Transcript Unit 3 Power Point

Ethanol, C2H6O
NaCl, salt
Buckyball, C60
Just to Review…
Please convert 12.4 fluid ounces to cm3
1 fl. oz. = 0.0295735 L
1 cm3 = 1 mL
Convert 4.5 centuries to seconds.
The Carbon-12 Scale
The atomic mass of an element indicates how
heavy, on average, an atom of an element is
when compared to an atom of another element
– Unit is the atomic mass unit (amu)
The standard scale is based on the
carbon-12 isotope
• Mass of one 12C atom = 12 amu (exactly)
• Note that 12C and C-12 mean the same thing
Isotopic Abundance
•
•
•
Most elements exist in nature as a
mixture of two or more isotopes.
To determine the mass of an element, we
must know the mass of each isotope and
the atom percent of the isotopes
(isotopic abundance)
The mass spectrometer can determine
the isotopic abundance & atomic mass.
Figure 3.1 – Mass Spectrometer
• A mass spectrometer is used to determine
atomic masses
Figure 3.1 – Mass Spectrometer
Avogadro’s Number
• A sample of any element with a mass equal to its
atomic mass contains the same number of
atoms, NA, regardless of the identity of the
element.
– NA = 6.022 X 1023  Avogadro’s #
• It represents the number of atoms of an
element in a sample whose mass in grams is
numerically equal to the atomic mass of the
element.
Amadeo Avogadro &
Avogadro’s Number
Avogadro’s Number
Amadeo Avogadro
1776-1856
6.02214199 x
23
10
(6.022 x 1023 to 4 s.f.)
There is Avogadro’s number of
particles in a mole of ANY substance.
Examples:
• 6.022 x 1023 H atoms in 1.008 g
atomic mass H = 1.008 amu
• 6.022 x 1023 S atoms in 32.07 g
atomic mass S = 32.07 amu
Mole: A Perspective on Size
The Green Pea Analogy:
A Dramatic Reading
Examples of Mole Quantities
1 mole of stars in the universe = 6.022 x 1023 stars
1 mole of pennies
= 6.022 x 1023 pennies
(beats the lottery!)
1 mole of glucose molecules = 6.022 x 1023 molecules
1 mole of helium atoms
= 6.022 x 1023 atoms
1 mole of potassium ions (K+) = 6.022 x 1023 ions
EOS
Molecular Mass
Molecular mass: the sum of the atomic
masses of all atoms in a molecular formula.
- (units are amu or u)
1 Oxygen atom
Example:
water - H2O
2(1.01 amu) + 16.00 amu
= 18.02 amu
2 Hydrogen atoms
Sample Problem: Glucose
Glucose - C6H12O6
= 6(12.01 u) + 12(1.01 u) + 6(16.00 u)
= 180.18 u
EOS
Formula Mass
Formula mass is the sum of the masses of the
atoms or ions present in a formula unit – the
unit for an ionic formula
ClNa+
Cl-
Na+
Cl-
Na+
Cl-
Na+
Crystal of
sodium chloride
One Na+ and one Cl– make a
formula unit for sodium chloride
The mass of one formula unit is:
= 22.99 amu + 35.45 amu
= 58.44 amu
Sample Problem
Example 3.1 Determine the formula or
molecular mass for each of the following:
CaI2
(NH4)2S
Al(NO3)3
C6H12O6
Mole--Definition
Chemistry is a
quantitative
science—we need a
“counting unit” aka
the MOLE!
1 mole = the amount of
substance that contains as
many particles (atoms,
molecules, formula units) as
there are in 12.0 g of 12C.
518 g of Pb,
2.50 mol
One-mole Amounts
Analogies
We can group items and count by grouping:
12 eggs = 1 dozen eggs like
6.02 x 1023 items = 1 mole
We can also group items and
count by weighing:
Grass seed and nails —Purchased by
the POUND, not by the item.
The molar mass, MM, in grams/mole,
is numerically equal to the sum of the
masses (in amu) of the atoms in the
formula
***Molar mass is the mass of one mole of a
particular substance.
Molar Masses of Some
Substances
MOLECULAR WEIGHT VS.
MOLAR MASS
Molecular weight = sum of the atomic
weights of all atoms in the molecule (in amu)
Molar mass = molecular weight in grams/mole
We will use molar mass in all problems in this
chapter!!!!
Equivalencies
1 mole of any substance contains Avogadro’s
number of particles (and the mass on the periodic
table expressed in grams).
