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Chapter 7
Quantitative Composition of Compounds
Making new chemicals is much like following a recipe from a cook book...
1 cup flour + 2 eggs
+ ½ tsp baking powder  5 pancakes … except you
don’t get to lick
the spoon!
What if you want to make more (or less)?
Suppose you have plenty of flour and baking powder,
but only 8 eggs. How many pancakes can you make?
You can solve it using conversion factor:
8 eggs x
5 pancakes
2 eggs
= 20 pancakes
5 pancakes
2 eggs
Solve it in your head:
2 eggs makes 5
pancakes, so four
times more eggs
makes 20 (5x4)
pancakes.
Practice using the following mouthwashing, diet-buster recipe:
3 blocks cream cheese + 5 eggs + 1
cup sugar = 1 cheese cake.
How many cheese cakes can we make out of 15 eggs?
1 cake
15 eggs x 5 eggs = 3 cheese cakes
How much sugar do we need for 5 cheese cakes? (5)
Suppose you want to ‘whip’ a batch of hydrogen
iodide, following the balanced chemical equation:
H2 + I2  2 HI
How much H2 and I2 should you use to make 10 g of HI?
A common mistake is that H2 and I2 react in one-to-one mass ratio so:
5 g H2 + 5 g I2  10 g HI
The coefficients balancing the
equation refer to number of
atoms, not masses.
Introducing the mole. The mole is like a dozen, but much, much more.
The mole is Avogadro’s Number of items.
1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 1023.
We need the mole because the mass of an atom is too small to
be measured on a balance. Remember: 1 amu = 1.6 x 10-24 g.
1 mole of
anything:
donuts, pancakes,
is always 6.022 x
atoms, molecules, ions… 1023 of that thing.
1 mole of soft drink cans is enough to cover the surface of
the earth to a depth of over 200 miles.
If we were able to count atoms at the rate of 10 million
per second, it would take about 2 billion years to count
the atoms in one mole.
The mole translates between the number of atoms (or
molecules, ions) and grams of atoms (molecules, ions).
It is defined as the mass of Avogadro’s number of
atoms 126C, which, in turn, weights exactly 12 g.
A mole of atoms weighs the same number of
grams as the atomic mass.
One mole of H atoms weighs 1.0079 g.
One mole of C atoms weighs 12.011 g.
H2
Amadeo Avogadro
The mole
Atomic mass refers to: the sum of protons and
neutrons in a single atom, weighted average mass
of all isotopes of an element and also to the
number of grams in one mole of atoms.
+
I2
=
2 HI
1 molecule
2 H atoms
1 molecule
2 I atoms
12 molecules
12 molecules
6.022 x 1023
molecules
1 mole
2.0158 g
6.022 x 1023
molecules
1 mole
253.81 g
2 molecule
2 x (1 atom H, 1 atom I)
24 molecules
1.204 x 1024
molecules
2 mole
255.8258 g
or any number of
molecules
1 mole of H2 weighs 2 x 1.0079 g = 2.0158 g
Conversion factors:
1 mole
6.022 x 1023 species
1 mole
molar mass
Mole - mass - atoms
conversions
Q1: How many atoms in 0.5 mole Au?
6.022 x 1023 atoms Au
0.5 mole Au x
1 mole Au
= 3.011 x 1023 atoms Au
Q1a: How many moles in 7.12 x 1024 atoms of
Cu?
1 mole Cu
7.12 x 1024 atoms Cu x
6.022 x 1023 atoms Cu
= 11.8 mol Cu
Q2: What is the mass of 0.5 mol Au?
196.967 g Au
0.5 mole Au x 1 mole Au
= 98.4835 g Au
Q3: How many atoms
in 15.00 g Au?
6.022 x 1023 atoms Au
1 mole Au
x
= 4.59 x 1022 atoms Au
15.00 g Au x
1 mole Au
196.987 g Au
Percent Composition
Percent composition is % mass that each
element in a molecule contributes to the
total molar mass of the compound.
