Chapter 6
Chemical Quantities
Homework

Recommended



“Questions” 1-15 (odd)
Required

“Problem” 17-97 (odd)

“Cumulative Problems” 99-123 (odd)
Recommended

“Highlight Problems” 125, 127
Counting Particles By Weighing


If a person requests 500 quarter inch
hexagonal nuts for purchase
How would you count 500 hex nuts?
a) You could count the hex nuts individually
b) Or, count the number of hex nuts by weight

First, find the average weight by weighing out 10 hex nuts
and obtaining the total weight (10 hex nuts weigh 105 g)
105g
10.5 g / hexnut
10 hexnuts
Average
Weight
Counting Particles By Weighing


What size (wt.) will
contain 500 hex nuts?
Calculate what weight
will contain 500 hex
nuts
10.5 g
500 hex nuts 
 5250g
hex nut

So, weigh out 5.25 kg
of the hex nuts
Counting Atoms by the Gram





Many chemical calculations require counting atoms and
molecules
There are two common methods for specifying the
quantity of material in a sample of a substance: in terms
of mass and in terms of amount
It is difficult to do chemical calculations in terms of
atoms or formula units
Since atoms are so small, extremely large numbers are
needed in calculations
Need to use a special counting unit just as used for
other items
 A ream of paper
 One dozen donuts
 A pair of shoes
Counting Atoms by the Gram



It is more convenient to use a special counting unit
for such large quantities of particles
Mole: A unit that contains 6.022 х 1023 objects
It is used due to the extremely small size of atoms,
molecules, and ions



6.022x1023 particles in 1 mole
Called Avogadro’s Number
Periodic Table



The average atomic mass in amu (one atom)
The weight of 1 mole of the element in grams
Avogadro’s number provides the connecting
relationship between molar masses and atomic
masses
Converting Between Moles and
Number of Atoms


Avogadro’s number relates the number of
particles present to moles present
Use when solving chemical calculations
involving atoms, molecules, or ions
present in a given amount of a substance.
mole
21moles
10
moles
Moles of
substance
Avogadro’s
Number
24
23
1.204
6.022 × 1024
Number of Atoms
(Particles)
Converting Between Moles and
Number of Molecules

How many molecules of bromine are present in 0.045
mole of bromine gas?
Given: 0.045 mol Br2
Equality:
Avogadro’s number
Find: molecules of Br2
1 molBr2  6.0221023 moleculesBr2
6.0221023 moleculesBr2
mol Br2
Conversion factors:
and
mol Br2
6.0221023 moleculesBr2
Solution:
0.045 molBr2
6.0221023 moleculesBr2


mol Br2
2.7 1022 moleculesBr2
Converting Between Grams and Moles of an Element:
The Atomic Mass Unit


Atoms are too small to conveniently count individually
The usual standards of mass (g or lbs.) are not
convenient for use with atoms
The mass of one atom of uranium-238 = 3.95 × 10-22 g



To avoid such small numbers scientists use a mass value
for atoms relative to a standard instead of an actual
mass value
The standard is the atomic mass unit (amu) with all
masses measured relative to carbon-12
The mass of this isotope is set at 12.00 amu
Converting Between Grams and Moles of an Element:
The Atomic Mass Unit



One amu is defined as
the mass of one-twelfth
the mass of a carbon12 atom (nearly equal
to one proton)
The masses of all other
atoms are determined
relative to carbon-12
The masses of these
particles in amu brings
the mass of the proton
and neutron close to 1.
24
1 amu  1.66 10
g
Converting Between Grams and Moles of an Element:
The Atomic Mass Unit



Atomic masses are
determined on a
relative scale
The standard scale
references the
carbon-12 isotope =
12.00 amu
All other atomic
masses are
determined relative to
carbon-12
Converting Between Grams and Moles of an Element:
The Atomic Mass Unit




Atoms of the same
element do not
necessarily have exactly
the same mass
Each element exists as a
mixture of isotopes so
use an average mass
To calculate the average
mass of a sample of
atoms
 Use a “weighted
average” for the
atomic mass
Number on the bottom of
each square in the
periodic table is the
average weight of the
element (in amu)
Converting Between Grams and Moles of an Element:
The Atomic Mass Unit





