hydrazine (N2H4) and dinitrogen tetraoxide (N2O4) ignite to form nitrogen gas and
water vapor. How many grams of nitrogen gas will form when 1.00 x 102 grams of N2H4 and
2.00 x 102 grams of N2O4 are mixed? Which reactant is the limiting reagent? Is the limiting
reagent the same if we calculated the amount of water vapor formed instead? Prove your
answer mathematically.
2N2H4 (l) + 1N2O4 (l)  3N2 (g) + 4H2O (g)
Excess reagents:
Now that you have determined the amount of product formed, as well as identified the
Limiting Reactant, we now need to figure out how much of the other reactant or reactants we
have left over.
When determining this, NEVER re-use calculated numbers unless you absolutely have to (e.g.
molar masses). Here’s how we will approach excess reactants (yes, there are other ways, but
you CANNOT use calculated amounts to perform the calculations. If you do, you will NOT
receive full credit, even if your answer is correct!)
You have identified the Limiting Reactant (LR). Now let’s examine the exact same problem in a
different way. The question you now want to ask yourself is “self, how much of the excess
reactant would react with this given amount of limiting reactant?”
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So, using the reaction stoichiometry, perform the calculations.
Step 1: Using the starting amount of the Limiting Reactant, determine how much of the excess
reactant would be used/consumed/reacted with the LR.
Step 2: Set up your conversion factor, beginning with the LR and solving for a final answer of
your other reactant (s)
Step 3: The answer in step 2 is called the amount of excess reactant USED.
Step 4: Take the starting amount of the excess reactant given in the calculation and subtract the
amount of reactant USED. This equals the amount of excess reactant!
When you perform chemical reactions, ideally you will end up with a yield of 100%. In reality,
it just does not happen. Along they way, as chemical reactions proceed, product is lost. The
theoretical yield is the amount of product expected. You will CALCULATE the theoretical
yield. What we have been doing, calculating the amount of product given a certain number of
moles of reactant or grams of reactant IS calculating the theoretical yield. However, the actual
yield is determined by experimentation, and thus the value will be given to you and you will
calculate % yield, or the % yield would be given and you would be asked to calculate the actual
yield.
% yield =
actual yield
x 100
theoretica l yield
Concept Test:
In the above problem you calculated the theoretical amount of N2 gas. What is the % yield if
you actually obtained 122 grams N2 instead?
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