Balancing Chemical Equations and Stoichiometry
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What happens in a chemical reaction?
e.g., for the reaction 2 H2 + O2  2 H2O reaction
(2 g + 16 g  18 g); (8 g + 64 g 172 g); (16g+128g  144 g)
We have starting materials (reactants) that are converted into different chemical substances (the
products).
We could describe, in words, the reaction and its materials, but it is generally more useful to
write down a chemical equation
 NOTE: Formula of starting materials, the reactants, appear on LHS. Formula of
products appears on RHS.
 the LHS and RHS and separated by an arrow, which means "yields" or gives.
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Chemical equations are usually written balanced (and they better be for the quizzes and exams).
Balanced equations have the same number of atoms of a given element on the LHS and the RHS
because of the law of conservation of mass.
All the reactants and products must be identified, we must know their formulae, and generally, their
physical states (solid, liquid, or gas!).
Chemical equations report the results of experimentation
Steps in Balancing a Chemical Equation
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Rocket propulsion; reactants: hydrazine N2H4 and dinitrogen tetraoxide (N2O4) with N2 and H2O
as the products
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Step 1 :
Step 2 -
N 2 H 4 + N 2O4  N 2 + H 2O
skeletal equation
use the law of conservation of mass to balance the reaction. Let's start with O
4 O atoms (LHS)
H atoms 4 x 2 = 8 (RHS)
N atoms LHS 2x2+2 = 6
Balanced equation
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 need 4 H2O on RHS
 need 2 N2H4 on LHS
 need 3 N2 on RHS
2 N 2 H 4 + N 2 O4  3 N 2 + 4 H 2 O
Another example
C2 H 5OH + O2  CO2 + H 2 O
1. 2 C atoms on LHS  need 2 atoms on RHS
2. 6H atoms on LHS  need 3 H2O molecules on RHS
3. 7O atoms on RHS  need 3 O2 molecules on LHS
2
C2 H 5OH + 3 O2  2 CO2 + 3 H 2 O
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Balanced Equation
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It is usually beneficial to know the physical state of the reactants and the products i.e., whether they
are solid, liquid, gas, or solutions.
Types of Reactions
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A combustion reaction can be written for any compound containing C, H, O or C, H, S etc. O2 (g)
is present as a reactant and is usually INXS.
C2 H4 + 3 O2  2 CO2 + 2 H2O
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We can also have combination and decomposition reactions
Combination reaction – two or more substances combine to form one product
2Mg(s) + O2(g)  2 MgO(s)
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Decomposition reaction, i.e., one substance breaks down to form two or more substances
MgCO3(s)  MgO(s) + CO2(g)
NH4Cl(s)  NH3(g) + HCl(g)
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Decomposition of sodium azide, NaN3
2NaN3(s)  2N(s) + 3N2(g)
 this is the reaction that is important in auto safety.
 How do we use the atomic masses of elements to determine the molecular masses of molecules?
 Simple. All we have to do is add the masses of the constituent elements in the molecule.
 What is the mass of a molecule of methane, CH4 ?
 Molecular mass = 1 atom C x 12.011 amu C/atom C + 4 atoms H x 1.0079 amuH/atomH
= 16.043 amu's
 Calculate the molecular mass of sodium nitrate, NaNO3
 Molecular mass = 1 atom Na x 22.990 amu / atom Na + 14.007 amu /atom N x 1 atom N +
15.999 amu / atom O x 3 atoms O = 84.967 amu / molecule
3
 For H2SO4
 Molecular mass = 2x(1.008 amu) + (32.066 amu) + 4 x (15.999 amu) = 98.078 amu
Molar Mass + Avogadro’s Number
 NOTE: 1 amu = 1.6605 x 10-24 g (how could we handle that?)
 1 mole (SI definition) - the amount of substance that contains as many elementary particles (atoms,
12
molecules, ions) as there are atoms in 12 grams (exactly) of 6 C . This number is called
Avogadro's # (experimentally determined)
 1 mole = 6.022  1023 particles
 1 mole of
12
6
C has a mass of 12.00000 .... g exactly!
 the molar mass of any element in g/mole is the same numerically as its mass in atomic mass
units!
