1
Chapter 6 Chemical Calculations
Submicroscopic
Macroscopic
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Chapter Outline
1.
2.
3.
4.
5.
Formula Masses (Ch 6.1)
Percent Composition (supplemental material)
The Mole & Avogadro’s Number (Ch 6.2)
Molar Mass (Ch 6.3)
Chemical Formula & Mole Calculations
(Ch 6.4 & 6.5)
6. Empirical & Molecular Formula
(supplemental material & Lab Exp 8)
7. Chemical Equations & Stoichiometric
Calculations (Ch 6.6, 6.7, 6.8)
8. Heat of Reaction (supplemental material)
9. Percent Yield (supplemental material)
10. Limiting Reagent (supplemental material)
3
1. Formula Masses (Ch 6.1)
Molecular mass/weight – sum of atomic masses of
the atoms in a molecule.
Example:
________
M.W. = 46.0 amu
C: 2 x 12.0 amu = 24.0 amu
H: 6 x 1.0 amu = 6.0 amu
O: 1 x 16.0 amu = 16.0 amu
46.0 amu
4
Formula mass/weight – sum of atomic masses of
atoms in an ionic substance (formula unit, NOT a
molecule).
Example: Ammonium sulfide _____________
F.W. = 68.1 amu
N: 2 x 14.0 amu = 28.0 amu
H: 8 x 1.0 amu = 8.0 amu
S: 1 x 32.1 amu = 32.1 amu
68.1 amu
Due to its offensive smell, ammonium
sulfide it is the active ingredient in a
variety of foul pranks, including the
common stink bomb.
5
2. Percent Composition (supplemental material)
% Composition = # of g of each element
in 100 g of a compound
= mass of element x 100
total mass
Example: Ammonium sulfide (NH4)2S
F.W. = 68.1 amu
%N =
2 x 14.0 amu N
68.1 amu (NH4)2S
x 100 = 41.111 % N
%H =
8 x 1.0 amu H
68.1 amu (NH4)2S
x 100 = 11.747 % H
%S =
32.1 amu S
x 100 = 47.136 % S
68.1 amu (NH4)2S
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3. The Mole & Avogadro’s Number (Ch 6.2)
1 dozen = 12 objects
1 ream = 500 sheets
1 mole = _________ objects
6.0221415 x 1023
Avogadro’s Number
7
Sample problem:
How many He atoms are in
2.55 moles of He?
information x
given
_________
factor
=
information
sought
2.55 mole He x 6.02 x 1023 He atoms = __________ He atoms
1 mole He
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4. Molar Mass (Ch 6.3)
1 mole 12C = exactly 12 g 12C
demo sample
Molar mass of 12C = __________
We now have three conversion factors to relate moles,
number of atoms, and mass of Carbon:
6.02 x 1023 C atoms
1 mole
12 g C
1 mole
6.02 x 1023 C atoms
12 g C
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Figure 6.5
1 mole of S, Zn, C,
Mg, Pb, Si, Cu, Hg
(start counterclockwise from
yellow S and Hg in center)
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C is our standard and when we compare it to other
elements, we find that there are Avogadro’s number of
atoms of any element in a sample whose mass in grams is
numerically equal to its atomic weight.
Mg = 24.31 amu
1 mole Mg = 24.31 g Mg = 6.02 x 1023 atoms Mg
Pb = ________ amu
1 mol Pb = _______ g Pb = 6.02 x 1023 atoms Pb
10
Now we can do the same for ionic compounds as well as
for molecules because the molar mass is the mass (in
grams) of a substance that is numerically equal to the
substance’s formula mass.
Ammonium sulfide (NH4)2S formula mass = 68.1 amu
68.1 g (NH4)2S = 1 mole
(molar mass or formula weight, F.W.: ____________)
68.1 g (NH4)2S = 6.02 x 1023 formula units of (NH4)2S
Carbon dioxide CO2 formula mass = 44.01 amu
44.01 g CO2 = 1 mole
(molar mass or molecular weight, M.W.: _____________)
44.01 g CO2 = 6.02 x 1023 molecules of CO2
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Sample Problem:
If 7.50 moles of ammonia, NH3, are required for a certain
experiment, what mass of ammonia is needed?