Assume that you have one mole of the
compound.
What is the % composition of CH2O?
Total mass = 12.01 g + 2.016 g + 16.00 g
= 30.026 g
12.01 g
%C = 30.026 g x 100
%C = 40.00 %
%H = 6.71 %
Practice: What is the % composition of glucose?
Check your answer: it is the same as in CH2O!
Types of Formulas
+
%O = 53.29 %
100.00 %
CH2O is the empirical formula for glucose, C6H12O6
Empirical Formula: the formula of a
compound that expresses the
smallest whole number ratio of the
atoms present.
Formulas describe the relative
number of atoms (or moles) of each
element in a formula unit. It’s always
a whole number ratio.
If we can determine the relative
number of moles of each element in a
compound, we can determine a
formula for the compound.
Molecular Formula: the formula that states
the actual number of each kind of atom found
in one molecule of the compound.
1 molecule of C9H8O4
= 9 atoms of C, 8
atoms of H and 4
atoms of O.
1 mole of C9H8O4 = 9
mol of C, 8 mol of H
and 4 mol of O atoms.
Dr. Ent burned
0.5 g of the
sample and
obtained the
total of over 1
g of products.
How is that
possible?
Oxygen from
air is a
reactant!
From the mass
of the products
(water and
carbon dioxide)
we determine
the number of
moles of C, H,
and O, and from
them obtain the
empirical
formula of the
compound.
1. Determine the mass in grams of
each element present, if necessary.
Remember, % means “out of 100”.
2. Convert grams of CO2 and H2O
(or C and H) into moles of C and H
atoms.
3. Convert moles of C into grams
of C. Do the same for H.
4. Add masses for C and H and
subtract the sum from the mass of
the sample to obtain mass of O.
Convert the mass into moles of O.
5. Divide all number of moles with
the smallest to obtain the subscripts
of the empirical formula.
6. Divide the molar mass of the
compound by the molar mass of
the empirical formula. To find the
molecular formula, multiply all
subscripts in the empirical formula
by this product.
Note: steps 3, 4 apply only for finding
formulas from combustion analysis.
Combustion analysis shown 0.300 g H2O and
0.733 g CO2 from 0.500 g of sample. Find the
empirical and molecular formula if the molar mass
of the compound is 180.15 g/mol.
2 mol H atms
1 mol H2O
x
0.3 g H2O x 18.01 g H
1 mol H2O
2O
2.
= 0.0333 mol H at.
1 mol C atms
1 mol CO2
x
0.733 g CO2 x 44.01 g CO
1 mol CO2
2
= 0.0166 mol C at.
0.0333 mol H x
3.
1.008 g H
1 mol H = 0.0336 g H
12.01 g C
= 0.199 g C
1 mol C
g O = 0.5 – (0.0336 + 0.199) = 0.267 g O
1 mol O at.
0.267 g O x 16.00 g O = 0.0169 mol O at.
Empirical Molar mass
H: 0.0333 / 0.0166 = 2
formula
emp. form.
C: 0.0166 / 0.0166 = 1
CH2O
30.026
O: 0.0169 / 0.0166 ~ 1
0.0166 mol C x
4.
5.
6.
Molar mass sample
Molar mass emp. formula
There is 6 CH2O units
in the compound.
=
180.15
=6
30.03
Molecular formula: C6H12O6.
Find the empirical and molecular formulas if the % composition is 40.0% C,
6.70% H, 53.3% O, and the molar mass of the compound is 180.155 g/mol.
1. Assume that you have 100.00 g sample; the mass of each
element is equal to the % composition. 40.0 g C, 6.70 g H, 53.3 g O.
2.
1 mol C
1 mol H
1 mol O
40.0 g C x 12.01 g C 6.70 g H x1.008 g H 53.3 g O x 16.00 g O
= 3.33 mol O
= 3.33 mol C
= 6.65 mol H
1. Determine the mass in grams of
each element present, if necessary.