Using Atomic Mass to Count Atoms
Calculating the number of atoms in a specific mass
If you have a sample of an element, can calculate the
number of atoms in that sample
From the atomic mass per one atom a conversion
factor can be made
For example: One nitrogen atom has an atomic mass
of 14.01 amu
1 N atom  14.01 amu
14.01 amu
1 N atom
and
1 N atom
14.01 amu
Calculating Atomic Mass in amu
(Sample Calculation)

Calculate the mass (in amu) of 1.0 х 104
carbon atoms
4
2) Plan: Convert from atoms to amu
1) Given: 1.0  10 atoms C
Find: total mass of C atoms (amu)
3) CF 1 atom C = 12.01 amu
12.01 amu
1 C atom
and
1 C atom
12.01 amu
4) Set Up Problem
12.01 amu
1.0 10 C atom
 1.2105 amu
1 C atom
4
Converting Between Grams and
Moles of an Element





The molar mass of an element (monatomic) is
the mass of one mole of atoms of the element
The molar mass of one mole of carbon-12
atoms 12.01 g
The molar mass (in grams) of any element
has the same numerical value as the atomic
mass (in amu) of the same element
This mass contains 6.022 х 1023 particles of
that element
Use the periodic table to obtain the molar
mass of any element
Converting Between Grams and
Moles of an Element

When the number of grams (weighed out) of a
substance equals the formula mass of that
substance, Avogadro’s number of atoms or
molecules of that substance are present
Formula
mass
Molar mass
of water
molecule
of water
==18.02
18.02amu
g
Formula mass of
Molar mass of
I molecule
I22 = 253.8 g
= 253.8 amu
Converting Between Grams and
Moles of an Element

The three quantities most often calculated
in chemical problems




Number of particles
Number of moles
Number of grams
Using molar mass as a conversion factor is
one of the most useful in chemistry

Can be used for gram to mole and mole to
gram conversions
Converting Between Grams and
Moles of an Element


The molar mass provides a relationship between
the number of grams of an element and the
number of moles of the same element
The molar mass becomes part of the conversion
factor used to convert from grams to moles or
moles to grams
Solution map
Grams of
Element
Molar Mass
Moles of
Element
Converting Between Grams of an Element
and Number of Particles


The grams, moles, and number of particles of a
substance are interrelated through the conversion
factors of molar mass and Avogadro’s number
These two relationships can be combined into a single
diagram which is useful to convert grams of a substance
to the number of particles (atoms) of a substance
Molar Mass
Grams of
substance
Moles of
substance
Moles of
substance
Avogadro’s
Number
Particles of
substance
Converting Between Grams of an Element and
Number of Atoms (Sample Calculation)

How many lead atoms are there in a lead
weight with a mass of 1.2 g?
Given: 1.2 g Pb
Conversion Factors
Find: Pb atoms
1 mol Pb= 207.19 g
1 mol Pb= 6.022 × 1023 Pb atoms
Solution Map
1.2 g Pb
Molar mass
mol Pb
Avog. Number
number of Pb atoms
Solution
1.2 g Pb 1 mol Pb 6.0221023 Pb atoms

207.19g Pb
1molPb
3.5 × 1021 Pb atoms
Converting Between Grams and Moles
of a Compound


The molar mass provides a relationship between
the number of grams of a substance and the
number of moles of the same substance
The molar mass of any substance is always
numerically equal to its formula mass in amu
Solution map
Grams of
substance
Molar Mass
Moles of
substance

Converting Between Grams and Moles
of a Compound
Calculate the mass (in grams) of 0.555 mol of
Al2O3
Given: 0.555 mol Al2O3
Find: g Al2O3
Conversion Factors
1 mol Al2O3 = 101.96 g
Solution Map
mol Al2O3
Molar mass
g Al2O3
Solution
0.555mol Al2O3
101.96g
 56.6 g Al2O3
1 mol Al2O3
Converting Between Grams of an Compound
and Number of Molecules



