Quick Review: The Mole
A very large counting number =
6.022 x 1023 (6.022 followed by 23 zeros)
Also known as Avogadro’s number
The number shown under an element on the
periodic table is the mass of 1 mole of that element
(from the weighted average of naturally occurring
isotopes).
1 mole of Li has a mass of 6.941 grams.
6.022 x 1023 atoms of Li has a total mass of 6.941
grams.
A mole of Cl has a mass of 35.45 grams.
The molar mass of chlorine is 35.45 g/mole
What is the mass of 6.022 x 1023 atoms of Al?
From the periodic table we can see that the molar
mass of Al is 26.98 g/mol
Use conversion factors to convert 10.0 g Al to the
correct number of moles
•
•
•
•
Grams to moles
Given
End Unit
Conversion factor
10.0 g Al x 1 mol Al
= 0.371 mol Al
26.98 g Al
Given
End Unit
Periodic
Table
How many grams of Al are there in 10.0 mol of Al?
Invert the conversion factor (Mol to Gram)
10.0 mol Al x 26.98g Al= 269.8 g Al
mol Al
How many atoms of Al are in the sample?
(Mol to atoms)
10 mol Al x 6.022 x 1023 atoms Al
mol Al
= 6.022 x 1024 atoms of Al.
Review: Molar Mass of compounds (Molecular
Weight)
Compounds are made up of two or more elements
The molar mass of the compound is the sum of the molar
masses of each element in the compound.
Molar mass of O2 (the element, oxygen) is 2 x molar
mass of the O atom.
2 x 16.00 grams/mol = 32.00 grams/mol
The molar mass for the element O2 is 32.00 g/mol
6.022 x 1023 molecules of O2 weigh 32.00 g
In the molecule CO2, there is one atom of carbon.
The subscript on the symbol for oxygen tells us
that there are 2 atoms of oxygen in the molecule.
There is one mole of carbon in CO2
There are two moles of oxygen in CO2
The molar mass of CO2 is
1 x 12.01 g C / mol CO2
2 x 16.00 g O / mol CO2
44.01 g / mol CO2
What is the molar mass of SO2?
a. how many moles of S?
b. how many moles of O?
1 mol S x 32.06 g/mol S
= 32.06 g/mol
2 mol O x 16.00 g/mol of O = 32.00 g/mol
64.06 g/mol
1 mol SO2 = 64.06 g
This is an equivalence statement
1 mol SO2
64.06 g SO2
=1
64.06 g SO2 = 1
1 mol SO2
This is the same type of equivalence statement as
2.54 cm = 1 in and can be used as conversion
factors in calculations.
Calculate the number of moles of CO2 in
2.25 x 102 g of the gas.
(Grams to moles)
•Calculate the molar mass of CO2.
•1 x 12.01 g/mol C
•2 x 16.00 g/mol O
44.01 g/mol CO2
•1 mol CO2 = 44.01 g
•Second use conversion factors to convert grams
to moles.
•2.25 x 102 g CO2 x 1 mol CO2 =
44.01 g CO2
5.11 mol CO2
•Mass percent composition of compounds (molecules
or ionic compounds)
•Expressed by identifying the elements present and
giving the mass percent of each.
mass fraction, = mass of element in 1 mol compound
given element
mass 1 mol of compound
Mass % element A = molar mass of element A x 100
molar mass of compound
Mass percent of elements in ethanol, C2H5OH
The molar mass of ethanol is 46.08 g/mol
In each mole of ethanol there are 2 moles of carbon
The mass percent of carbon in ethanol is
2 x 12.01 = 24.02 g C in each mole of ethanol
24.02 g C
x 100 = 52.13%
46.08 g Ethanol
52.13% of the mass of ethanol is from the mass of
carbon in the molecule.
What is the mass percent of hydrogen in ethanol,
C2H5OH?
Total molar mass = 46.08 g.
(1 mol ethanol = 46.08 g)
6 moles of hydrogen in each mole of ethanol x 1.01
g/mol = 6.06 g/mol
6.06 g/mol hydrogen /46.08 g/mol ethanol x 100 =
13.2%
Do the two add up to 100%???
Where is the difference?
