AP Chemistry
Mr. Ferwerda - Tecumseh High School
AP Chemistry - Notes - Chapter 3 : Stoichiometry
5/20/08
A. Stoichiometry - the study of quantities of substances consumed and produced in chemical reactions.
1. Atomic mass - the average mass of an atom of an element in atomic mass units (amu)
a. Atomic masses are based on 12C ("carbon twelve"), which is assigned a value of exactly 12 atomic
mass units.
b. The atomic masses of other elements are determined by comparison to 12C.
c. Mass spectrometer - currently best method for determination of atomic masses of atoms
- Procedure :
- atoms or molecules are placed in abeam of high-speed electrons (knocks electrons off the
atoms or molecules giving them a positive charge - cations)
- an electric field is then applied accelerating these cations which causes each cation to create
a magnetic field
- cations pass through an applied magnetic field
- cations with the least mass are deflected more than heavier cations and masses are
determined by comparing amounts of deflection
e.g. If , in a mass spectrometer, 13C is found to have a mass 1.0836 times that of 12C, then the
atomic mass of 13C = 1.0836 (12 amu) = 13.003 amu
A(n) _______________ ________________ ____________ (_____) is exactly one-twlfth of a carbon-12
atom.
_____ (T/F) The relative masses of other atoms are determined experimentally.
_____ (T/F) The assigned mass of an atom is determined by the ratio of the mass of that atom to the mass of
the C-12 atom.
If oxygen has a mass that is 1.33325 times greater than C-12, what is its mass in amu?
A fictitious element has only two isotopes and an atomic mass of 28.23 amu. One isotope has a mass of 29.06
amu and the other has a mass of 28.02 amu. What are the frequencies of the two isotopes?
The _________________ _______________________ is the tool that most accurately compares the masses of
atoms.
d. The mass spectrometer can also be used to determine the isotopic composition of an element. When
a sample of an element is placed in a mass spectrometer, a mass spectrum can be obtained which
indicates the relative amounts of the various isotopes present in the sample.
e. The average mass of an atom (or mole) of an element is the sum of the fractions of each isotope
times their mass.
These are __________________________.
What do these have to do with carbon having an atomic amss of 12.01 amu.
1
AP Chemistry
Mr. Ferwerda - Tecumseh High School
Naturally occurring carbon is composed of three isotopes, 12C (98.89%),
Determine the average mass of the element carbon.
5/20/08
13
C (1.11%) and C-14 (negligible %)
2. The Mole (abbrev. mol) - SI unit for the amount of a substance and is equal to the number of atoms in
exactly 12 grams of pure 12C.
- Mass spectrometry has determined this number to be 6.02214 x 1023 (Avogadro's number) (We
will use 4 sig.fig : 6.022 x 1023).
a. Calculations involving the mole
- calculate mass from atoms and vice versa
How many atoms are there in 4.44 grams of sodium?
- calculate moles from mass and vice versa
How many moles are there in 13.56 grams of carbon?
- calculate volume of gases at STP
What is the volume of 3.46 x 1025 molecules of a gas at STP?
3. Molar mass - the mass in grams of one mole of a substance( equals the atomic mass in grams)
- be able to calculate the molar mass of atoms, ions or molecules or numbers of atoms, ions or
molecules from moles or vice versa
4. Mass percent composition - the percent composition of a substance by mass
Sample Problem : What is the percent composition of glucose (C6H12O6)
mass
of
carbon
72.0
mass
percent
of
carbon