Remember, % means “out of 100”.
2. Convert grams of CO2 and H2O (or
C and H) into moles of C and H atoms.
3. Convert moles of C into grams of
C. Do the same for H.
4. Add masses for C and H and
subtract the sum from the mass of
the sample to obtain mass of O.
Convert the mass into moles of O.
5. Divide all number of moles with the
smallest to obtain the subscripts of
the empirical formula.
6. Divide the molar mass of the
compound by the molar mass of the
empirical formula. To find the
molecular formula, multiply all
subscripts in the empirical formula
by this product.
Skip steps 3 and 4, they apply for combustion analysis only.
5. C: 3.33 / 3.33 = 1
H: 6.65 / 3.33 = 2
O: 3.33 / 3.33 = 1
Empirical formula CH2O.
Emp. Formula mass = 30.026
Molar mass sample
180.155
=
=6
Molar mass emp. formula
30.026
Thus, there are 6 (CH2O) units. Molecular formula: C6H12O6.
6.
Practice (answer in parenthesis):
1. A compound has an empirical formula of NO2. The colorless
liquid used in rocket engines has a molar mass of 92.0 g mole-1.
What is the molecular formula of this substance? (N2O4)
2. A sample of a brown gas, a major air pollutant, is found to contain
2.34 g N and 5.34 g O. Determine an empirical formula for this
substance. (NO2)
3. Percent composition of a compound is found to be 43.2% K,
39.1% Cl, and some O. Find the empirical formula. If the molar mass
of the compound is 90.550 g mol-1, find the molecular formula. (KClO)
Chapter 9
Calculations from Chemical Equations
The molar mass of an element is its atomic mass in grams.
It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.
The molar mass of a compound is the sum of the atomic masses of
all its atoms. For instance: molar mass of NaCl is 22.99 + 35.45 = 65.44 g
For calculations of mole-mass-number_of_particle relationships:
Remember me?
Conversions go through moles.
1. Use balanced equation.
D
2 Al + Fe2O3  Al2O3 + 2 Fe
2 mol
1 mol 1 mol
2 mol
2. The coefficient in front of a
formula represents the number of
moles of the reactant or product.
To quantitatively convert from one quantity
to another we introduce mole ratio:
1 mol Fe2O3
2 mol Al
1 mol Fe2O3
1 mol Al2O3
Mole ratio is found from the coefficients
of the balanced equation.
moles of desired substance
Mole ratio =
moles of starting substance
Which conversion factor will be used depends on starting and desired substance
A mole of a compound weighs the sum of all atoms in the compound.
Mole – Mole Conversions - Molecules
Example 1:How many moles of NaCl
result from the complete reaction of
3.4 mol of Cl2? Assume that there is
more than enough Na.
2 Na(s) + Cl2(g)  2 NaCl(s)
desired substance
3.4 moles Cl2 x
1 mole
2 moles
2 moles NaCl
= 6.8 moles NaCl
1 mole Cl2
starting substance
The following examples refer to the equation:
Ca5(PO4)3F(s) + 5H2SO4(aq) 3H3PO4(aq) + HF(aq) + 5CaSO4(s)
1 mole
5 moles
3 moles
1 mole
5 moles
Example 2: Calculate the number of moles of phosphoric acid (H3PO4) formed
by the reaction of 10 moles of sulfuric acid (H2SO4) on phosphate rock:
10 moles H2SO4 x
3 moles H3PO4
5 moles H2SO4
= 6 moles H3PO4
Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce
6 moles of H3PO4.
6 moles H3PO4 x
1 mole Ca5(PO4)3F
3 moles H3PO4
= 2 moles Ca5(PO4)3F
Mass – Mole conversion
Example 4: Calculate the number
of moles of H2SO4 necessary to
yield 784 g of H3PO4.
g
Molar mass of H3PO4 = 97.994 mole
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mole
5 moles 3 moles 1 mole 5 moles
1. Convert the starting
1 mole H3PO4
784 g H3PO4 x
substance into moles.