Formula mass = 3 x 1.0 (H) + 14.0 (N) = 17.0 amu
1 mole NH3 (molar mass) = 17.0 g NH3
7.50 moles NH3 x 17.0 g NH3 = _______ g NH3
1 mole NH3
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5. Chemical Formula & Mole Calculations
(Ch 6.4 & 6.5)
The subscripts in a chemical formula give the number of
moles of atoms present in 1 mole of the substance:
Example: Ammonium sulfide (NH4)2S
For N:
2 moles of N atoms
or 1 mole (NH4)2S
1 mole (NH4)2S formula units
___ moles N
For H:
8 moles of H atoms
or 1 mole (NH4)2S
1 mole (NH4)2S formula units
___ moles H
For S:
1 mole of S atoms
or 1 mole (NH4)2S
1 mole (NH4)2S formula units
___ mole S
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Sample calculation:
How many H atoms are in 35.6 g of (NH4)2S?
(NH4)2S formula mass = 68.1 amu
1 mole (NH4)2S = 68.1 g
1 mole (NH4)2S = ___ moles H atoms
1 mole H atoms = 6.02 x 1023 H atoms
Strategy:
mass (NH4)2S → moles (NH4)2S → moles H → atoms H
35.6 g (NH4)2S x 1 mol (NH4)2S x 8 mol H x 6.02 x 1023 H
1 mol (NH4)2S
1 mol H
68.1 g (NH4)2S
= ______ x 10___ H atoms
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Fig 6.7 “Transitions” allowed in solving chemical-formula bases problems:
Drill problem:
How many g of (NH4)2S are required to obtain ___ moles of NH4+?
(NH4)2S = 68.1 g/mol (molar mass)
moles NH4+ →
mole (NH4)2S → g (NH4)2S
0.50 moles NH4+ x 1 mole (NH4)2S x 68.1 g (NH4)2S
1 mol (NH4)2S
2 moles NH4+
= ______ (NH4)2S
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6. Empirical & Molecular Formula
(supplemental material & Lab Exp 8)
_______
C6H12O6
CH2O
Molecular Formula
Empirical Formula
The Empirical Formula (E.F.) is the simplest ratio of atoms
in a compound.
_____ acid C2H4O2
CH2O
Molecular Formula
Empirical Formula
Both compounds have the same composition:
40.0 % C, 6.7 % H, 53.3 % O
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If we are given the experimentally determined composition of a
compound, we can calculate the E.F.
Sample calculation. Composition:
40.0 % C, 6.7 % H, 53.3 % O
Step 1: assume a 100 g sample and convert to ________
For C: 40.0 g C x 1 mole C = 3.33 moles C
12.0 g C
For H:
6.7 g H x 1 mole H = 6.7 moles H
1.0 g H
For O: 53.3 g O x 1 mole O = 3.33 moles O
16.0 g O
Step 2: divide each number of moles by the smallest of the
numbers to obtain mole ratios = E.F.
For C: 3.33/3.33 = 1.0
For H: 6.7/3.33 = 2.0
For O: 3.33/3.33 = 1.0
Empirical Formula = _______
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Drill problem:
Composition of Borazole = 40.28%B, 52.20%N, 7.52%H
Molar mass = 80.5 amu
Calculated the molecular formula
B: 40.28g x 1 mole = 3.726 moles = 1.0
10.81 g
3.726
N: 52.20g x 1 mole = 3.726 moles = 1.0
14.01
3.726
H:
7.52g x 1 mole = 7.45 moles = 2.0
1.01 g
3.726
E.F.= _____ E.F. mass: 10.81 + 14.01 + (2x1.01) = 26.84
Molar mass = ______ amu
E.F. mass
______ amu
= 3
Molecular Formula = 3 x (BNH2) = B3N3H6 Borazole
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Mole ratios must be within 0.1 of a whole number. If they
are not, each result must be multiplied by the same
multiplication factor until every value is + 0.1 of a whole
number.