100%
x

100
x

39
mass
of
compo
180.
mass
of
hydro
12.1
mass
percent
of
hydroge

100%
x

100
x

6.7
%
mass
of
compo
180.
mass
of
oxyge
96.0
mass
percent
of
oxygen

100%
x

100
x

53
%
mass
of
compo
180.
5. Determining the Formula of a Compound
a. The _________________________ formula of a compound is the lowest whole number ratio of the
atoms in a compound.
- Determination from percent composition data :
Step 1 : Determine mass composition (if not given) from percent composition (assume 100g
and then percents convert to grams)
Step 2 : Determine moles of each element in the compound.
Step 3 : Determine the lowest whole number ratio of each element in the compound (empirical
formula)
2
AP Chemistry
Mr. Ferwerda - Tecumseh High School
5/20/08
What is the empirical formula of ascorbic acid (vitamin C - 40.92% C, 4.58% H and 54.50% O)?
b. The __________________________ is the actual ratio of the elements in a compound - a whole number
multiple of the empirical formula (empirical formula)n = molecular formula.
n=
MW
EFW
What is the molecular formula of ascorbic acid if it has an empirical formula of C3H4O3 and a molecular mass
of 176.14?
- Determination of empirical formula from combustion data :
- Use stoichiometry to determine the grams of C and H from the balanced chemical equation
(if the combusted substance contains oxygen, determine the mass of oxygen from the mass
of the original compound)
1. A compound contains only carbon and hydrogen. When 12.00 g of the compound are burned, 36.33 g of
carbon dioxide and 18.596 g of water are produced. What is the empirical formula of this compound?
A compound contains C, H and O. When twenty grams of the compound is burned, 16.360 g of water and
39.95 g of carbon dioxide are produced. What is the empirical formula of this compound?
3
AP Chemistry
Day 2 :
Mr. Ferwerda - Tecumseh High School
5/20/08
6. Chemical Equations - represent what happens in a chemical reaction
a. reactant  products (read "Reactants yield products")
b. conservation of atoms (mass) - atoms are neither created nor destroyed in chemical reactions, they
are recombined to form different substances
- mass is neither created nor destroyed chemical reactions (as opposed to nuclear reactions)
- chemical reactions must therefore be balanced - have same kinds and numbers of atoms on both
sides of the yields sign ()
c. The physical states of the substances involved in a chemical reaction are represented by symbols :
(aq) - aqueous (dissolved in water)
(g) - gas
(l) - liquid
(s) - solid
7. Balancing chemical equations - usually by inspection
- it helps to do the most complex substance first
- never change the correct formulas of reactants or products (change the amounts of the substances
represented, do not change the substances)
- balance by placing coefficients in front of the reactant or product species
Diatomic elements?
8. Stoichiometric Calculations : Amounts of Reactants and Products
a. The following diagram represents the basic conversions used in stoichiometry :
- Calculations between moles and mass and moles and numbers of particles have already been
covered (moles and volume conversion will be covered later). The only new thing here is the use
of the mole ratio in a balanced equation to convert moles of one substance (known) to moles of
another (unknown)
Ex : Solid lithium hydroxide is used to absorb and remove carbon dioxide from the atmosphere inside space
vehicles (water and lithium carbonate are formed). How much gaseous carbon dioxide can 1.00 kg of LiOH
remove?
.
9. Calculations Involving a Limiting Reactant.
a. Stoichiometric quantities - the exact amounts of reactants needed so that no amount of any reactant
will be left unused
4
AP Chemistry
Mr. Ferwerda - Tecumseh High School
5/20/08
b. Limiting Reactant - a reactant that there is proportionally less of so that it is used up first and limits
the amount of product that can be produced
- you must first determine which reactant is the limiting reactant when doing stoichiometry
problems involving limiting reactants
- to determine which reactant is limiting arbitrarily choose one reactant and use the mole ratio of
reactants to see if you have an excess or not of the other reactant.
From the balanced equation for the Haber process (N2 + 3H2  2 NH3) we see that one mole of nitrogen is
needed for every three moles of hydrogen present. What is the limiting reactant between 12.2 L of nitrogen gas
and 36.2 L of hydrogen gas?
For the combustion of methane, what is the liming reactant if 1.0 moles of oxygen gs reacts with 2.0 moles of
methane gas?
What is the limiting reactant for a reaction between 100.0 g of hydrogen and 200.0 grams of nitrogen and you
are going to combine them to form ammonia according to the Haber Process :
N2 + 3H2  2NH3
Multiple equations :
1. 2A + 3B --> 2C
2. C + D --> 2E
3. 3E + F --> 2G
How many moles of G can be produced from 3 moles of A?
How many moles of B are needed to produce 4 moles of G?
Aluminum reacts with oxygen to form an aluminum oxide coating. If 22.00 grams of aluminum oxide are
formed from 20.00 grams of aluminum, what would the percent yield be? (assume an excess of oxygen)
5
AP Chemistry
Review :
Mr. Ferwerda - Tecumseh High School
5/20/08
A compound is 88.9% O and the rest of it is hydrogen. What is the empirical formula of the compound?
If 2.50 g of the above substance reacts with calcium metal to form calcium hydroxide and hydrogen gas, what
fraction of moles of the hydrogen ends up in each of the products?
2.50 g of sodium oxide reacts with another substance to form 4.28 g of a compound which is composed of Na,
O and C. If the product compound is 45.3% O, what is the empirical formula of the compound?
Alternate method for determining limiting reactant :
Pb + PbO2 + 2H2SO4 ---> 2PbSO4 + 2H2O
What is the limiting reactant if 12.0 g of Pb reacts with 24.0 g of lead(IV) oxide and 30.0 g of sulfuric acid?
c. Excess reactant -opposite of limiting reactant - some will remain at the completion of a reaction
d. Theoretical yield - the amount of product a reaction should yield if 100% of the limiting reactant is
converted into the product
e. Actual yield - the amount of product that is actually produced when the reaction takes place
f. Percent yield - a measure of the efficiency of a chemical reaction or process
( )
actual
yield
Percent
yield