16.00 g/mol O / 46.08 g/mol ethanol x 100 = 34.7 %
•Percent composition makes it possible to
determine a compound’s
•EMPIRICAL formula, the smallest whole
number ratio of one element to others in the
compound that agrees with the percent
composition.
•CH2O is the empirical formula for formic acid,
acetic acid (C2H4O2) and glucose (C6H12O6).
•What is the molar mass of each compound?
CH2O
32.05 g / mol
C2H4O2
64.06 g / mol
C6H12O6
192.3 g / mol
Steps - empirical forumula
•Assume 100 g of compound
•Divide the mass percent of the element by the molar
mass of the element to get the number of moles of
each element.
•Divide the # moles of each element by the smallest #
moles to get the ratio of the elements, one to another.
•If any result is far from a whole number multiply
through by a common factor that converts each
number of moles to integers (or close)
•If each is close to a whole # round off each to nearest
integer
An unknown clear solution was analyzed
and found to contain 84.1% carbon and
15.9% hydrogen by mass.
If we assume 100 grams of the substance,
the compound would contain 84.1 g carbon
atoms and 15.9 grams of hydrogen atoms.
Taking 84.1 g carbon and using the atomic mass and
dimensional analysis, we can determine the number
of moles of the two elements in the substance.
84.1 g C x 1 mol C = 7.00 mol C
12.01 g C
15.9 g H x 1 mol H = 15.8 mol H
1.01 g H
divide moles of each component by smallest number
of moles
15.8 mol H = 2.26 mol H
7.00 mol C 1 mol C
The C-H mole ratio is 1:2.26
We can write C1H2.26 as a temporary
formula. This is obviously not a true
chemical formula (cannot have 2.26 atoms –
violates Dalton’s law).
We must multiply the formula by a number
which converts the fractional number of
atoms of hydrogen into a whole number. In
the above instance, if we multiply by four, we
get C4H9.
Work Time
• Finish warm-up sheet
• 11.4 notes
• Section assessment questions page 337 (5861)
C4H9
This is an empirical formula. The true chemical
formula could be a whole number multiple of the
empirical formula. The chemical formula of the
substance analyzed could be C8H18 or any other
whole number multiple.
Molar mass of empirical formula
4 moles C x 12.01 g/mol C = 48.04 g C
9 moles of H x 1.01 g/mol H = 9.09 g H
57.13 g total
Vitamin C (ascorbic acid) contains 40.92% C,
4.58% H, and 54.50% O by mass. What is
the empirical formula of ascorbic acid?
Assume 100 g.
40.92 g C x 1 mol C = 3.41 mol C
12.01 g
4.58 g H x 1 mol H
1.01 g
= 4.53 mol H
54.50 g O x 1 mol O = 3.41 mol O
16.00 g
40.92 g C x 1 mol C
12.01 g
4.58 g H x 1 mol H
1.01 g
54.50 g O x 1 mol O
16.00 g
= 3.41 mol C
= 4.53 mol H
= 3.41 mol O
To get empirical formula divide the all by the smallest
# moles
3.41 mol C / 3.41 = 1
4.53 mol H / 3.41 = 1.33
3.41 mol O / 3.41 = 1
C1H1.33O1
???
Must multiply subscripts of the derived formula,
C1H1.33O1, by some factor to convert the fractional
atom into a whole number. This is done by trial and
error, but the multiples are usually small, whole
numbers.
x 2 = C2H2.66O2
x 3 = C3H4O3 molar mass = 88.07 g / mol
This is the empirical formula, which may or may not
be the molecular formula. It tells us only the relative
numbers of atoms.
What is the molecular formula for ascorbic acid
whose molar mass is 176.13
C3H4O3 molar mass = 88.07 g / mol
molar mass compound = 176.14 g / mol
176.14 / 88.07 = 2
molecular formula ascorbic acid = C6H8O6
The composition of ibuprofen is 75.7% C,
8.8% H and 15.5% O, by mass.
Determine the mass percentage of each
element and from that data determine the
empirical formula.
