Transcript Example - Course

Chemistry 1210:
General Chemistry
The Mole and Stoichiometry
Dr. Gina M. Florio
13 September 2012
Jespersen, Brady, Hyslop, Chapter 4
Conversion Factors
Conversion Factor
– relates one quantity to another
– used to convert between two units in chemistry
What is my height in centimeters (cm) if I am 5 feet 4 inches tall?
1. How many inches are in a foot?
12 inches = 1 foot
2. How many inches are in a centimeter?
1 inch = 2.54 cm
Factor Label Method
The factor-label method, or dimensional analysis lets us treat a
numerical problem as one involving a conversion from one kind of
units to another using conversion factors.
Example:
A recipe calls for 32 grams of cheese. My kitchen scale only displays weight
in ounces. How many ounces of cheese do I need to use?
Conversion
Factor
Factor Label
Method
1 ounce = 28.349 523 125 grams
1 ounce
32 grams 
 1.3 ounces
28 grams
Atomic Mass and Molecular Mass
Atomic Mass – the mass of an individual atom measured in atomic mass
units, u
– listed in the periodic table beneath the element symbol
Molecular Mass – the mass of an individual molecule measured in u
– the sum of the atomic masses of the atoms in the
molecular formula
Example:
The molecular mass of water, H2O, is twice the mass of
hydrogen (2 x 1.008 u) plus the mass of oxygen (15.999 u)
= 18.015 u
Ch 4.1
Formula Mass
Formula Mass – the mass of an individual formula unit measured in atomic
mass units, u
– used for ionic compounds
– calculated the same as a molecular mass
Example:
The formula mass of calcium oxide, CaO, is the mass of
calcium (40.08 u) plus the mass of oxygen (15.999 u)
= 56.08 u
Ch 4.1
The Mole: Connecting the macroscopic & molecular
The mole is a conversion factor
– relates mass to the number of atoms or molecules for a
chemical substance (element, molecular compound, ionic
compound)
– one mole of a substance has a mass (g) equal to its
formula (or molecular or atomic) mass (u)
Example
H2O
Molecular Mass = 2(H) + (O)
= 2(1.0 g) + 16.0 g
= 18.0 g
1 mole H2O = 18.0 g H2O
Ch 4.1
How many water molecules are in a mole of water?
1 mole H2O = 18.0 g H2O
Recall:
1 molecule H2O = 18.0 u (molecular mass)
Need the conversion factor between mass (g) and u:
1 u = 1.66 x 10-27 kg
Need the conversion factor between g and kg:
1 kg = 1000 g
Answer:
Ch 4.1
Avogadro, how many ______ are in a mole?
6.023 x 1023 ______ are in a mole.
Always.
The “blank” can be anything:
Flowers
Cats
Grains of sand
Molecules
This conversion factor (6.023 x 1023 objects/mole) is
known as Avogadro's number.
Ch 4.1
Avogadro’s Number
Two important points:
1. Notice the magnitude of Avogadro's number (6.023 x 1023):
23
~10
Why is this number so
HUGE?
Because atoms and molecules are so
.
tiny
A huge number of atoms or molecules are needed
to make a lab-sized sample.
2. Avogadro’s number links moles and atoms, or moles and molecules,
and provides an easy way to link mass and atoms or molecules.
Ch 4.1
Example Problem
Assuming that pennies are 100% Cu, how many Cu atoms are found in a
penny weighing 3.00 g?
1. Convert mass (g) to moles using molar mass
2. Convert moles to number of atoms (or molecules)
using Avogadro’s number
Ch 4.1
Mole-to-Mole Ratios: Chemical Formulas
Recall:
Water (molar mass 18.015 g/mol) as an example:
1 mole H2O  6.023 x 1023 molecules H2O
1 mole H2O  18.015 g H2O
18.015 g H2O  6.023 x 1023 molecules H2O
Mole-to-Mole Ratio:
Within chemical compounds, moles of atoms always combine in the
same ratio as the individual atoms themselves.
1 mole H2O  2 moles H
1 mole H2O  1 mole O
Ch 4.2
Stoichiometry
Stoichiometry – the study of the mass relationships in chemical compounds
and reactions
– relates the masses of reactants needed to make a
compound
– MOLE to MOLE ratios
Ch 4.2
Example Problem
How many grams of iron are in a 15.0 g sample of iron(III) oxide?
ANALYSIS: 15.0 g Fe2O3  ? g Fe
KNOWN:
SOLUTION:
Ch 4.2
1 mol Fe2O3  2 mol Fe
1 mol Fe2O3  159.7 g Fe2O3
1 mol Fe  55.85 g Fe
Percent Composition
Percentage composition (or percentage composition by mass)
The percentage by mass is the number of grams of the element in
100 g of the compound and can be calculated using:
% element 
mass of element  100%
mass of wholesample
From our previous example:
10.5 g Fe
% Fe 
 100%  70.0% Fe by mass
15.0 g Fe2O3
Ch 4.2
Percent Composition, Why Should I Care?
Hmmm…what in the
world is this stuff?
Detective, send that
sample back to the lab for
analysis!