= 8.00 moles H3PO4.
97.994 g H3PO4
2. Convert moles of
starting substance into
5 moles H2SO4
8.00
moles
H
PO
x
= 13.3 moles H2SO4.
3
4
moles of desired
3 moles H3PO4
substance.
3. Convert moles of desired substance into the units specified in the problem.
done.
Ex. 5: Calculate the mass of phosphate rock, Ca5(PO4)3F needed to yield 200. g of HF.
Molar masses: Ca5(PO4)3F = 504.31 g/mol; HF = 20.008 g/mol
Step 1, 200. g HF x 1 mole HF = 10.0 moles HF x 1 mole Ca5(PO4)3F = 10.0 moles
20.008 g HF
1 mole HF
Ca5(PO4)3F
Step 2
504.3 g ph.r. = 5.00 kg Ca (PO ) F.
Step 3:
5
4 3
10.0 moles Ca5(PO4)3F x
1 mole ph.r
Step_by_step:
Mass – mass conversion
Ex. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4.
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mole
5 moles
3 moles
1 mole
5 moles
Molar mass H3PO4 = 97.994 g
mole
g
Molar mass H2SO4 = 98.086 mole
1. Convert the starting substance into moles.
1 mole H3PO4
392 g H3PO4 x
= 4.00 moles
97.994 g H3PO4
5 moles H2SO4
2. Convert moles of starting substance
4.00 moles H3PO4 x
3 moles H3PO4
into moles of desired substance.
= 6.67 moles
3. Convert moles of desired substance into the units specified in the problem.
6.67 moles H2SO4 x
392 g H3PO4 x
98.086 g
= 654 g H2SO4.
1 mole H2SO4
Combined steps:
1 mole H3PO4 x 5 moles H2SO4 x 98.086 g
= 654 g H2SO4.
97.994 g H3PO4 3 moles H3PO4 1 mole H2SO4
Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO2,
assuming that there is more than enough water to react with all the CO2. Molar
masses are 44.01 g (CO2) and 180.16 (glucose).
sunlight
6 CO2(g)  6 H 2O(l)  
 6 O 2(g)  C6 H12O6(aq)
58.5 g CO2 x
1 mole CO2
1 mole glucose 180.16 g glucose = 39.9 g glucose
x
x
44.01 g CO2
6 moles CO2
1 mole glucose
Conversion – General Case
Mass to moles
of starting compound
Step 1
Moles of starting compound
to moles of desired compound
Step 2
Moles of desired comp.
to units desired.
Step 3
Mass – mass: All 3 steps
Example 8: Calculate the mass of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2  2NH3
grams H2  moles H2  moles NH3  grams NH3
Molar masses: H2: 2.016 g/mol; NH3: 17.034 g/mol
1 mole H2 2 moles NH3 17.034 g NH3
x
x
= 1420 g NH3 = 1.42 kg NH3.
2.016 g H2 3 moles H2
1 mole NH3
Starting
Step 1
compound
result
Step 2
Step 3
Moles – moles: Step 2 only
Example 9: Calculate the moles of NH3 formed by the reaction of 1.5 moles of H2.
2 moles NH
1.50 moles of H2 x 3 moles H 3 = 1.00 mole NH3.
2
112 g H2 x
Starting
compound
Step 2
result
Moles – mass: Step 2 and Step 3 only
Example 10: Calculate the mass of NH3 formed by the reaction of 1.50 moles of H2.
17.034 g NH3
2 moles NH
1.50 moles of H2 x 3 moles H 3 x 1 mole NH
= 17.0 g NH3.
3
2
Starting
result
Step 2
compound
Step 3
Conversion – General Case (cont’d)
Mass – moles: Step 1 and Step 2 only
Example 11: Calculate the moles of NH3 formed by the reaction of 150. g H2.