 The molar mass is the mass of 1 mole of a compound, i.e., the molar mass of a compound (in
grams) is equal to its molecular mass (in amu). Above example molar mass of H2SO4 = 98.09 g
H2SO4 / mol.
 Avogadro’s number is very large.
Example
 If Keith Richards drank 10 shots (oz.) of Rebel Yell/day (an underestimate) how long would it take
him to consume of Rebel Yell shots?
Consumption (neglecting leap years)
10 shots Rebel Yell/day x 365 days / year = 3650 shots Rebel Yell/year
1 mole Rebel Yell shots = 6.022 x 1023 Rebel Yell shots
 number of years = 1 year/3650 Rebel Yell shots x 6.022 x 1023 Rebel Yell shots = 1.65 
1020 years
Example #2
239
Doc Brown needed 0.002043 g of Pu-239 ( 94 Pu ) to power the flux capacitor in his time
machine in Back to the Future. How many moles of Pu -239 is this? How many atoms?
4
Number of moles of
239
94
Pu
239
= 0.002043g x 1 mol/239g = 8.55 x 10-6 mol of 94 Pu
239
94
239
10-6 moles 94
239
1018 atoms of 94
Pu
Number of atoms of
= 8.55 
= 5.15 
Pu
x 6.022 x 1023 atoms moles
Pu
Example #3
Ne is a gas used in lighting (neon lights)
How many moles of Ne are contained in 32.47g of the gas? How many atoms are contained in
this amount of the gas?
Answer
Number of moles of Ne
= 32.47g of Ne  1mol Ne/20.180 g Ne
= 1.609 mol Ne
Number of atoms of Ne
= 1.609 mol  6.022 x 1023 /mol
= 9.689  10 23 Ne atoms.
5
Masses of Anions and Cations
Mass of ions
 Since the electron mass is small, the mass of e.g. Na+ is essentially that of Na and the mass of Cl- is
essentially that of Cl, i.e., the masses of ions are virtually unchanged from the unchanged atoms from
what they are derived.
EXAMPLE
How many formula units of NaF are present in 27.73g of NaF; note - NaF is composed of the Na+
cation and the F- anion. Therefore, the molar mass would simply be the molar masses of the elements
Na and F = 41.988g/mole (18.998 + 22.990)
Answer:
27.73 of NaF  1 mole NaF / 41.988 NaF  6.022  1023 FU's of
NaF/1mol NaF = 3.977  1023 formula units of NaF
Percent Composition
 percent composition is the percent by mass of each element in a compound
e.g.
formula H2CO; molecular mass = 2  mol H (1.008g/mol) + 1 mol C x
12.011 g C/mol C + 1 mole O  15.999 g O / 1mole O
= 30.026 g H2CO/1 mol H2CO
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Percent composition % C = 12.011 g / 30.026 g of H2CO of C
This represents the mass of C in 1 mole H2CO = 12.011 g / 30.026 g = 0.40002 x100% = 40.002 %.
in the same fashion, the % H = 6.714 % and the % O = 53.284 %
%sum = 6.714 % + 40.002 % + 53.284 % = 100.000 %
 Another example, "iso-octane" (2,2,4 - trimethylpentane); molar mass = 114.232 g / mole.
 % C = 8 mol C/1mol iso-octane  12.011 g C/ mole C x
1mol iso-octane/ 114.232 g  100% = 84.117 % C.
 Similarly, % H = 18 mole H / 1 mole iso-octane  1.008 g H / 1 mole H  1 mole iso-octane/
114.232  100% = 15.883 % (adds to 100%)
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 Question if we have the percent composition of the compound, can we calculate its empirical
formula?
 cetylpyridinum chloride (CPC) is a surfactant found in many brands of mouthwash!
percentage composition : 74.19% C; 10.43%Cl 11.26% H; 4.12%N
Take exactly 100.00g of CPC
nc = 74.19 g C x 1 mole H/12.011 g C = 6.1693 moles of C
nH = 11.26 g H x 1 mole H/1.008 g H = 11.170 moles of H
nCl = 10.43 g Cl x 1 mole Cl/35.453 g Cl = 0.29419 moles of Cl
nN = 4.12 g N x 1 mole N/14.007 g N = 0.2941 moles of N
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Moles of atoms in 100g of CPC
6.1693 moles C; 11.170 moles H; 0.29419 mole Cl; 0.2941 mole N
 We need whole numbers
 We divide by the smallest number of moles of substance present. In this example, the
substances present in the fewest number of moles are Cl and N.
