Transcript Chapter 4

Chapter 3
Chemical Reactions and Earth’s
Composition
The Development of Earth
• Matter in the Universe condensed into
planets.
• The planets closer to the sun have
different chemical compositions than the
rest of the universe.
 Many of the volatile chemicals were lost
from these planets.
The Composition of Compounds
• The law of definite proportions states
that a specific chemical compound
obtained from any source always
contains the same proportion by mass
of its elements
• H+ + OH- ---> H2O
• 2 H2 + O2 ---> 2 H2O
The Composition of Compounds
• The law of multiple proportions states
that the masses of element Y that
combine with a fixed mass of elements
X to form two or more different
compounds are in the ratios of small
whole numbers.
• Examples: NO, NO2, N2O, N2O5, etc.
Chemical Equations
• Reactants ↔ Products
Chapter 4 Definitions
Chemical Reaction - A process in which substances are changed
into other substances through rearrangement, combination, or
separation of atoms.
Chemical Equation - A written representation of a chemical
reaction, showing the reactants and products, their physical states,
and the direction of the reaction.
Reactants - The starting material in a chemical reaction or
equation.
Products - The substances formed in a chemical reaction or
equation.
Physical States - solids (s), liquids (l), gases (g) and aqueous (aq).
Balanced Chemical Equation - A written representation of a
chemical reaction that gives the relative amounts of the reactants
and products, their physical states, and the direction of the
reaction.
Balanced Chemical Equations
1. The reactants appear on the left and the products appear on the
right. The adjoining arrow shows the direction of the reaction.
2. The phase of the reactant or product is written after the
chemical symbol and is in parentheses.
3. An integer precedes the chemical formula of each substance.
This number, known as the stoichiometric coefficient, is the
smallest integer that allows the equation to be balanced.
4. Matter and charge are conserved in balanced chemical
equations.
Chemical Reactions
Avogadro number and the mole.
NA = 6.0221367 E23 mol-1
A mole is the amount of a substance that contains as
many elementary particles (atoms, molecules, or
whatever) as there are in exactly 12 g of the carbon12 isotope.
1 mol = 6.0221367 E23 particles
Problem
Express the following estimates for the year 2010 in
nanomoles.
a. USA. 298. million
b. China. 1.34 billion
c: US Debt (to China): 3 Trillion
Mass ↔ moles ↔ molecules ↔ atoms
Problem 39
Aluminum, silicon, and oxygen form minerals known as
aluminosilicates. How many moles of aluminum are in
1.50 moles of:
a. pyrophyllite, Al2Si4O10(OH)2
b. Mica, KAl3Si3O10(OH)2
Problem 44. How many moles of O2- ions are in 0.55
mol of Aluminum oxide? What is their mass in grams?
Answers: 1.65 mol O2-, 26.4 grams
Conversions
• Converting between a number of particles
and an equivalent number of moles (or
vice versa) is a matter of dividing (or
multiplying) by Avogadro’s number.
Molar Mass
• The average mass of an atom of helium
is 4.003 amu, and the mass of a mole of
helium (6.022 x 1023 atoms of He) is
4.003 g.
• The molar mass (M) of helium is 4.003
g/mol.
Molar Mass
• A substance’s molar mass is the mass
in grams of one mole of the compound.
CO2 = 44.01 grams per mole
C
+
12.01
+
2O
2(16.00) = 44.01g/mol
Problem. Calculate the molar masses of the following:
a. sucrose, C12H22O11
Answer: 342.31
d. fructose, C6H12O6
Answer: 180.158
Mole Calculations
• How many moles of Ca atoms are
present in 20.0 g of calcium?
• How many Cu atoms are present in
15.0 g of copper?
Mole Calculations
•
How many grams are present in 3.40
moles of nitrogen gas (N2)?
•
How many molecules are present in
5.32 moles of chalk (CaCO3)?
A. How many oxygen atoms are present in
this sample?