Start with 100 grams
75.7 g from C, 8.8 g from hydrogen, 15.5 g from O
75.7 g C x 1 mol C
12.01 g C
= 6.30 mol C
8.8 g H x 1 mol H
1.01 g H
= 8.70 mol H
15.5 g O x 1 mol O
16.00 g O
= 0.969 mol
DIVIDE EACH NUMBER OF MOLES BY THE
SMALLEST TO GIVE THE FOLLOWING.
6.30 mol C / 0.969 mol O = 6.50 mol C / mol O
8.7 mol H / 0.969 mol O = 9.0 mol H / mol O
0.969 mol O / 0.969 mol O = 1 mol O / mol O
Empirical formula = C6.50H9O
Elemental Analysis
Combustion
A compound of unknown composition is
decomposed by heat. The elements are
carefully trapped and the number of moles of
each are analyzed.
A sample of a compound composed of carbon
oxygen and hydrogen are combusted in a
stream of O2 to produce CO2 and H2O. The H2O
and CO2 are trapped and the masses of each
measured.
The sample has a mass of 0.255g. When the
reaction is complete, 0.561 g of CO2 and
0.306g of H2O are produced. What is the
empirical formula of the compound?
1. Determine the mass of C in the sample. For
each mole of CO2, there is one mole of C.
Convert moles of C to grams of C.
0.561 g CO2 x (1 mol CO2 / 44.01 g CO2) x
(1 mol C / 1 mol CO2) (12.01 g C/ mol C) =
0.153 g C
There are 2 moles of hydrogen per mole of H2O.
0.306 g H2O x (1 mol H2O / 18.0 g H2O) x
(2 mole H / mole H2O) x (1.01 g H / mol H)
= 0.0343 g H
Mass O = mass sample - mass H - mass C
Mass sample = 0.255 g
Mass O = 0.255 - 0.153 - 0.0343 = 0.068 g O
To get empirical formula, convert g back to moles
0.153 g C x ( 1 mol C / 12.01 g C ) = 0.0128 mol C
0.0343 g H x (1 mol H / 1.01 g H) = 0.0340 mol H
0.068 g O x (1 mol O / 16.0 g O) = 0.0043 mol O
Divide each by 0.0043 to get ratio of each element to O
C
0.0128 mol C / 0.0043 mol O = 2.98 ~ 3
There are 3 moles of carbon for each mole of oxygen
H
0.0340 mol H / 0.0043 mol O = 7.91 ~ 8
There are 8 moles of hydrogen per mole of oxygen
Empirical Formula
C3H8O
Chemical Equations
Short hand describing a chemical change using
symbols and formulas to represent the elements
and compounds involved in the change.
Carbon reacts with molecular oxygen to yield
carbon dioxide
C + O2
CO2
The equation tells us that 1 mole of C reacts with 1
mole of O2 to yield 1 mole of CO2
Stoichiometric Calculations:
Stoicheion (element) ; metron (measurement)
Relating mass or moles of reactants in a chemical
reaction to mass or moles of products.
EQUATION MUST BE BALANCED
Reaction of C2H4 (g) + HCl(g)
C2H5Cl(g)
If we have 15.0 g of C2H4 how many moles of HCl
are needed to carry out the reaction to completion?
C2H4(g) + HCl(g)
C2H5Cl(g)
The balanced equation tells us it takes 1
mole of HCl for each mole of C2H4 reacted.
We must begin by converting the number of
grams of C2H4 which has a molar mass of
28.08 g/mol, to moles
15.0 g C2H4 x 1 mol C2H4 = 0.534 mol C2H4
28.08 g C2H4
C2H4(g) + HCl(g)
C2H5Cl(g)
The coefficients in the balanced equation tells
us that one mole of HCl is needed for each
mole of C2H4 (ethylene).
It therefore takes 0.534 mol of HCl to
consume all of the ethylene.
What mass of HCl is needed to carry the
reaction through to completion?
C2H4 (g) + HCl(g)
C2H5Cl(g)
To determine the mass of HCl needed, use
molar mass.
Begin with what is asked.
0.534 mol HCl x 36.5 g HCl = 19.6 g HCl
mol HCl
C2H4(g) + HCl(g)
C2H5Cl(g)
How many moles of product are made when
15.0 g of C2H4 is reacted with an excess of
HCl?
Pay close attention to what is being asked.