moles of C/moles of Cl = 6.1693 / 0.29419 = 21.00;
moles of H/moles of Cl = 11.170 / 0.29419 = 37.98;
moles of N/moles of Cl = 0.29419 / 0.29419 = 1.00;
moles of Cl/moles of Cl = 0.2941 / 0.29419 = 1.000;
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empirical formula C21 H38 N1 Cl1
C21 H38 N1 Cl1 is the simplest whole number ratio for CPC; can we get the molecular formula for
this compound?
 Yes, provided we have information about the molecular mass from other sources (e.g., the
colligative properties).
Chemical Analysis
 For any sample containing carbon, hydrogen, and oxygen that is burned in excess O2
(C,H,O)  CO2 + H2O
l mole of C  1 mole CO2
2 moles of H  1 mole H2O
Example
7
 Wile E. Coyote has to take his "super-duper" ACME vitamin pills to catch the roadrunner. After
numerous attempts (and several falls off a cliff) he decides to analyze his pills. Wile E. found that
when 50.00 g of the pills were burned, he obtained 30.00 g water and 73.28 g CO2. He knew the
compound contained C, H and O. What was Wile E.’s "super-duper" vitamins?
Note that the compound contains C, H, and O. However, by combustion analysis, we can only
get the C and H content of the compound. How do we obtain the mass of the other element in
the compound?
 First, find out how much C, H was obtained.
g of C in compound = 73.28 g CO2 x 12.01 g C / 44.009 g CO2 = 20.00 g C
g of H in compound = 30.00 g H2O x 2.02 g H / 18.015 g H2O = 3.35g H
 We then assume that the rest of the compound is O!
50.00 g = mass of compound = 3.35 g H + 20.00 g C + g O
 g O = 50.00 g - 20.00 g - 3.35 g = 26.65 g O.
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 Now we get the molar ratio for the calculation of the empirical formula
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number of moles of C = 20.00 g C x 1 mole C/12.01g
 moles of C = 1.6653;
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number of moles of H = 3.35 g H x 1 mole H/1.008 g H
 moles of H = 3.323;
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number of moles of O = 26.65 g O x 1 mole O/16.00 g O
 moles of O = 1.6656.
 1.6653 mol C; 3.323 mol H; 1.6656 mol O
 Again, we need the simplest whole number ratio; therefore, we divide by the smallest moles
of substance (here, moles of C).
 1.6653 mol C/ 1.6653 mol = 1.000 C;
 3.323 mol H/ 1.6653 mol = 2.00 H;
 1.6656 mol O/ 1.6653 mol = 1.001 O.
Empirical Formula = CH2O
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From boiling point elevation measurements, we know that the molecular mass of the super vitamin
is 180.2 g/mol. Give the empirical formula above, calculate the molecular formula.
Solution
Note from above
(Empirical formula)N = molecular formula.
How do we calculate N?
N x (empirical formula mass) = molecular mass
 Empirical formula mass = 1 mol C x 12.01 g C/1molC + 2 mol H x 1.00 g H/1 mol H+ 1
mol O x 16.00 g O/ mol O = 30.03 g/mole
 N = molecular mass/empirical formula mass
= (180.2 g/mol) / (30.03 g/mol) = 6
 The molecular formula is N x (C1H2O1) or C(Nx1) H (Nx2) O(Nx1) =
 C6H12 O6 (i.e., glucose or fructose!!!)
 worked through a P and O containing compound 0.67moles of P; 1.67 moles of O  PO 2.5 => x
2 to get P2O5
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Simplest whole number ratio
P2O5 (but we have molecular mass information)
 N = 2; molecular formula = 2 x (P2O5)
= P(2x2) O(5x2)
= P4O1O
Another Combustion Example
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1.367g of a sample containing was burned and 3.002 g CO2 and 1.640 g H2O If the original
compound contained only C, H, and O, what is the empirical formula of the compound.