100%
x
theoretica
l
yield
When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 1.50 g
copper was heated with excess sulfur to yield 1.76 g copper(I) sulfide. What is the theoretical yield? What is
the percent yield?
6
AP Chemistry
Mr. Ferwerda - Tecumseh High School
5/20/08
Summary problems : Use these as models to help you work through the Free Response packet. KNOW
THESE WELL.
S1. Phosphorus combines with chlorine gas to form phosphorus trichloride and phosphorus pentachloride. A
sample of phosphorus is mixed with an inert substance and reacted with 82.0 mL of chlorine gas at 622K
and a pressure of 722 mm Hg. If the mass of the mixture is 0.450 g and the ratio of PCl3:PCl5 is 2:1, what is
the mass percent of phosphorus in the mixture?
Answer :
P(s) + 5.5 Cl2 --> 2 PCl3 + PCl5
PV (722/760)0.082L
=
=0.001526 mol
RT (.08206 x 622K)
3P
30.97g
mass P = 0.001526 mol Cl 2(
)(
)= 0.0258g
5.5 Cl 2 mol P
0.0258 g
Percent mass =
x 100% = 5.73%
0.450 g
mol Cl 2 =
S2. Element A combines with chlorine to form ACl2 and ACl4. If 4.63 g of A combines with 2,748 mL of
chlorine gas at 22.0 C and 755 mm Hg to form an equal number of moles ACl2 and ACl4, what is the molar
mass of A. What is the identity of A?
Answer : A + 3Cl2 --> ACl2 + ACl4
mol Cl 2 =
PV (755/760) x 2.784L
=
= 0.114 mol Cl 2
RT (0.08206 x 295K)
æ 1 mol A ö
mol A = 0.114 mol Cl 2 ç
÷ = 0.0380 mol
è 3 mol Cl 2 ø
g
4.63g
M=
=
= 122 g/mol = Sb (antimony)
mol .0380 mol
S3. The following three step process is called the Ostwald process and is used to produce nitric acid.
1. 4NH3(aq) + 5O2(g) --> 4NO(g) + 6 H2O(g)
2. 2NO(g) + O2(g)
--> 2NO2(g)
3. 3NO2(g) + H2O(l) --> 2HNO3(aq) + NO(g)
(a) How many moles of NH3 are needed to produce 8.2 x 103 L of 10.0 M HNO3?
æ 10.0 mol ö
mol HNO 3=8200 L ç
÷=82000 mol HNO 3
L
è
ø
æ 3 mol NO 2 öæ 1 mol O 2 ö æ 4 mol NH 3 ö
82000 mol HNO 3 ç
֍
֍
÷ = 4920 mol NH 3
è 2 mol HNO 3 ø è 2 mol NO 2 ø è 5 mol O 2 ø
(b) Would you need a greater or lesser amount of NH3 to get the 8.2 x 103 L of 10.0 M HNO3 if some of the
NO2(g) is lost in step 2?
- greater
7
AP Chemistry
Mr. Ferwerda - Tecumseh High School
5/20/08
(c) If step one is 87% efficient, step 2 is 93% efficient and step 3 is 98% efficient, what is the overall
efficiency and how much NH3 would be actually needed for (a) above.
0.87 x .97 = .84 or 84% efficient overall.
S4. When 26.72 g of solid compound composed of Ca, Si and oxygen combines with a gas which is 5.05% H
and the rest fluorine, it forms calcium fluoride, water and silicon tetrafluoride gas.
(a) Determine the empirical formula of the gas.
Assuming 100. grams of the gas :
æ 1mol ö
5.05 g H ç
÷ = 5.00 mol H
è 1.01g ø
æ 1mol ö
95.95 g F ç
÷ = 5.00 mol F
è 19.00g ø
Since the mole ratio of H:F = 1:1, the empirical formula is HF.
(b) If 17.96 grams of calcium fluoride are formed and 5.66 L of silicon tetrafluoride gas at 25 ºC and 755 mm