The molar mass provides a relationship between the number
of grams of a substance and the number of moles of the
same substance
Avogadro’s number provides a relationship between the
number of particles of a substance and the number of moles
of the same substance
These two relationships can be combined into a single
diagram which is useful to convert grams of a substance to
the number of particles (molecules) of a substance
Molar Mass
Moles of
substance
Grams of
substance
Moles of
substance
Avogadro’s
Number
Particles of
substance
Converting Between Grams of an Element and
Number of Molecules (Sample Calculation)

How many molecules of CO2 are in a 5.61 g
sample of CO2?
Given: 5.61 g CO2
Conversion Factors
Find: CO2 molecules
1 mol CO2= 44.01 g
1 mol CO2 = 6.022 × 1023 CO2 molec
Solution Map
5.61 g CO2
Molar mass
mol CO2
Avog’s No.
number of CO2 molecules
Solution
5.61g CO 2 1 molCO 2 6.0221023 CO2 molecules

44.01g CO2
1molCO2
7.68 × 1022 CO2 molecules
Converting Between Grams of an Element and
Number of Molecules (Sample Calculation)

Ethylene glycol has the chemical formula: C2H6O2. Find
the mass of 1.24 × 1024 molecules of ethylene glycol
(antifreeze)
Find: g of C2H6O2
Given: 1.24 × 1024 molecules C2H6O2
Conversion Factors
1 mol C2H6O2= 62.05 g
1 mol C2H6O2= 6.022 × 1023 C2H6O2 molec
Solution Map
Avog No.
molecules C2H6O2
molar mass
mol C2H6O2
g of C2H6O2
Solution
1.24 1024 moleculesC2 H6O2
1 molC2 H6O2
62.05g C2 H6O2

23
6. 02210 moleculesC2 H6O2 1molC2 H6O2
128 g C2H6O2
Chemical Formulas as Conversion Factors




The subscripts in a chemical formula indicate the
number of atoms of each element present in a
single molecule or formula unit of a compound
The subscripts in a chemical formula can also
indicate the number of moles of atoms of each
element present in one mole of a compound
e.g. In one molecule of glucose (C6H12O6) there
are 6 atoms of carbon, 12 atoms of hydrogen,
and 6 atoms of oxygen
e.g. In one mole of glucose (C6H12O6) there are
6 moles of carbon, 12 moles of hydrogen, and 6
moles of oxygen
Converting Between Moles of a Compound
and Moles of a Constituent Element


Some problems require a calculation of the moles of a particular
element in a chemical formula of the compound
The subscript of the particular element in the compound becomes
part of the conversion factor used to convert moles of compound to
moles of element within the compound
Solution map
Moles of
compound
Formula
subscript
Moles of
Element
(within compound)
Converting Between Moles of a Compound and Moles of a
Constituent Element (Sample Calculation)

How many moles of carbon atoms are present in 1.85
moles of glucose?
Given: 1.85 mol C6H12O6
Conversion Factors
Find: mol C
1 mol C6H12O6= 6 mol C
Solution Map
mol C6H12O6
Formula
Subscript
mol C
Solution
1.85molC6 H12 O6
6 molC

1 molC6 H12 O6
11.1 mol C
Converting Between Grams of a Compound and Moles of a
Constituent Element (Sample Calculation)

International Foods Coffee contains 3.0 mg of sodium
chloride per cup of coffee. How many moles of sodium
are in each cup of coffee?
Find: mol Na
Given: 3.0 mg NaCl
Conversion Factors
1 mg NaCl = 10-3 g NaCl
1 mol NaCl= 58.44 g NaCl
1 mol Na =1 mol NaCl
Solution Map
10-3 g NaCl
1 mg NaCl
mg NaCl
1 mol Na
1 mol NaCl
1 mol NaCl
58.44 g NaCl
g NaCl
mol NaCl
mol Na
Solution
3.0 mg NaCl 10-3 g NaCl 1 mol NaCl 1mol Na

1mg NaCl 58.44g NaCl 1 mol NaCl
5.13 × 10-5 mol Na
Converting Between Grams of a Compound and Grams of a
Constituent Element