C, H, O  CO2 + H2O
1.367g 3.002g 1.640g
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mass of C = 3.002g CO2 x12.01 C / 44.01 g CO2 = 0.8192 g C
mass of H = 1.640g H2O x 2.016 g H /18.02 g H2O = 0.1835gH
assume the balance is O
g O = 1.367g compound - 0.8192 g C - 0.1835 g H = 0.364 g O
 need to find molar ratios (divide by respective molar masses)
0.8192 g C 1 mole C / 12.01g C = 0.0683 mole C;
0.364 g O x 1mole O/ 16.00 g O = 0.0228 mole O;
0.1835 g H x 1 mole H / 1.008 g H = 0.1820 mole H.
Divide by smallest to get molar ratio
C: 0.0683 mole /0.0228 mole = 3 C
O: 0.0228 mole/0.0228 mole = 1 O
H: 0.1820mole/0.0228 mole = 8 H
 C3O1H8 empirical formula
Laws of Chemical Combination
 Law of Definite Proportions  all samples of a pure substance contain the same elements in the
same proportion. (Proust. 1799 -> 8 yrs. before Dalton)
 Law of Multiple Proportions  if two elements can form more than one compound, the masses of
one element that combine with a fixed mass of the other element are in the ratio of small whole
numbers (Dalton's work again)
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e.g., PCl3, PCl5
PCl3  mass of Cl that combines with 1 mol (31g) of P is 106.35 (i.e., 3x35.45 g Cl/mol)
PCl5  mass of Cl that combines with 1 mol (31g) of P is 177.25 (i.e., 5x35.45 g Cl/mol)
 mass of Cl in PCl5 / mass of Cl in PCl3 = 177.25 g / 106.35 g = 5/3
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QUANTITATIVE INFORMATION FROM CHEMICAL REACTIONS
A+B C+D
 Normally, we start out with certain quantities of reactants and we wish to know what quantity of
product we can expect. We may also want to know how much reactant we must start with to obtain
a certain amount of product.
 we need to know
 Mass of reactants  mass of products
 Volume of reactants  Volume of products (gases).
Example Problem
Ammonia is used to make fertilizer. Produced from N2 and H2
N2(g) + 3H2(g)  2 NH3(g) balanced equation (Haber Process)
This equation says that 1 mole N2(g) reacts with 3 moles of H2 (g)
(molecular level) 1 molecule N2(g) + 3 molecules H2(g)  2 molecules of NH3 (g)
Calculate the amount of N2 required to form 75.0 g of NH3 (g)?
Solution
 75.0g NH3 = 4.40 moles NH3

# of moles of N2 = 2.20 moles N2 x 28.02 1 mole
= (2.20 x 28.02)g = 61.6g of N2 needed
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How much NH3 can be produced from 37.5 g of N2?
Solution
Number of moles of N2 = 37.5 g / (28.02 g/mole)
= 1.34 mole of N2
# moles of NH3 produced = 1.34 moles of N2 x 2 mole NH3/1 mole N2
= 1.34 x 2 moles = 2.68 moles of NH3
g of NH3 = 2.68 moles of NH3 x 17.04g NH3/1mol NH3
= 45.7g NH3
in one step
NH3 = 37.5g N2 x 1 mole N2/28.02g N2 x 2 moles NH3/1mole N2 x 17.04 g
NH3/1mole NH3
= 45.6g of NH3
(b)
How much N2 is required to form 100.0 kg NH3 (industrial level preparation)?
100.kg NH3 x 1 mole NH3/17.04gNH3 x 1000g/1kg = 5868 moles of NH3
 we need how many moles of N2?
# moles of N2 = 5868 moles NH3 x 1 mol N2/NH3 = 2934 mol N2
# g of N2 = 2934 mol N2 x 28.02g/1mol = 8.222 x 104 g of N2
convert to kg => # kg of N2 = 8.272 x 104 g N2 x 1 kg N2/ 1000gN2
= 82.22kg of N2
(1 step process) # kg N2 = 100.0 kg NH3 x 1000 NH3/1kg NH3 x 1 mol NH3/28.02g
= 88.22 kg of N2
=1 moleN2/2molNH3 x 28.02gN2/1molN2 x1.00kgN2/1000N2
Limiting Reagent
Chemical Equation
NOTE 
gives the molecular or molar ratio of combination, i.e., the ratio of reactant
consumption or product produced
Reactants will react in the proper ratio until one of them is consumed completely! The
reaction then stops.
