1
Stoichiometry
Isotopes
Same atomic number (number of protons or electrons)
Different atomic weights (atomic masses)
The Mol and Avogadro’s Number
1 mol = 6.02 x 1023 (Avogadro’s number)
1 mol = 1 molecular mass (or 1 mw)
1 atomic mass of an element = 1 mol of that element is the same as the atomic weight of
that element (except for diatomic molecules)
Ex.
1 mol Na = 23 g = 1 at. mass of Na = 6.02 x 1023 Na atoms
1 mol Ag = 108 g = 1 at. mass of Ag = 6.02 x 1023 Ag atoms
1 mol of hydrogen (H2) = 2 g (1 x 2) = 6.02 x 1023 hydrogen molecules
1 mol of oxygen (O2) = 32 g (16 x 2) = 6.02 x 1023 oxygen molecules
1 mol of chlorine (Cl2) = 71 g (35.5 x 2) = 6.02 x 1023 chlorine molecules
1 mol of any compound = 6.02 x 1023 molecules
Ex.
1 mol CO2 = 44 g CO2 = 6.02 x 1023 CO2 molecules
1 mol SO2 = 64 g SO2 = 6.02 x 1023 SO2 molecules
1 mol C2H6 = 30 g C2H6 = 6.02 x 1023 C2H6 molecules
2 mol C2H6 = 60 g C2H6 = 2 x 6.02 x 1023 C2H6 molecules
Calculations using Avogadro’s Number
Calculate the number of CO2 molecules in 2 g of CO2.
44 g CO2 6.02 x 1023 CO2 molecules
2 g CO2 x
2 g CO2 x 6.02 x 1023 molecules
44 g CO2
= 2.7 x 1022 CO2 molecules
2
Calculate the mass of 3.01 x 1024 molecules of CH4.
16 g CH4 6.02 x 1023 CH4 molecules
x 3.01x 1024 CH4 molecules
16 g CH4 x 3.01x 1024 CH4 molecules
6.02 x1023 CH4 molecules
= 80 CH4 molecules
Calculate the mass of 1.2 x 1020 Na atoms.
23 g Na 6.02 x 1023 Na atoms
x 1.2 x 1020 Na atoms
23 g Na x1.2 x1020 Na atoms
6.02 x1023 Na atoms
= 4.6 x 10-3 g Na
Percents of Elements in Compounds
C2H6
%C =
2 x12
24
x 100 =
x 100 = 80 %C
30
(2 x12) (6 x1)
%H = 100 – 80 = 20 %H
CH3OH
12
x 100 = 37.5 %C
32
16
%O =
x 100 = 50 %O
32
%C =
%H = 100 – (37.5 +50) = 100 – 87.5 = 12.5 %H
Balancing Equations
Given:
Balanced:
Fe + O2 FeO
2Fe + O2 2FeO
Given:
Balanced:
Or
C5H11OH + O2 CO2 + H2O
C5H11OH + 7.5O2 5CO2 + 6H2O
2C5H11OH + 15O2 10CO2 + 12H2O (x 2 to get rid of the half mol)
Thinking process:
Step 1:
C5H11OH + O2 5CO2 + H2O (now carbon is balanced)
Step 2:
C5H11OH + O2 5CO2 + 6H2O (now hydrogen is balanced)
Step 3:
C5H11OH + 7.5O2 5CO2 + 6H2O (now oxygen is balanced)
Given:
Balanced:
C10H8 + O2 C8H6O3 + CO2 + H2O
C10H8 + 4O2
C8H6O3 + 2CO2 + H2O
Thinking process:
Step 1:
C10H8 + O2
Step 2:
C10H8 + O2
C8H6O3 + 2CO2 + H2O (now carbon is balanced)
C8H6O3 + 2CO2 + H2O (now hydrogen is balanced)
3
Step 3:
C10H8 + 4O2
C8H6O3 + 2CO2 + H2O (now oxygen is balanced)
Given:
Balanced:
CaH2 + H2O Ca(OH)2 + H2
CaH2 + 2H2O Ca(OH)2 + 2H2
Thinking process:
Step 1:
CaH2 + 2H2O
Step 2:
CaH2 + 2H2O
Ca(OH)2 + H2 (now oxygen is balanced)
Ca(OH)2 + 2H2 (now hydrogen is balanced)
Calculations Using Mol Equivalencies in Equations
ALWAYS BALANCE EQUATIONS FIRST!! (These equations are already balanced.)