In chemical problems, the three quantities most often calculated for
a substance are number of particles (atoms, molecules, or formula
units), number of moles, number of grams
All of these quantities are interrelated and the conversion factors
which deal with these relationships are Avogadro’s number, molar
mass, and chemical formula subscripts
The three concepts can be combined into a single diagram that is
useful for problem solving
mol NaCl
1 mol Na =
1 mol NaCl
1 mol = 58.44 g NaCl
0.003 g NaCl
mol Na
Converting Between Grams of a Compound and Grams of a
Constituent Element
 Throughout Chapter 6, there are calculations for which the
initial information is given: moles, grams, or particles.
 Additional information is then requested concerning a
component of that same substance (find): moles, grams,
or particles
 Now, use concepts illustrated in previous problems to solve
problems in which a given mass of a sample substance is
given and we need to calculate the mass of a constituent
element
Grams of
compound
Molar
Mass
Formula
subscript
Molar
mass
Moles of
compound
Moles of
element
Grams of
element
Converting Between Grams of a Compound and
Grams of a Constituent Element

Caffeine is the stimulant found in coffee and tea. It has the
chemical formula C8H10N4O2. How many grams of carbon are
present in a 50.0 g sample?
Given: 50.0 g C8H10N4O2
Conversion Factors
Find: g of C
1 mol C8H10N4O2= 194 g C8H10N4O2
8 mol C= 1 mol C8H10N4O2
Solution Map
50.0 g C8H10N4O2
1 mol C= 12.01 g C
Molar
Mass
Mol-mol
ratio
mol C8H10N4O2
Molar
Mass
g of C
mol of C
Solution
50.0g C8 H10 N 4 O 2 1 molC8 H10 N 4 O 2
8 molC
12.01g C

194g C8 H10 N 4 O 2 1 molC8 H10 N 4 O 2 1 molC
24.8 g C
Mass Percent Composition of
Compounds


The mass percent of an element may be
obtained by:
Decomposition of a sample of a known mass



Then, the masses of the constituent elements present
in each sample are then obtained
Each element’s mass, and the original sample mass,
is used to calculate the percent composition
Mass data obtained from the combination of
elements which combine to form a compound

Each element’s mass, and the product’s mass, is used
to calculate the percent composition
Mass Percent Composition of
Compounds



The composition of compounds is determined by their
formulas (e.g. the formula of carbon dioxide tells us it’s
composed of carbon and oxygen (2 oxygen atoms and 1
carbon atom)
Another way to describe the composition of compounds
is in terms of the mass percent of each element in the
compound
The mass percent composition is the percent by mass of
each element in a compound
Mass percentof element X 
Mass of X in a sampleof the compound
 100%
Mass of sample of thecompound
Mass Percent Composition from a
Chemical Formula


The mass percent composition is the percent by mass
of each element in a compound
To determine a compound’s mass percent
composition:
 its chemical formula is needed
 the molar mass of each element that composes
the compound
 the subscript numbers in the formula are equal to
the number of moles of each element in one mole
of the compound (its molar mass)
Mass percentof element X 
Mass of element X in 1 mol of compound
 100%
Massof 1 mol of compound
Mass Percent Composition of Compounds


Percent numbers are always in terms of parts
per hundred
For example, the percent composition of water
is 88.81% oxygen and 11.19% hydrogen



The percent composition for each element is
determined by dividing the mass contributed by each
element, by the molar mass and multiplied by 100
The sum of the percents of all of the elements in a
compound equals 100%
If the mass percent of a particular element in a
compound is known, you can calculate the amount of
the element present in any sample size of that
compound
Mass Percent Composition from a
Chemical Formula

Calculate the mass percent composition of each
element in NH4OH?
Mass percent of element X 
Mass of X in 1 mol of compound
 100%
Mass of 1 mol of compound
Determine the mass of 1 mol of compound
(114.01g)  (5 1.008g)  (116.00g)  35.05g Molar mass
14.01 g N
5.040 g H
16.00 g O
Determine the contribution of each element
N:
14.01g
100%  39.97% N
35.05g
H:
5.040g
100%  14.38% H
35.05g
O:
16.00g
 100%  45.65% O
35.05g
Calculating Empirical Formulas for
Compounds


Chemical formulas are determined by calculation
using experimentally obtained information
Depending on the amount of experimental data
available, two types of formulas can be obtained


Empirical formulas (simplest formula)
Molecular formulas (true formula)
Calculating Empirical Formulas for
Compounds