Note from balanced eqn that there is one
mole of C2H4 is needed to make one
molecule of C2H5Cl.
Note also that the question only refers to the
reactant C2H4 and product, not HCl.
C2H4(g) + HCl(g)
C2H5Cl(g)
To determine moles from mass use dimensional
analysis.
Molar mass of C2H4 (ethylene) = 28.08 g/mol
The problem states that you have 15.0 g C2H4
15.0 g C2H4 x 1mole C2H4 = 0.534 mol C2H4
28.06 g C2H4
Since the stoichiometry of the balanced equation
indicates that 1 mol of C2H5Cl is produced from
each mole of C2H4, the ratio is 1:1.
C2H4 (g) + HCl(g)
0.534 mol
C2H5Cl(g)
0.534 mol
(molar mass C2H5Cl = 64.51 g/mol)
Mass of product: 0.534 mol product x 64.51
g/mol = 34.448 g = 34.5 g in correct # sig fig
(we started with 15.0 g reactant. Answer must
be in 3 sig.fig).
Calculations of product masses when the
stoichiometric coefficients are not the same.
N2(g) + 3 H2(g)
2 NH3(g)
Note from the balanced equation that two
moles of product are yielded when 1 mole of
N2 reacts with 3 mole H2.
To determine the mass of the product, we
follow the same steps, but now we must
consider another factor, the mole ratio.
N2(g) + 3H2(g)
2NH3(g)
If there is enough N2 in the mixture and we
have 4.8 mol H2, how many moles of NH3 are
produced?
The relationship is now 3:2 and we must take
that into consideration.
Again, begin with what you are given. You
have 4.8 mol H2.
N2(g) + 3H2(g)
2NH3(g)
4.8 mol H2 x 2 mol NH3 = 3.2 mol NH3
3 mol H2
Mass of NH3 is determined by calculating mass
from moles using molar mass as before.
N2 + 3H2
2NH3
1. Assuming there is enough hydrogen
for the reaction to go to completion how
many grams of nitrogen are needed to
yield 0.430 moles of NH3?
2. For the reverse reaction, how many
grams of nitrogen are produced by the
decomposition of 3.24 grams of
gaseous NH3?
Limiting Reactants
Suppose you do not have enough of one
reactant to carry the reaction to completion?
To determine the moles or masses of products,
we need to know which reactant will run out first
and which is in excess.
We cannot use coefficients or masses to make
the determination. As usual, everything is
measured in moles.
Limiting Reactants
1. Write the balanced equation.
2. Convert masses to moles.
3. The reactant producing the lowest
moles of product will be consumed
(limiting reactant).
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
If 0.30 mol Zn is added to 0.52 mol HCl,
how many moles of H2 will be produced?
Zn(s) + 2HCl(aq)
0.30 mol Zn x 1 mol H2
1 mol Zn
ZnCl2(aq) + H2(g)
= 0.30 mol H2
0.52 mol HCl x 1 mol H2 = 0.26 mol H2
2 mol HCl
The mole ratio conversions come directly from
the stoichiometric coefficients in the balanced
equation.
0.26 mol is the maximum amount of H2 produced
when 0.30 mol of Zn and 0.52 mol of HCl are
mixed.
Grams of A
Use molar mass as a conversion factor
Moles of A
Use coefficients from balanced equation
as a conversion factor
Moles of B
Use molar mass as a conversion factor
Grams of B
Percent Yield
In most reactions not all reactants go forward to
form products. Some are lost to byproducts and
some do not react.
The actual yield is usually less than the anticipated,
theoretical yield.
Percent yield = actual yield
x 100
theoretical yield
Zn(s) + 2 HCl(aq)
ZnCl2(aq) + H2(g)
Theoretical yield is 0.26 mol H2.
If 0.19 moles are produced, what is the percent
yield?
(0.19/0.26) x 100 = 73%
Balanced eqn
Masses to moles
Mole ratio of
limiting reactant
To products
Moles
product
Mol ratios from
coefficients
Compare mol
ratios, which is
limiting?
Mass
product
You must practice these types of
conversions to proceed any further in the
class.
We will be using these skills through the
balance of the semester.
Go through sample exercises while they are
still fresh in your mind.