1 mole H
2 moles NH
N2 + 3H2  2NH3
150. g H2 x 2.016 g H2 x 3 moles H 3 = 49.6 g NH3.
2
2
Starting
Step 1
Step 2
result
compound
Mass – particles: All 3 steps
Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2.
1 mole H2 2 moles NH3 6.022 x 1023 molecules NH3 = 2.23 x 1025
112 g H2 x
x
x
2.016 g H2 3 moles H2
1 mole NH3
molecules NH3.
Starting
Step 1
Step 3
result
compound
Step 2
Limiting Reactant and Yield Calculations
The amount of the product(s) depends on the reactant
that is used up during the reaction, i.e. limiting reactant.
One bicycle needs 1 frame, 1 seat and 2 wheels,
therefore not more than 3 bicycles can be made.
The number of seats is the limiting part (reactant);
one frame and two wheels are parts in excess; 3
bicycles is the yield.
Limiting Reactant and yield Calculations (cont’d)
Example 13: How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe
with 10.0 g H2O? Which substance is the limiting reactant? Which substance is in
excess? How much of the reactant in excess remains unreacted?
Strategy:
1. Write and balance equation.
2. Calculate the number of moles of product for each reactant;
3. The reactant that gives the least moles of (the same!) product is the limiting reactant.
4. Find the amount of reactant in excess needed to react with the limiting reactant.
Subtract this amount from the starting quantity to obtain the amount in excess.
5. Find the yield from the limiting reactant.
Balanced equation: 3 Fe (s) + 4 H2O (g)
Yield
1 mol Fe x
from Fe:
16.8 g Fe x
55.85 g Fe
limiting reactant
1 mol H2O
x
From H2O: 10.0 g H2O 18.02 g H O x
2
Reacted
H2O
16.8 g Fe x
Excess:
10.0 g – 7.01 g = 2.99 g H2O.
1 mol Fe x
55.85 g Fe
Fe3O4 (s) + 4 H2 (g)
1 mol Fe3O4
yield
=
0.100
mol
Fe
O
.
3 4
3 mol Fe
Least moles Fe3O4?
1 mol Fe3O4
4 mol H2O = 0.139 mol Fe3O4.
18.02 g H2O
4 mol H2O
= 7.01 g H2O.
x
1 mol H2O
3 mol Fe
Answer: Yield is 0.100 mol Fe3O4, Fe is the limiting reactant, 2.99 g H2O is in excess.
Percent Yield
Calculations done so far assumed that the reaction gives maximum (100%) yield.
Many reactions (especially organic) do not give the 100% yield, due to:
side reactions, reversible reactions, product losses due to human factor.
Theoretical yield: Amount calculated from the chemical equation.
Actual yield: Amount obtained experimentally.
Actual yield
Percent yield:
x 100 %
Theor. yield
Strategy:
Find limiting reactant.
Calculate theoretical yield.
Calculate percent yield.
Example 14: If 65.0 g CCl4 was prepared by
CS2 + 3 Cl2
CCl4 + S2Cl2
reacting 100. g CS2 and 100. g of Cl2, calculate
the percent yield. Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.81 g/mol
1 mol CS2
1 mol CCl4
100. g CS2 x
x
= 1.31 mol CCl4.
76.15 g CS2 1 mol CS2
1 mol Cl2
1 mol CCl4
x
= 0.470 mol CCl4.
70.90 g Cl2
3 mol CS2
Limiting reactant
153.81 g CCl4
0.470 mol CCl4 x
= 72.3 g CCl4.
1 mol CCl4
Theoretical yield
65.0 g CCl4
65.0 g CCl4
HW, Chp. 7: 1, 5, 15, 26, 33
x 100 % = 89.9 %
Actual yield
72.3 g CCl4
Chp. 9: 3, 7, 13, 15, 23, 29
Percent yield
100. g Cl2 x