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Transcript
•The chemical formula for water is H2O.
•How many atoms of hydrogen and oxygen are
there in one water molecule?
H2O
2 hydrogen atoms
1 oxygen atom
•How might you calculate the mass of a water
molecule, given the atomic masses of hydrogen
and oxygen?
•In this section, you will learn how to carry out
these and other calculations for any compound.
Using Chemical Formulas
•As well as indicating the elements and the
relative numbers of atoms or ions of each
element in a compound, chemical formulas can
also be used to calculate formula mass, molar
mass, and the percent composition of a
compound.
Formula Mass
•The formula mass of any molecule, formula
unit, or ion is the sum of the average atomic
masses of all the atoms represented in its
formula.
•The units for formula mass are atomic mass
units (amu).
•The mass of a water molecule, H2O, can be
referred to as a molecular mass.
•The mass of one formula unit of an ionic
compound, such as NaCl, is not a molecular
mass.
Example: Find the formula mass for H2O.
H2O has 2 hydrogen atoms and 1 oxygen atom.
Atomic mass of hydrogen = 1.01 amu
Atomic mass of oxygen = 16.00 amu
2 hydrogen atoms x 1.01 amu = 2.02 amu
1 oxygen atom x 16.00 amu = 16.00 amu
Total
Formula mass of H2O
18.02 amu
Molar Masses
•The molar mass of a substance is equal to the
mass in grams of one mole, or 6.02 x 1023
particles, of the substance.
•Units for molar mass are g/mol.
•The molar mass of a compound is calculated by
summing the masses of the elements present in
a mole of the molecules or formula units that
make up that compound.
•A compound’s molar mass is numerically equal
to its formula mass.
Example: Find the molar mass for KClO3.
KClO3 has 1 potassium (K) atom, 1 chlorine
atom (Cl), and 3 oxygen (O) atoms.
Atomic mass of potassium = 39.10 amu
Atomic mass of chlorine = 35.45 amu
Atomic mass of oxygen = 16.00 amu
1 potassium atom x 39.10 amu = 39.10 amu
35.45 amu
1 chlorine atom x 35.45 amu =
48.00 amu
3 oxygen atoms x 16.00 amu =
Total 122.55 amu
Since a compound’s molar mass is numerically
equal to its formula mass (KClO3 = 122.55 amu)
the molar mass of KClO3 would be
Numerically
equivalent to
formula mass but
units are different.
This tells us that if we have one mole of KClO3
we would have 122.55 g of KClO3 or
6.02 x 1023 formula units of KClO3.
Molar Mass as a Conversion Factor
•The molar mass of a compound can be used
as a conversion factor to relate an amount in
moles to a mass in grams for a given
substance.
•To convert moles to grams, multiply the amount
in moles by the molar mass.
•Amount in moles x molar mass (g/mol) = mass
in grams
Example:
What is the mass in
grams of 2.50 mol of
oxygen (O2) gas?
Given: 2.50 mol O2
Unknown: mass of O2 in grams
Solution: mol O2 → grams O2
mol O2 x molar mass (g / mol )  g O2
Determine the molar mass of O2.
O2 has 2 oxygen atoms.
Atomic mass of O is 16.00 amu.
2 oxygen atoms x 16.00 = 32.00 amu
Remember that molar mass is numerically
equal to the formula mass so the molar mass
of O2 = 32.00 g/mol.
2.50 mol O x 32.00 g O2
= 80.0 g O2
2
1 mol O2
Ibuprofen, C13H18O2, is the active ingredient in
many nonprescription pain relievers. Its molar
mass is 206.31 g/mol.
a. If the tablets in a bottle contain a total of 33 g
of ibuprofen, how many moles of ibuprofen are
in the bottle?
33 g C13H18O2 x
1 mol C13H18O2
= 0.16 mol C13H18O2
206.31 g C13H18O2
b. How many molecules of ibuprofen are in the
bottle? Remember we have 0.16 mol.
0.16 mol C13H18O2 x
6.02 x 1023 molecules
1 mol
= 9.6 x 1022
molecules
C13H18O2
c. What is the total mass in grams of carbon in
33 g of ibuprofen?
To start, determine the number of moles of
carbon per 1 mole of C13H18O2. There are 13
moles of carbon per 1 mole of C13H18O2.
Two conversion factors are needed to solve
problem: number of moles of carbon per mole
of C13H18O2 and the molar mass of carbon.
13 moles C
12.01 g C
1 mol C13H18O2
1 mol C
13 mol C
0.16 mol C13H18O2 x
1 mol C13H18O2
12.01 g C
x
1 mol C
= 25 g C
Problem:
Consider a sample of 10.0 g of the gaseous
hydrocarbon C3H4 to answer the following
questions.
a. How many moles are present in the sample?
b. How many molecules are present in the C3H4
sample?
c. What is the total mass in grams of carbon in
the sample?
Additional Problems:
1. How many molecules of aspirin, C9H8O4, are
there in 0.165 moles of aspirin?
2. If we have 6.54 x 108 molecules HCN, how
many moles of HCN are there?
3. What is the total mass of nitrogen in 1.25
moles of Zn(NO3)2?
Percent Composition
•It is often useful to know the percentage by
mass of a particular element in a compound.
•For example, suppose the compound
potassium chlorate, KClO3, were to be used as
a source of oxygen. It would be useful to know
the percentage of oxygen in the compound.
•To find the mass percentage of an element in a
compound, the following equation can be used.
mass of element in sample of compound
x 100
mass of sample of compound
•The mass percentage of an element in a
compound is the same regardless of the
sample’s size.
•The percentage of an element in a compound
can be calculated by determining how many
grams of the element are present in one mole of
the compound.
•The percentage by mass of each element in a
compound is known as the percentage
composition of the compound.
mass of element in 1mol of compound
x 100  % element in compound
molar mass of compound
Percentage Composition of Iron Oxides
Example: Find the percentage composition of
copper(I) sulfide, Cu2S.
1) Find the molar mass of the compound.
Cu2S has 2 copper atoms and 1 sulfur atom.
The atomic mass of copper is 63.55 amu
and the atomic mass of sulfur is 32.07 amu.
2 copper atoms x 63.55 amu = 127.1 amu
32.07 amu
1 sulfur atom x 32.07 amu =
The formula mass of Cu2S = 159.2 amu.
The molar mass of Cu2S = 159.2 g/mol.
2) Use the mass of each element present in one
mole of the compound to calculate the mass
percentage of each element.
2 atoms of Cu = 127.1 g Molar mass of Cu S
2
1 atom of S = 32.07 g
= 159.2 g.
mass of element of 1mol of compound
x 100
molar mass of compound
127.1g Cu
x 100  79.85% Cu
159.2 g Cu 2S
32.07 g S
x 100  20.15% S
159.2 g Cu 2S
A good check is to see if the results add up to
about 100%. (Because of rounding, the total
may not always be exactly 100%).