Hg, what is the empirical formula of the solid reactant?
æ 1mol ö æ 40.08 g Ca ö
Ca : 17.96 g CaF2 ç
÷ = 9.22 g Ca = 0.230 mol Ca
֍
è 78.08g ø è Mol CaF2 ø
Si : n =
PV æ ( 755 / 760 ) x5.66 L ö
=ç
÷ = 0.230 mol SiF4 = 0.230 mol Si = 6.46 g Si
RT è (.08206 x 298 K ø
O : 26.72 g - (9.22 g Ca + 6.46 g Si) = 11.04 g O = 0.690 mol O
Empirical formula :
Ca 0.230Si 0.230O 0.690= CaSiO 3
0.230
0.230
0.230
(c) If the solid reactant compound has only one atom of silicon per mole, and is 24.18% Si, what is its molar
mass and molecular formula?
æ 28.09 g Si ö
ç
÷ x 100% = 24.18 ; x = 116.17 g/mol
x
è
ø
The empirical formula mass is equal to the molecular mass, so the molecular formula is CaSiO3.
8
AP Chemistry
Mr. Ferwerda - Tecumseh High School
(d) What percent of the fluorine from the gas ends up in the calcium fluoride?
5/20/08
æ 1mol ö æ 38.00 g F ö
Mass of F in CaF2 : 17.96 g CaF2 ç
÷ = 8.74 g F
֍
è 78.08g ø è Mol CaF2 ø
æ 4 mol F ö æ 19.00 g F ö
Mass of F in SiF4 : 0.230 mol SiF 4 ç
֍
÷= 17.48 g F
è 1 mol SiF4 ø è mol F ø
æ
ö
8.74 g
Percent F in CaF2 : ç
÷ x 100% = 33.3%
è (8.74 g + 17.48 g) ø
(e) Write a balanced chemical equation for this reaction.
CaSiO3 + 6 HF --> CaF2 + 3 H2O + SiF4
S5 A 1.255 g sample of a mixture of calcium hydroxide and sodium hydroxide are thermally decomposed.
1,220 mL of water vapor were produced at 762 mm Hg and 240 C.
(a) What mass of water vapor was produced?
n=
PV
(762/760) x 1.22L
æ 18.02g ö
=
= 0.0291 mol ç
÷ = 0.524 g H 2O
RT
(0.08206 x 513 K)
è mol ø
(b) Write the balanced equations for the decomposition of both hydroxides.
Ca(OH)2 ---> CaO + H2O
2 NaOH ---> Na2O + H2O
(c) By a separate analysis it was determined that the original 1.255 g sample contained 366 mg of Ca. What
percent of the mixture was Ca(OH)2?
ö
æ 1 mol Ca ö æ 1 mol Ca(OH) 2 ö æ
74.10g
366 mg = 0.366 g Ca ç
ç
÷ = 0.677 g Ca(OH) 2
÷
ç
÷
è 40.08 g Ca ø è 1 mol Ca ø è 1 mol Ca(OH) 2 ø
æ 0.677g ö
%Ca= ç
÷ x 100% = 53.9 %
è 1.255g ø
(d) How many grams of the sodium containing compound were present in the dry product mixture?
Grams NaOH = 1.255 - 0.677 g = 0.578 g NaOH
æ 1 mol NaOH ö æ 1 mol Na 2O ö æ 61.98 g ö
0.578 g NaOH ç
÷ = 0.448 g Na 2O
֍
֍
è 40.00 g ø è 2 mol NaOH ø è 1 mol Na 2O ø
9
AP Chemistry
Mr. Ferwerda - Tecumseh High School
S6 Three volatile compounds contain a certain element E as listed below
Compound
Percent by mass of
Molecular weight
element E
A
70.21%
?
B
83.48%
84.93
C
89.09%
119.37
5/20/08
(a) When 12.2 grams of compound A is vaporized at 766 mm Hg and 100.0 ºC it occupies a volume of 7,338
mL. What is the molar mass of A?
n=
PV (766 / 760) x 7.338 L
=
= 0.242 mol
RT
(0.08206 x 373 K)
molar mass = g/mol = 12.2 g / 0.242 mol = 50.4 g/mol
(b) Determine the mass of E in each of the three compounds.
Compound
Mass of element E
A
(0.7021 x 50.4) = 35.4
B
(0.8348 x 84.93) = 70.90
C
(0.8909 x 119.37) = 106.3