2Na + 2H2O
2 mol Na
2NaOH + H2
2 mol H2O
2 mol NaOH
1 mol H2
Calculate the mol O2 needed to react with .75 mol CH4.
CH4 + 2O2
CO2 + 2H2O
1 mol CH4 2 mol O2
.75 mol CH4 x
.75 mol CH4 x 2 mol O2
= 1.5 mol O2
1 mol CH4
Calculate the mol O2 formed from decomposing 2.5 mol KClO3.
2KClO3
2KCl + 3O2
2 mol KClO3 3 mol O2
2.5 mol KClO3 x
2.5 mol KClO3 x 3 mol O2
= 3.75 mol O2
2 mol KClO3
Calculate the g O2 formed from decomposing 4 g KClO3.
2KClO3
2KCl + 3O2
2 x 122.6g KClO3 3 x 32 g ox.
4 g KClO3 x
4 g KClO3 x 3 x 32 g ox.
= 1.5 g ox.
2 x 122.6g KClO3
Calculate the g NaOH needed to react with 1.5 g HCl.
4
HCl + NaOH
NaCl + H2O
36.5g HCl 40 g NaOH
1.5 g HCl x
1.5 g HCl x 40 g NaOH
= 1.64 g NaOH
36.5g HCl
Calculate the g CO2 formed from the reaction of .5 mol C3H8 in excess of ox.
C3H8 + 5O2
3CO2 +4H2O
1 mol C3H8 3 x 44 g CO2
.5 mol C3H8 x
.5 mol C3H8 x 3 x 44 g CO2
= 66 g CO2
1 mol C3H8
Or
1 mol C3H8 3 mol CO2
.5 mol C3H8 x
.5 mol C3H8 x 3 mol CO2
= 1.5 mol CO2
1 mol C3H8
mol =
g
mw
g = mol x mw
= 1.5 mol CO2 x
44 g CO2
= 66 g CO2
1 mol CO2
The loss in the weight of a test tube containing KClO3 is .45 g. Calculate the weight of
KClO3 in the test tube.
2KClO3 (s)
2KCl (s) + 3O2 (g)
2 x 122.6g KClO3 3 x 32 g ox.
x .45 g
2 x122.6g KClO3 x .45g ox.
= 1.2 g KClO3
3 x 32 g ox.
A test tube contains KClO3 and NaCl. The weight of the mixture before heating is 3.75 g.
After heating the weight of the mixture in the test tube is 3.62 g. Calculate (a) the weight
of NaCl in the mixture and (b) the % of KClO3 in the mixture.
2KClO3 + NaCl
(a)
2KCl + 3O2
2 x 122.6g KClO3 3 x 32 g ox.
x .13g ox.
.13g ox. x 2 x 122.6g KClO3
= .33 g KClO3
3 x 32 g ox.
(Note: The .13 g ox. is the loss in weight of test tube after heating.)
(b)
NaCl KClO3 3.75g
KClO3 .33g
.33
x 100 = 8.8 %KClO3
3.75
Calculate the mass of 80% pure S needed to prepare 6 g SO2.
5
S + O2
SO2
32 g S 64 g SO2
x 6 g SO2
6 g SO2 x 32 g S
=3gS
64 g SO2
100 g impureS 80 g pureS
x 3 g pureS
100 g impureS x 3 g pureS
= 3.75 g imp. S
80 g pureS
Excess and Limiting Reagents in Calculations
(Hint: Problems will give the masses of both the reagents.)