Example for a hypothetical set of mole ratios obtained from
% composition:
2.247 for C
1.98 for H
1.000 for O
The result for C is not + 0.1 of a whole number; therefore,
each result must be multiplied by an integer until all of the
values are. For the above example, the multiplication
factor that works is 4:
2.247 x 4 = 8.988 for C
1.98 x 4 = 7.92 for H
1.000 x 4 = 4.000 for O
Empirical Formula = __________
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7. Chemical Equations & Stoichiometric Calculations
(Ch 6.6, 6.7, 6.8)
Summary of submicroscopic and macroscopic levels of a
chemical equation:
2 H2O → 2 NaOH(aq)
+ H2(g)
2 Na(s)
+
2 atoms
+ 2 molecules → 2 formula units + 1 molecule
2 moles
+ 2 moles
→ 2 moles
+ 1 mole
2x23=46g + 2x18=36g → 2x40=80g
______ reactants
→
+ 1x2=2g
______ products
Law of Conservation of Mass
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The coefficients in a balanced equation give the numerical
relationships among formula units consumed or produced
in a chemical reaction.
Keys to the calculations = _______________
N2
+
3 H2
→
2 NH3
1 mole N2
3 moles H2
3 moles H2
1 mole N2
2 moles NH3
1 mole N2
1 mole N2
2 moles NH3
3 moles H2
2 moles NH3
2 moles NH3
3 mole H2
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Figure 6.9 In solving stoichiometric calculations, only the
following “transitions” are allowed:
Sample calculation:
How many g O2 are needed to convert 45 g glucose
(C6H12O6) into CO2 and H2O?
First write a balanced equation and calculated the molecular masses of
glucose and oxygen:
M.W.
C6H12O6 + 6 O2 →
180 amu + 32 amu
6 CO2
+ 6 H 2O
Then set up the calculations according to Figure 6.9
45 g glu x 1 mol glu x ___mol O2 x 32 g O2
180 g glu
1 mol glu
1 mol O2
= ________ O2
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Drill Problem: Figure 6.10
The chemical equation for the
deployment of airbags is
2 NaN3(s) → 2 Na(s) + 3 N2(g)
How many g NaN3 would have to
decompose on order to generate
253 million molecules of N2?
F.W. NaN3 = 65.0 amu
Avogadro’s # = 6.02 x 1023
molecules N2 → moles N2 → moles NaN3 → g NaN3
2.53x108 molecules N2 x
1 mole N2
= 4.203 x 10-16 moles N2
6.02 x 1023molecules N2
4.203 x 10-16 moles N2 x ___ moles NaN3 x 65.0 g NaN3
___ moles N2
___mol NaN3
= _______ x 10-14 g NaN3
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8. Heat of Reaction (Ch 9.5 & supplemental material)
When the chemical energy stored
in reactants is greater than that
stored in the products, energy is
released by the reaction, and it is
termed exothermic.
The change in energy is called the
enthalpy change and is represented
by ∆H; the value is negative for an
exothermic reaction.
Example: combustion of propane is exothermic.
The heat released can be represented with energy on the
product side:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) + 530 kcal
However, the reaction is also represented by placing the
enthalpy change to the right of the equation:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H = ________
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An endothermic reaction is a
chemical reaction in which a
continuous input of energy is
needed for the reaction to
occur. Energy is a reactant.
Example: photosynthesis is
endothermic
∆H = positive
6 CO2(g) + 6 H2O(g) + 678 kcal → C6H12O6(aq) + 6 O2(g)
6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)
∆H = _________
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Sample Problem:
How much energy is produced when
0.50 g of butane, C4H10, is burned in
a butane lighter?
C4H10 = 58.1 g/mole
2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l) + 1365 kcal
The equation shows that 1365 kcal of heat are produced
when 2 moles of butane undergo combustion.
0.50g C4H10 x 1 mol C4H10 x 1365 kcal = _____ kcal
58.1 g C4H10
___ moles C4H10
Is this reaction exothermic or endothermic?
What is the sign for ∆H?
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9. Percent Yield
(supplemental material)
The theoretical yield in a reaction
is the amount of product that
could be obtained if a given
reactant reacted completely. In
most cases the actual yield is
smaller that the theoretical yield because of side reactions,
incomplete reaction, or other experimental limitations. The
discrepancy between the theoretical yield and the actual yield is
reported as the percent yield, which is calculated as shown:
% yield =
actual yield
theoretical yield
x 100
The theoretical yield is calculated from the given amount of the
specified reactant. The actual yield is identified in the problem.
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Pure iron can be produced from
iron oxide in a blast furnace.
Sample Problem. When 884 g of
Fe2O3 was reduced with excess
carbon, 507 g of Fe were obtained.
What was the percent yield?