 The reactants left over are in excess and the amount of product formed is limited by the amount of
reactant that is completely consumed, the limiting reagent.
 The amount of product produced assuming complete consumption of the limiting reagent is the
theoretical yield!
Example
13
ethanol can be produced by the hydration of ethylene
i.e. C2H4 + H2O  C2H5OH
how much ethanol can be prepared from 6.74g of C2H4 and 14.02g of H2O?
# of moles of C2H4 = 6.74g C2H4 x 1 mol/28.05g = 0.2402 moles
# of moles of H2O = 14.02g of H2O/18.02 g H2O 1 mole H2O = 0.7780 moles of H2O
 since 1 mole C2H4 reacts with 1 mole of H2O
all the C2H4 will react until consumed
i.e., 0.2402 moles of C2H4 => 0.2402 moles of water (but we have 0.778 moles of H2O)
water is INXS (all the C2H4 with disappear)
Calculate the theoretical yield of C2H5OH
limiting reagent C2H4 = 0.0240 moles
Theoretical yield of ethanol = # of moles of C2H5OH produced
= 0.240 moles of C2H4 x 1 mole C2H5OH / 1mole C2H4
= 0.240 moles of C2H5OH = 11.1 g ethanol
(b) What is the % yield if we obtain only 0.160 moles of C2H5OH?
% yield = actual yield/theoretical yield x 100%
= 0.160moles/0.240moles x 100% = 66.7%
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Example 2
2 Al (s) + 3 I2 (s)  2 AlI3
1.20g
2.40g
A.
Determine limiting reagent and the theoretical yield of AlI3!
# moles of Al = 1.20g/27.0g . 1 mole = 4.44 x 10-2 moles
# moles of I2 = 2.40g x 1mole/253.8g (9.46 x 10-3 moles/3)
Assume Al is the limiting reagent = 3.15 x 10-3 moles and calculate the number of moles of I2 reacting
with this many moles of Al.
# of moles of I2 reacting = 4.44 x 10-2 moles of Al x 3 moles of I2/2 moles of Al
= 6.66 x 10-2 moles of I2 (if all the Al is consumed completely)
but we have only 9.46 x10-3 moles of I2
 I2 is the limiting reagent (Al is INXS)
theoretical yield of (I3 = # of moles of limiting reagent x 2.41 g / 372 g/mole)
= 9.46 x 10-3 moles if I2 x 2 moles Al I3/3 moles I2
= 6.31 x 10-3 moles of Al I3
# of g of Al I3 = 6.31 x 10-3 moles of Al I3 x 406.7g/1mole AlI3
# of g of Al I3 = 2.57g
15
The Easier Way to do this calculation
# of moles of I2 = 9.46 x 10-3 ; # of moles of Al = 4.44 x 10-2 moles
now divide through by the stoichiometric coefficients from the balanced equation
# of moles of I2 = 9.46 x 10-3 moles / 3 = 3.15 x 10-3 moles
# of moles of Al = 4.44 x 10-2 moles / 2 = 2.22 x 10-2 moles
These are the stoichiometric coefficients from the
balanced chemical equation.
B.
How much Al is left behind ??
# moles of Al reacting with I2 = 9.46 x 10-3 moles of I2 x 2 moles Al/3moles AlI3
= 6.31 x 10-3 moles of Al reacted
1 mole Al  1 mole AlI3
 # of moles of Al remaining = 4.44 x10-2 moles (initial) - 6.31 x 10-3 (consumed)
= # of moles of Al left over
= 3.81 x 10-2 Al moles x 27.0g/1mole Al = 1.03 g of Al remaining
C.
What if we recovered only 62.4% of the expected product (AlI3).
Calculate the # of moles of AlI3 recovered
% yield = 62.4% = actual # of moles / theoretical # of moles x 100%
= 0.624 x theoretical # of moles = actual # of moles
# moles AlI3 (actual) = 3.94x10-3 moles of AlI3
= 1.61 g of AlI3 (after converting to grams using molar mass)