Sample Problem
A sample was analyzed and found to contain 0.1417 g nitrogen and
0.4045 g oxygen. What is the percentage composition of this compound?
ANALYSIS:
Find sample mass and calculate %
KNOWN:
Mass of the whole sample
WholeSample  0.1417g N  0.4045g O
 0.5426g
SOLUTION:
Ch 4.2
Molecular Formula & Empirical Formula
Hydrogen peroxide consists of molecules with the formula H2O2.
This is called the molecular formula.
However, the simplest formula for hydrogen peroxide is HO and is called
the empirical formula.
The empirical formula thus contains the simplest mole-to-mole ratio of
the atoms in the compound.
We can calculate the empirical formula for a compound from mass data.
Ch 4.3
Example Problem
A 2.012 g sample of a compound contains 0.522 g of nitrogen and 1.490 g of
oxygen. Calculate its empirical formula.
ANALYSIS: We need the simplest whole number mole ratio
between nitrogen and oxygen
SOLUTION:
Convert to moles
0.522g N  1 mol N  0.0373mol N
14.01g N
1.490g O  1 molO  0.0931molO
15.999g O
Find the simplest
whole number
mole-to-mole ratio
Ch 4.3
N 0 .0 3 7 3O 0 .0 9 3 1 N1.00O2.50
0 .0 3 7 3
0 .0 3 7 3
N1.00 2 O2.50 2  N2.00O5.00  N2O5
Empirical Formula (a few final notes)
The formula for an ionic compound is the same as the empirical formula.
For molecules, the molecular formula and empirical formula are usually
different.
The molecular formula will be a common multiplier times all the coefficients
in the empirical formula.
Example:
The empirical formula of hydrazine is NH2, and its molecular mass is 32.0
g/mol. What is its molecular formula?
Ch 4.3
Indirect Analysis of Empirical Formula
Combustion Analysis: an indirect method used to determine the empirical
formula of a unknown compound containing only C, H, and O.
When a compound made only from C, H, and O burns completely in pure
oxygen (O2) only carbon dioxide (CO2) and water (H2O) are produced:
___Cx H yOz  ___O2 ( g)  ___CO2 ( g)  ___ H2O
We can collect and quantify all of the carbon dioxide (CO2) and water (H2O)
produced to find the empirical formula of the original compound.
Ch 4.3
Example Problem
The combustion of a 5.217 g sample of a compound of C, H, and O gave
7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the
compound.
ANALYSIS:
This is a multi-step problem.
First determine the mass of each element.
Then use the masses of the elements to
calculate the empirical formula of the
compound.
Ch 4.3
Example Problem
The combustion of a 5.217 g sample of a compound of C, H, and O gave
7.406 g CO2 and 4.512 g of H2O. Calculate the empirical formula of the
compound.
SOLUTION:
Ch 4.3
Stoichiometry & Chemical Equations
The coefficients of a balanced chemical equation provide the mole-to-mole
ratios for the substances involved in the reaction.
Whenever a problem asks you to convert between different substances, the
calculation usually involves a mole-to-mole relationship.
Example:
If 0.575 mole of CO2 is produced by the combustion of propane, C3H8,
how many moles of oxygen are consumed? The balanced equation is:
C3H8 + 5 O2  3 CO2 + 4 H2O
Ch 4.4
Balancing Chemical Equations
Chemical equations provide quantitative descriptions of chemical reactions.
Conservation of mass is the basis for balancing equations.
To balance an equation:
1. Write the unbalanced equation.
2. Adjust the coefficients to get equal numbers of each kind of atom
on both sides of the arrow
Ch 4.4
Balancing Chemical Equations: some tips
Guidelines for Balancing Equations:
1. Balance elements other than H and O first
2. Balance as a group any polyatomic ions that appears unchanged on
both sides of the arrow
3. Balance separately those elements that appear somewhere by themselves
4. As a general rule, use the smallest whole-number coefficients when
writing balanced chemical equations
Example:
_ NH3 + _ O2  _ NO + _ H2O
Element
Ch 4.4
Reactants
Products
_ NH3 + _ O2  _ NO + _ H2O
Element
N
N
O
O
H
H
Reactants
Products
Limiting Reactants (aka Limiting Reagent)
All reactions eventually use up a reactant and stop.
The reactant that is consumed first is called the limiting reagent (reactant)
because it limits the amount of product that can form.
Any reagent that is not completely consumed during the reactions is said to
be in excess.
The computed amount of product is always based on the limiting reagent.
Ch 4.5
Example Problem
Example:
How many grams of NO can form when 30.0 g NH3 and 40.0 g O2 react
according to:
4 NH3 + 5 O2  4 NO + 6 H2O
Ch 4.5
Percentage Yield
The amount of product isolated from a chemical reactions is almost always
less than the calculated, or maximum, amount.
The actual yield is the amount of the desired product isolated.
The theoretical yield is the amount that would be recovered if no loss
occurred (the calculated, maximum amount).
The percentage yield is the actual yield as a percentage of the theoretical
yield:
percentageyield
Ch 4.6
actualyield
theoretical yield
100%