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Mole-Concept With Solutions

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0% found this document useful (0 votes)
9 views53 pages

Mole-Concept With Solutions

Uploaded by

mrkeil0984val
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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The Mole Concept:

Mass Relationship and


Percentage Composition
Table of Contents

01 02 03 04 Percent
Atomic Molecular Avogadro’s Number and
Composition of
the Molar Mass of an
Mass Mass Element Compounds
01
Atomic Mass
Introduction
❑ The mass of an atom depends on
the number of protons, and
neutrons it contains

❑ atomic mass (sometimes called


atomic weight) is the mass of the
atom in atomic mass units (amu).
Introduction

❑ One atomic mass unit is


defined as a mass exactly
equal to one-twelfth the mass
of one carbon-12 atom
For Example:
Element Atomic Mass Atomic Mass Unit

Hydrogen (H) 1.0078 g/mol 1.0078 amu

Nitrogen (N) 14.007g/mol 14.007amu

Iron (Fe) 55.845g/mol 55.845 amu

Bromine (Br) 79.904g/mol 79.904 amu


Average Atomic Mass
❖ It is the measure of the
average atomic mass of the
element.

❖ the average mass of the


naturally occurring mixture
of isotopes.
Average Atomic Mass
❖ For example, the natural
abundances of carbon-12 and
carbon-13 are 98.90 percent and
1.10 percent, respectively. The
atomic mass of carbon-13 has
been determined to be 13.00335
amu. Thus, the average atomic
mass of carbon can be calculated
as follows:
Average Atomic Mass
Average Atomic Mass of the
element
= (0.9890)(12amu) + (0.0110)(13.00335
amu)
=12.01

❖ It is important to understand that when we


say that the atomic mass of carbon is 12.01
amu, we are referring to the average value.
Example:
02
MOLECULAR MASS
MOLECULAR MASS
❖The molecular mass (sometimes called molecular
weight ) is the sum of the atomic masses (in amu)
in the molecule.

❖In general, we need to multiply the atomic mass of


each element by the number of atoms of that
element present in the molecule and sum over all
the elements.
Example

❖For example, the molecular mass of H2O is


Example
Sample problem

What is the molecular mass of methanol (CH4O)?.


03
Avogadro’s Number and the
Molar Mass of an Element
Collection Terms

A collection term states a


specific number of items.
•1 dozen donuts = 12 donuts
•1 ream of paper = 500
sheets
•1 case = 24 cans
Collections of items
include dozen, gross,
and mole.
The Mole

Collections of items include


dozen, gross, and mole.
❖In the SI system the
mole (mol) is the amount
of a substance that
contains as many
elementary entities
(atoms, molecules, or
other particles) as there Italian scientist
are atoms in exactly 12 g Amedeo Avogadro
(or 0.012 kg) of the
carbon-12 isotope.
❖The actual number of atoms in 12 g of carbon-
12 is determined experimentally.

❖This number is called Avogadro’s number (NA )

NA= 6.0221415 x 1023

Generally, we round Avogadro’s number to


NA= 6.022 x 1023
❖ From the molecular mass we can determine
the molar mass of a molecule or compound.

❖ The molar mass of a compound (in grams)


is numerically equal to its molecular mass
(in amu).
❖ For example, the molecular mass of water
is 18.02 amu, so its molar mass is 18.02 g.
Note that 1 mole of water weighs 18.02 g
and contains 6.022x1023

❖ H2O molecules, just as 1 mole of


elemental carbon which contains
6.022x1023 carbon atoms
❑Note that the molar mass of carbon-
12 (in grams) is numerically equal to
its atomic mass in amu.

❑If we know the atomic mass of an


element, we also know its molar mass.
❑Note that the molar mass of carbon-
12 (in grams) is numerically equal to
its atomic mass in amu.

❑If we know the atomic mass of an


element, we also know its molar mass.
For example, we know the molar mass of carbon-12 is
12.00 g and there are 6.022 x 1023 carbon-12 atoms in 1
mole of the substance; therefore, the mass of one carbon-
12 atom is given by:
SAMPLE PROBLEM 1
SAMPLE PROBLEM 2
SAMPLE PROBLEM 3
SAMPLE PROBLEM 4
SAMPLE PROBLEM 5
04
PERCENT COMPOSITION OF
COMPOUND
Percent Composition of
Compound
❑ The percent composition by mass is the percent by mass
of each element in a compound.
❑ Percent composition is obtained by dividing the mass of
each element in 1 mole of the compound by the molar
mass of the compound and multiplying by 100 percent

❑ where n is the number of moles of the element in 1 mole


of the compound.
For example, in 1 mole of hydrogen peroxide (H2O2) there
are 2 moles of H atoms and 2 moles of O atoms. The
molar masses of H2O2, H, and O are 34.02 g, 1.008 g,
and 16.00 g, respectively. Therefore, the percent
composition of H2O2 is calculated as follows:

The sum of the percentages is 5.926% + 94.06% =99.99%


Note:The small discrepancy from 100 percent is due to the way we rounded off the molar
masses of the elements.
Example 1
Empirical Formula of
Compound

Procedure for calculating the empirical


formula of a compound from its
percent compositions.
Example 2
Example 3

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