Fe2O3 + 3 C → 2 Fe + 3 CO
1. Calculate the theoretical yield:
g Fe2O3 → mol Fe2O3 → mol Fe → g Fe
884g Fe2O3 x 1 mol Fe2O3 x 2 moles Fe x 55.85g Fe = ___g Fe
159.7g Fe2O3 1 mol Fe2O3 1 mol Fe
2. Calculate the % yield:
% yield =
actual yield x 100 = 507 g x 100 = _______ %
theoretical yield
____ g
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10. Limiting Reagent (supplemental material)
Two batteries are required for
these flashlights to work.
So, if you have 10 flashlights and 17 batteries, how many working
flashlights do you have?
The batteries are the LIMITING “REAGENT”.
The flashlights are IN EXCESS.
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Reaction A: Stoichiometric amounts of reactants are used.
NaOH
1 mol
+
+
HCl → NaCl + H2O
1 mol → 1 mol + 1 mol
Reaction B: One reactant is limiting.
NaOH + HCl →
NaCl +
H2O
0.6 mol + 0.3 mol → ___ mol + ___ mol
Which is the limiting reagent?
_______
Which reagent is in excess? _______
What are the amounts of each product?
How much of each compound is present at the end of the
reaction?
NaOH
HCl
NaCl
H2O
=
=
=
=
Please take note that you must calculate the theoretical yield of any
reaction only from the limiting reagent!
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Sample Problem:
If you have 25 moles of N2 reacting with 45 moles of H2
according to the following reaction, which is the limiting
reagent?
N2(g) + 3 H2(g) → 2 NH3(g)
moles of A x mole ratio = moles of B
available
needed
25 moles N2 x 3 moles H2 = ____ moles H2
1 mole N2
For 25 moles N2 we would need ____ moles H2 but only 45
moles H2 are available.
H2 = limiting reagent
You can do the same analysis starting with H2
45 moles H2 x 1 moles N2 = ____ moles N2
___ moles H2
We have more N2 than we need.
N2 = _________
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Here is another way to finding the limiting reagent.
N2(g)
25 moles
+
3 H2(g) → 2 NH3(g)
45 moles
The limiting reagent has the lowest mole-to-coefficient
ratio:
25 moles N2 = 25
1 mole N2
45 moles H2 = 15
3 moles H2
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Demo: Identify the limiting reagent
HC2H3O2(aq) + NaHCO3(s) → H2O(l) + CO2(g) + NaC2H3O2(aq)
A
0.19 mol
0.19 mol
0.19 mol
0.19 mol
+
+
+
+
B
0.048 mol
0.095 mol
0.19 mol
0.38 mol
in Rxn I
in Rxn II
in Rxn III
in Rxn IV
Reagent A and B are mixed and CO2 evolution is measured.
Which is the limiting reagent?
Rxn I
Rxn II
Rxn III
Rxn IV
B is limiting
B is limiting
stoichiometric amounts (balanced)
A is limiting
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The following drill problem summarizes the important
chemical calculations you have learned in this chapter:
When aqueous solutions of CaCl2 and
AgNO3 are mixed, a white precipitate
of AgCl forms.
If 3.33 g CaCl2 are reacted with 8.50 g
AgNO3 and 5.63 g AgCl are obtained,
what is the % yield?
CaCl2(aq) + 2 AgNO3(aq) →
•
•
•
•
•
2 AgCl(s) + Ca(NO3)2(aq)
Calculated formula masses
Convert g to moles
Check if you have a limiting reagent
Calculate theoretical yield
Calculate % yield
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CaCl2(aq) + 2 AgNO3(aq) →
3.33 g
111.0 amu
3.33 g CaCl2 x
8.50 g
169.9 amu
2 AgCl(s) + Ca(NO3)2(aq)
5.63 g
143.3 amu
1 mol = 0.0300 mol CaCl2
111.0 g
8.50 g AgNO3 x 1 mol = 0.0500 mol AgNO3
169.9 g
Mole-to-coefficient ratios:
0.0300 CaCl2 = 0.0300
0.0500 AgNO3 = 0.0250
______________
Theoretical yield:
0.0500 mol AgNO3 x 2 mol AgCl x 143.3 g AgCl = ___ g AgCl
2 mol AgNO3 1 mol AgCl
% Yield:
actual yield x 100 = 5.63 g x 100 = ______ %
theoretical yield
____ g