(c) Calculate the most probable atomic weight of E and determine its identity from the periodic table..
35.5 g - lowest common denominator - the element is chlorine
(d) Compound C contains only C, H and E. When 5.00 gram of compound C is oxidized all of the C and H are
converted to 1.84 g of carbon dioxide and 0.755 g of water. What are the empirical and molecular formulas
of C?
æ 1 mol CO2 ö æ 1 mol C ö
æ 12.01 g ö = 0.502 g C
C : 1.84 g CO2 ç
֍
÷ = 0.0418 mol C ç
÷
è 1 mol C ø
è 44.01 g CO2 ø è 1 mol CO 2 ø
æ 1 mol H 2O ö æ 2 mol H ö
æ 1.01 g ö = 0.0846 g H
H : 0.755 g H 2O ç
֍
÷ = 0.0838 mol H ç
÷
è 1 mol H ø
è 18.02 g H 2O ø è 1 mol H 2O ø
æ 1 mol Cl ö
E = Cl = 5.00g - (0.502 + 0.0846) = 4.413 g Cl ç
÷ 0.1245 mol Cl
è 35.45 g Cl ø
C 0.0418H 0.0838Cl 0.1245 =CH 2Cl 3
0.0418
0.0418
0.0418
Homework Hints :
10
AP Chemistry
Mr. Ferwerda - Tecumseh High School
5/20/08
WS 3-1
Alkali metal oxides and alkaline earth metal oxides are called basic anhydrides - form basic solutions when
combined with water (e.g. Na2O + H2O --> 2 Na(aq) + 2OH(aq)
Prob.16 Let x = percent of 151Eu then (1-x) = percent of 153Eu
Prob. 17 Br2 consists of two atoms of Br. What combination of isotopes add up to the given masses? Think of
punnett squares.
No calculator problems - get used to rounding to common fractions. For example - #30-0.48 mol is a little less
than one-half, so a mole will be 2 x 86 = 192 - some. Answer is C - glucose has a molar mass of 180.2 g.
#31. Change 0.24 to 24 = 0.5 mole of ozone. Readjust for 0.24 (0.5 --> .0050)
WS-3-2
#3 Transition metals form colored solutions.
#8 3 moles of O2 per mole glucose
#9 L x M = moles
#12 - Need to determine balanced chemical equation first to get mole ratios.
#16 Determine no. of moles of CO2 per mole CaCO3 when it decomposes.
#19 - Write the equation to calculate mass% of C for the compound, then substitute x for mass of compound in
this equation.
WS-3-3
#2 Nonmetallic oxides (CO2, NO2, P2O5 etc) are acid anhydrides - substances that react with water to form
acids (e.g. CO2 + H2O --> H2CO3 - carbonic acid).
#13 - Determine the formula of eh precipitate first - will determine which substances are taken out of solution.
#24 Molar mass has the units g/mol - so you need to get g and mol of M.
WS-3-4
#21 Let x = 85Rb, therefore 1-x = 87Rb. Solve for 85Rb, then 87Rb and then solve problem.
#22 Get g C and g H from combustion information. Use second experiment to calculate %N in unknown cpd
and use that to get g N in first sample. Grams of O can then be calculated by subtraction and empirical
formula can be calculated.
#29 Let x = mass of Cu2O and then 1.500-x =mass of CuO. Determine %Cu of each cpd. Percent Cu of each
times mass of each must add up to 1.252 g.
Hints for Stoichiometry free response problems :
1970
- Write balanced chemical equation first with a 2: 1 ratio of CO2 to CO (e.g --> 6 CO2 + 3 CO)