1 mol of C = 12.01 g of C= 6.022 x 1023 atoms of C
1 mol of O2 = 32.00 g of O2= 6.022 x 1023 molecules
1 mol of NaCl = 58.44 g of NaCl= 6.022 x 1023
formula units of NaCl
The Mole and Reactions
Example: consider the formation of carbon dioxide
At the molecular level ...
Problem: how does one mass out a single carbon atom?
Note that the mass in grams is ~2.00 x 10–23 g!
Answer: one doesn’t!
EOS
The Solution ...
Use a measurable amount – molar quantities
For carbon, mass out:
2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1
= 12 g C
EOS
Must memorize elements that
exist as diatomic elements!!
Remember HON17 !!!
Mole Conversions
Moles!
Use
Avogadro’s #
(6.02 x 1023)
Representative
particles
(atoms, molecules, formula
units, ions)
Use molar
mass
Mole
Conversions
Mass
Practice problems
1. Calcium carbonate, CaCO3, is the principal
mineral found in marble and limestone. How
many moles are in 188.0 g of CaCO3?
Practice problems
2. What is the mass, in grams, of 0.329 mol of
spearmint oil, C10H14O?
Practice problems
3. Find the mass of a single lead atom.
Practice problems
4. How many individual lead atoms are in a
1.000 g sample of this metal?
Practice problems
5. (a) Calculate the number of moles of
aluminum in a solid cube that measures 3.40
cm on a side. (d=2.70 g/cm3).
(b) How many atoms of aluminum are in the
same sample?
Practice problems
6. (a) How many molecules of oxygen, O2, are
in 0.00100 grams of this gas?
(b) How many atoms?
Percent Composition by Mass
Definition: Describes the proportion of
elements in a compound using a percent
Equal to the mass of each element
present in a 100 g sample of compound!
Mass of element
Percent composition of an element =
×100
Mass of compound
Example: Sodium carbonate is a compound
used in the manufacture of soap and glass.
Determine the percent composition by mass of
each element in this compound.
Example: Determine the percent by
mass of water in Al2(SO4)3∙18H2O.
Example: Magnetite, Fe2O3, is one of the
principal iron containing ores. How much
elemental iron can be obtained from a metric
ton (103 kg) of this ore, assuming 100 %
recovery? (hint: first find % iron in Fe2O3)
Determining Formulas
In chemical analysis we determine the % by
weight of each element in a given amount of
pure compound and derive the EMPIRICAL or
SIMPLEST formula.
PROBLEM: A compound of B and H is 81.10% B.
What is its empirical formula?
Empirical Formula Calculations
Percent to mass
Mass to moles
Divide by Small
Multiply ‘til whole
%  g  moles
Divide by
smallest
mole value
empirical formula
A compound of B and H is 81.10% B.
What is its empirical formula?
Percent to mass:
(always assume 100 g sample!)
A compound of B and H is 81.10% B.
What is its empirical formula?
Mass to moles:
1 mol
81.10 g B •
= 7.502 mol B
10.81 g
1 mol
18.90 g H •
= 18.75 mol H
1.008 g
A compound of B and H is 81.10% B.
What is its empirical formula?
Now, recognize that atoms combine in the ratio of
small whole numbers.
Find the ratio of moles of elements in
the compound by “dividing by small”
A compound of B and H is 81.10% B.
What is its empirical formula?
But we need a whole number ratio!
Must multiply each ratio by the smallest integer
available to obtain whole numbers
(multiply ‘til whole)
Multiply ‘til Whole Hints
Common Possible Endings:
.33 x 3
.25 x 4
.67 x 3
.50 x 2
Sample Problem
Example: Bicarbonate of soda is used in
products like Alka-Seltzer and generally
relieves an upset stomach. Determine the
empirical formula of this compound based on
the following percent composition: 27.36% Na,
1.200% H, 14.30% C, 57.14% O.
Sample Problem
Example: A 25.00 gram sample of an orange
compound contains 6.64 g of potassium, 8.84 g
of chromium, and 9.52 g of oxygen. Find its
empirical formula.
A compound of B and H is 81.10% B. Its
empirical formula is B2H5. What is its
molecular formula?
Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
B 2H 6
B2H6 is one example of this class of compounds.
We need to do an EXPERIMENT to find the
MOLAR MASS.
Here the experiment gives 53.3 g/mol
Compare with the mass of B2H5 = 26.66 g/unit
Find the ratio of these masses.
53.3 g/mol
2 units of B2H5
=
26.66 g/unit of B2H5
1 mol
Multiply all of the subscripts by the ratio and
obtain the molecular formula:
Molecular formula = B4H10
Sample Problem
Example : A certain compound has the empirical
formula C2H4O. Its molar mass is about 90 g/mol.