ALWAYS USE LIMITING REAGENT TO CALCULATE WEIGHT OF PRODUCT!!
HCl + NaOH NaCl + H2O
36.5 g HCl 40 g NaOH 58.5 g NaCl
18 g H2O
If 40 g HCl mixed with 40 g NaOH:
HCl not completely consumed – in excess.
NaOH consumed – limiting reagent.
If 2.2 g HCl is mixed with 2 g NaOH (a) Which is in excess? (b) Calculate the weight of
NaCl formed in the reaction.
(a) Step 1: Assume 2.2 g HCl was completely consumed to compare the mass of NaOH.
36.5g HCl 40 g NaOH
2.2 g HCl x
40 g NaOH x 2.2 g HCl
= 2.4 g NaOH
36.5g HCl
Since only 2 g NaOH was given, this is the wrong assumption.
Step 2: Assume 2 g NaOH was consumed.
36.5g HCl 40 g NaOH
x 2 g NaOH
2 g NaOH x 36.5g HCl
= 1.83 g HCl
40 g NaOH
Therefore, HCl is in excess by .37 g (2.2 – 1.83).
(b)
40 g NaOH 58.5g NaCl
2 g NaOH x
Or
2 g NaOH x 58.5g NaCl
= 2.93 g NaCl
40 g NaOH
6
36.5g HCl 58.5g NaCl
1.83g HCl x
58.5g NaCl x 1.83g HCl
= 2.93 g NaCl
36.5g HCl
Decomposing 5 g CaCO3 will give 2.2 g CaO. Calculate the % yield of the reaction.
CaCO3
CaO + CO2
100 g CaCO3 56 g CaO
5 g CaCO3 x
% yield =
5 g CaCO3 x 56 g CaO
= 2.8 g CaO
100 g CaCO3
actual yield
2.2
=
x 100 = 78.8%
theoretical yield 2.8
Calculate the weight of CaSO4 formed from the reaction of 10 g H2SO4 if % yield is 70%.
Ca(NO3)2 + H2SO4
CaSO4 + 2HNO3
98 g H2SO4 136 g CaSO4
10 g H2SO4 x
% yield =
actual yield
theoretical yield
10 g H2SO4 x 136 g CaSO4
= 13.8 g CaSO4
98 g H2SO4
.7 =
actual yield
13.8
.7 x 13.8 = actual yield = 9.7 g CaSO4
An ore of Cu contains 80 % CuS. Calculate the tons of Cu produced from 100,000 tons of
ore if the yield of the process is 60%.
CuS + O2
Cu + SO2
100 tons ore 80 tons CuS
100,000tons ore x
100,000tonsore x 80 tonsCuS
= 80,000 tons CuS
100 tonsore
96 tons CuS 64 tons Cu
80,000 tons CuS x
80,000tonsCuS x 64 tonsCu
= 53,333 tons Cu
96 tonsCuS
actual yield = % yield x theoretical yield
Simple/Empirical Formula
Shows the lowest ratio of elements in a compound.
x(mass of S.F) = mw
60
x 53,333 = 31,999.8 tons Cu
100
7
CH3 – CH3
2C 6H
E.F
Molecular Formula
x(CH3) = mw
x(CH3) = 30
15x = 30
x=
30
=2
15
% of each element
=x
at. mass of the element
Ex.
%A
= .25
at. mass A
=
.25
=1
.25
Divide
both by
smallest
%B
= .5
at. mass B
So E.F
is
AB2
=
.5
=2
.25
A 5 g sample of an oxide of lead contains 4.53 g Pb. Determine the Empirical Formula of
the compound.
4.53g Pb .022
=
=1
207
.02
So E.F. is PbO
.47 g ox. (5 - 4.53) .03
=
=1
16
.02