atm
L

.08206

) ; K = °+273; atm = mmHg/760
- Use ideal gas law to get mol O2 (R = 0

mol
K

- Use balanced equation to convert mole oxygen to grams carbon
- Calculate mass percent of carbon in sample
11
AP Chemistry
1993 B
Mr. Ferwerda - Tecumseh High School
5/20/08
a. Calculate moles oxygen using three balanced equations
ex : If I wanted to calculate mol Mn2+ from mol 1.00 mol I- :
b. L x M = mol to calc mol S2O32- Use mole ratio to get moles of oxygen
c.
d. Molarity is mol/L, so calculate molarity of oxygen (see b. above)
e.
1995 B
a. Use ideal gas law to solve for moles.
b. carbonates decompose into oxides and carbon dioxide
c. g Ca  mol Ca  mol CaCO3  g CaCO3, calc. mass percent CaCO3 of sample
d. Calc % of MgCO3 (If 60% CaCO3 …..)
- calc g MgCO3 from percent of original sample
- g MgCO3  mol MgCO3  mol MgO  g MgO
2000B
a.
b. (i) g BeC2O4·3H2O  mol BeC2O4·3H2O  mol BeC2O4 g BeC2O4
(ii) use stoichiometry to calc moles of water from 3.21 g BeC2O4·3H2O
- use ideal gas law to calculate volume
c. (i)reducing agent is species that gives electrons (is oxidized) – you must calculate oxidation states of all
species in chemical equation
(ii) L x mol = mol MnO4 - ; use mole ratio to get mol C2O42(iii) you just calculated number of moles in 20.0 mL of the 100 mL sample
(iv) mol C2O42-  mol BeC2O4  g BeC2O4 ; calc mass %
2001 B
a.
b. Use ideal gas law to get moles of CO2, then convert to g C (Need grams of C and g H to get g of O to
calc mol O).
c. Molar mass = g/mol, calculate mol AS acid from moles NaOH (1:1) ratio.
d. (i) -(ii) –
1991 B
a.
b. Calc. g. O in the moles of CO2 and H2O from above.
c. Step 1 : ΔT = 0.5 °C ; ΔT = Kbm; m = ΔT/Kb
Step 2 : m = mol/kg solvent ,  mol = kg x m
12
AP Chemistry
Mr. Ferwerda - Tecumseh High School
Step 3 : MW = g/mol
d.
5/20/08
1986 B

a. ideal gas law can be derived to get : MW
gRT
, where “g” is mass in grams
PV
b.
c. Look for smallest whole number ratio found in masses calc. in step b above, and than calculate the mass
from the above numbers. e.g. if masses are 12, 18 and 30 and the smallest whole number ratio was 2:3:5,
the at. wt. would be 6.0 amu.
d. Calculate moles C and H from g CO2 and g H2O. Grams of Q can be calc. from percent data. Calculate
empirical formula, and then molecular formula using data given in Table.
13