What is the molecular formula?
Sample Problem
Example: A hydrate of magnesium iodide has
the formula MgI2 ∙ X H2O. To determine the
value of X, a student heats a sample of the
hydrate until all the water is gone. A 1.628 g
sample of hydrate is heated to constant mass
of 1.072 g. What is the value of X?
Writing and Balancing
Chemical Equations
• All chemical reactions have two parts:
• Reactants - the substances you start with
(on left side of arrow)
• Products - the substances you end up
with (on right side of arrow)
• The reactants turn into the products.
Reactants  Products
In a chemical reaction…
• The way atoms are joined is changed.
• Atoms aren’t created or destroyed; they just
combine together in new ways.
• Can be described using sentences, symbols or
word equations:
Example:
Copper reacts with chlorine to form copper (II) chloride.
Copper + chlorine  copper (II) chloride
Cu + Cl2 CuCl2
Symbols Used in Equations
• The arrow separates the reactants from the
products; means “reacts” or “yields”
• The plus sign = “and”
• Subscripts are used to describe the
number of atoms in a FORMULA.
• Coefficients are used to describe the
number of molecules or formula units in
the REACTION. They are the only things
changed when balancing a reaction.
States of Matter
• Solid – (s) after the formula
– Precipitate -- a solid formed in a reaction
• Gas--(g) after the formula
• Liquid—(l) after the formula
• Aqueous – (aq) after the formula dissolved in water.
States of Matter
 used after a product indicates a gas
oSame as writing (g)
 used after a product indicates a solid
or precipitate
oSame as writing (s)
Other Symbols used in Equations
•
•
indicates a reversible reaction

heat
  ,   
show that
heat is supplied to the reaction
Pt
•  
is used to indicate a catalyst
used or supplied, in this case, platinum.
Chemical Equations
Example: consider the formation of water
H2(g) + O2(g)  H2O(g)
Law of Conservation of Mass
must be obeyed …
therefore, equations must be balanced.
EOS
Balancing Equations
Chemical “bookkeeping” of atoms
involved in the reaction:
H2(g) + O2(g)  H2O(g)
H–2
O–2
Reactants
H–2 O–1
Products
Note the imbalance in oxygen atoms
COEFFICIENTS must be added so reactant
atoms EQUAL product atoms!
Hints & Tips for Balancing Equations
• Take one element at a time, working from left to
right except for H and O. Save H for next to last
and O for last.
• IF EVERYTHING BALANCES EXCEPT FOR O,
and there is no way to balance O with a whole
number, double all the coefficients and try again.
(Because O is a diatomic element)
• (Shortcut) polyatomic ions that appear on both
sides of the equation can be balanced as
independent units!
Balancing Equations Practice
Balance the following chemical equation using
the appropriate coefficients:
____ Al(s) + _____ Br2 (l)  _____ Al2Br6 (s)
Balancing Equations Practice
Balance the following chemical equation using
the appropriate coefficients:
____ Na3PO4 + ____ Fe2O3  ____ Na2O + ____ FePO4
Types of Reactions
• Synthesis (combination) reaction
• Decomposition reaction
• Single replacement reaction
• Double replacement reaction
• Combustion reaction
Types of Reactions
Synthesis or Combination
Equation in Symbols: A + B  AB
Sample Equation:
2Cu (s) + O2 (g)  2 CuO (s)
Predicting Products:
Elements Compounds OR
Compounds 
More Complex Compounds
Types of Reactions
Decomposition
Equation in Symbols: AB  A + B
Sample Equation:
2 CuO (s)  2Cu (s) + O2 (g)
Predicting Products:
Compounds  Elements OR
More Complex Compounds 
Compounds
Types of Reactions
Single Replacement
Equation in Symbols: A + BC  AB + C
• Metal replacing metal
• Nonmetal replacing nonmetal
Sample Equation:
Mg (s) + CuCl2 (aq)  Cu (s) + MgCl2 (aq)
Predicting Products:
Cations replace cations
(can also have anions replacing anions)
Types of Reactions
Double Replacement – 2 ionic compounds
Equation in Symbols: AX + BY  BX + AY
Sample Equation:
2AgNO3(aq) + CuCl2 (aq)  Cu (NO3)2 (aq) + 2AgCl (s)
Predicting Products:
Cations switch places; solid formed
(must be driving force)
Types of Reactions
Combustion
Equation in Symbols: CxHy + O2 CO2 + H2O
Sample Equation:
CH4(g) + O2 (g)  CO2 (g) + H2O (l)
Predicting Products:
Hydrocarbons react to form CO2 and H2O
Stoichiometric Equivalence and
Reaction Stoichiometry
CS2 + 3O2  CO2 + 2 SO2
Interpretation in terms of moles:
1 mole of CS2 + 3 moles of O2 form:
1 mole of CO2 + 2 moles of SO2
Stoichiometric Equivalence and
Reaction Stoichiometry
CS2 + 3O2  CO2 + 2 SO2
Conversion factors extracted from balanced
equation:
1 mole of CS2
3 moles of O2
3 moles of O2
1 mole of CO2
etc.