The empirical formula gives the simplest ratio of
elements in a compound
It uses the smallest possible whole number ratio
of atoms present in a formula unit of a
compound
The subscripts in the formula cannot be reduced
any further to a simpler ratio by division with a
small integer
If the percent composition of a compound is
known, an empirical formula can be calculated
Calculating an Empirical Formula from
Experimental Data
To Determine the empirical formula:
Write down or calculate the given number of grams of
each element present in a sample of the compound
2) Convert the mass of each element to moles of element
using the appropriate molar mass
3) Using the calculated moles of each of the elements:

Write down a pseudoformula using the moles of each
element as a subscript
4) Find the lowest whole number mole ratio

Divide all the subscripts in the formula by the smallest
subscript
1)
Obtaining an Empirical Formula from Experimental Data
Converting Decimal Numbers to Whole Numbers




The subscripts in a “true” formula are expressed as
whole numbers, not as decimals
The resulting decimal numbers from a calculation
represent each element’s subscript
Arithmetic may need to be applied to convert decimals
to whole numbers
If the number(s) are NOT whole numbers, multiply
each number by the same small integer (2, 3, 4, 5, or
6) until a whole number is obtained for each subscript
number
Obtaining an Empirical Formula
from Experimental Data




Lactic acid has a molar mass of 90.08 g
and has this percent composition:
40.0% C, 6.71% H, 53.3% O
What is the empirical and molecular
formula of lactic acid?
Assume a 100.0 g sample size

Convert percent numbers to grams
Obtaining an Empirical Formula from
Experimental Data
Step 1: Mass of each element
in a 100 g
sample of the compound
40.0 g C
6.71 g H
53.3 g O
Step 2: Convert mass to moles
m olC
40.0 g C 
 3.33m olC
12.0 g C
Step 3: Pseudoformula
molH
6.71g H 
 6.66 molH
1.008g H
53.3 g O 
m olO
 3.33m olO
16.00 g O
C3.33H6.66 O3.33
Step 4: Divide by smallest subscript
3.33
C:
 1.00
3.33
H:
6.66
 2.00
3.33
O:
3.33
 1.00
3.33
Empirical
formula is
CH2O
Calculating Molecular Formulas for
Compounds




Molecular formulas are the formulas of
molecules
They illustrate all of the atoms of each element
in the molecule
It does not necessarily have the smallest set of
subscript numbers: It can be the same as or a
whole number multiplier of its empirical formula
To determine a molecular formula, the
compound’s molecular mass is needed
Relating Empirical and Molecular Formulas

Empirical Formula




Smallest possible set of subscript numbers
Smallest whole number ratio
All ionic compounds are given as empirical formulas
Molecular Formulas




The actual formulas of molecules
When an empirical formula is known, the molar mass
is needed to determine if the empirical formula and
the molecular formula are the same
It shows all of the atoms present in a molecule
It may be the same as the EF or a whole-number
multiple of its EF
Molecular formula = n х Empirical formula
Calculating Molecular Formulas for Compounds

n represents a whole number multiplier from 1
to as large as necessary
n


Molar m ass ( g / m ol)
Em pirical form ula m olarm ass( g / m ol)
Calculate the empirical formula and the mass of
the empirical formula
Divide the given molecular mass by the
calculated empirical formula molar mass
 Answer is a whole number multiplier
Relating Empirical and
Molecular Formulas


Multiply each subscript in the empirical formula
by the whole number multiplier to get the
molecular formula
For molecular compounds, the two types of
formulas may or may not be the same
Relating Empirical and Molecular Formulas

For ionic compounds


The empirical and molecular formula are the same
For molecular compounds
 The two types of formulas may be the same
but not necessarily
Calculating Molecular Formulas for
Compounds



What is the molecular formula of lactic acid?
Obtain the value of n (whole number multiplier)
Multiply the empirical formula by the multiplier
Empirical
formula is
CH2O
Empirical formula molar mass = 12.01 g + 2 (1.008 g) + 16.00 g = 30.03 g/mol
n
m olar m ass ( g / m ol)
90.08 g / m ol

3
em pirical form ula m olar m ass ( g / m ol)
30.03 g / m ol
Molecular formula = n х empirical formula
Molecular formula = 3 (CH2O)
C3H6O3
end