These ratios are called MOLE RATIOS!
Stoichiometric Equivalents
Coefficients from a balanced chemical equation show
molar equivalents of reactants and products
==> form conversion factors
2H2 + O2  2H2O
In the formation of water:
2 mol H2 = 1 mol O2
2 mol H2 = 2 mol H2O
1 mol O2 = 1 mol H2O
Example: Using the equation below, determine:
CS2 + 3O2  CO2 + 2SO2
•
the number of moles of oxygen required to react
with 1.38 mol of carbon disulfide
•
the number of moles of SO2 produced from 1.38
moles of carbon disulfide.
Example 3.13
For 2NH3 + H2SO4  (NH4)2SO4 determine:
a) the mass of product possible when 1.43 mol of
NH3 are reacted with an excess of sulfuric acid.
b) the mass of NH3 required to react completely with
35.00 g of sulfuric acid.
c) the mass of sulfuric acid required to form 1000
grams of product.
Stoichiometry Diagram
Volume
(liquids)
Volume
(liquids)
Known
Unknown
Substance A
Substance B
Mass
Volume
1 mole = 22.4 L @ STP
Mass
Use coefficients
from balanced
chemical equation
Mole
Mole
1 mole = 22.4 L @ STP
Volume
(gases)
(gases)
Particles
Particles
Example : How many milliliters of liquid water can be
produced by the combustion of 775 mL of octane with
oxygen? Assume that the volumes of the octane and
the water are measured at 20oC where the densities
are 0.7025 g/mL for octane and 0.9982 g/mL for water.
C8H18(l) + O2(g) 
CO2(g) +
H2O(l)
Reaction Yields
Theoretical yield – predicted
amount of product formed
from the limiting reagent,
based only on the
stoichiometry of the reaction
2H2(g) + O2(g)  2 H2O(g)
If all worked
perfectly ...
Example: 1 mol H2 will
produce 1 mol of water
Actual yield – amount of product produced
In practice, actual < theoretical: errors, poor technique, etc. ...
Limiting Reactants (Reagents)
Chemical reactant that is completely consumed in
a reaction and therefore limits the quantity of
product formed.
**Depends on stoichiometry of reaction
Excess Reactant = Reactant left over when
limiting reactant is used up
How many meals can be made from 105
sandwiches, 202 cookies, and 107 oranges?
1 meal
105 sandwiches x
 105 meals
1 sandwich
1 meal
202 cookies x
 101 meals
2 cookies
1 meal
107 oranges x
 107 meals
1orange
Cookies limit the
total number of whole
meals with excess
sandwiches and
oranges
Steps for Determining Limiting Reactant
1. Write a balanced equation.
2. Take first reactant, calculate theoretical
yield of desired product (in grams).
3. Repeat #2 for second reactant.
4. Compare results. Whichever reactant
gives the LEAST amount of the
product is the limiting reactant and
determines the theoretical yield. The
other is in excess.
Sample Problem
CS2(l) + 3 O2(g)  CO2(g) + 2 SO2(g)
Determine the theoretical yield of product (CO2)
in grams if one starts with 1.20 mol of CS2 and
3.83 mol O2
Sample Problem
Example: For the following reaction, determine
the theoretical yield of product (CO2) if one
starts with 105 g of CS2 and 145 g of O2
CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)
Percent Yield
Reaction yields are expressed as a ratio in the
form of a percentage:
actual yield
percent yield =
x100
theoretical yield
EOS
Example
For the reaction, determine the theoretical yield if one
starts with 1.20 g of antimony and 2.40 g of iodine.
2 Sb(s) + 3 I2(s)  2 SbI3(s)
Determine theoretical yield first:
Example
For the reaction, determine the theoretical yield if one
starts with 1.20 g of antimony and 2.40 g of iodine.
2 Sb(s) + 3 I2(s)  2 SbI3(s)
Theoretical Yield = 3.17 g SbI3
If 3.00 g of product are actually formed, what is the percent
yield?