Mole Calculations
The uranium used nuclear fuel exists in
nature in several minerals. Calculate
how many moles of uranium are found
in 100.0 grams of carnotite,
K2(UO2)2(VO4)2•3H2O.
ans: 902.176 g/mol; 0.4434 moles of U.
100 gram of K2(UO2)2(VO4)2•3H2O
Law of Conservation of Mass
• The law of conservation
of mass states that the
sum of the masses of the
reactants of a chemical
equation is equal to the
sum of the masses of the
products.
Chemical Change
• Chemical reactions follow the law of
conservation of mass.
Balanced Chemical Equations
• Chemical equations should be balanced
to follow the law of conservation of
mass.
 Total mass of each element on the
reactant side must equal the total mass of
each element on the product side.
 Total charge of reactant side must equal
the total charge of product side.
Combustion Reactions
• Reactions that occur between oxygen
(O2) and another element in a
compound.
 When the other compound is a
hydrocarbon, the products of complete
combustion are carbon dioxide and water
vapor.
• Hydrocarbons are molecular
compounds composed of only hydrogen
and carbon and are a class of organic
compounds.
Practice
Balance the following equations for the
following combustions reactions.
C3H8 + O2
C5H10 + O2
CO2 + H2O
CO2 + H2O
Stoichiometric Calculations
• Calculating the mass of a product from
the mass of a reactant requires:
 A balanced chemical reaction
 Molar mass of the reactant
 Molar mass of the product
Example
How much carbon dioxide would be formed if
10.0 grams of C5H12 were completely burned
in oxygen?
C5H12 + 8 O2 ---> 5 CO2 + 6 H2O
Percent Composition
• Mass percent of an element in a compound
Mass % = mass of element in compound x 100%
mass of compound
• Practice: Calculate the percent of iron in
iron(III) oxide, (Fe2O3).
Mass Percent and Empirical Formulas
Problem . A sample of an iron compound is 22.0% Fe,
50.2% oxygen, and 27.8% chlorine by mass. What is the
empirical formula of this compound.
Answer: FeO8Cl2
Simplified Carbon Cycle
Problem
Sodium carbonate (105.988 g/mol) reacts with
hydrochloric acid (36.461) to produce sodium
chloride, water, and carbon dioxide. How
much hydrochloric acid is required to produce
10.0 g of carbon dioxide (44.01g/mol)?
Combustion Analysis
CaHb + excess O2 ---> a CO2(g) + b/2 H2O
The percent of carbon and hydrogen in CaHb can be
determined from the mass of H2O and CO2 produced.
Percent Composition and
Empirical Formulas
1.
2.
3.
4.
5.
Assume there is 100 g of the sample, so the
percent composition will equal the number of grams
of each element.
Convert the grams of each element into the moles
of each element with their molar mass.
Divide the smallest number of moles of an element
into the moles of each element present.
Convert the fractional ratios for each element into
whole numbers by multiplying all the ratios by the
same number.
The resulting numbers are the subscripts for the
each element in the empirical formula.
Example
Asbestos was used for years as an insulating
material in buildings until prolonged exposure to
asbestos was demonstrated to cause lung
cancer. Asbestos is a mineral containing
magnesium, silicon, oxygen, and hydrogen.
One form of asbestos, chrysotile (520.27 g/mol),
has the composition 28.03% magnesium,
21.60% silicon, 1.16% hydrogen. Determine the
empirical formula of chrysotile.
Mass Spectrometry and
Molecular Mass
• All Mass spectrometers separate atoms
and molecules by first converting them
into ions and then separating those ions
based on the ratio of their masses to
their electric charges.
• Mass spectrometers are instruments
used to determine the mass of
substances.
Mass Spectrometer
Mass Spectra
Determining the Molecular
Formula
• The molecular formula can be
determined from the percent
composition and mass spectral data.
Example
A combustion analysis of an unknown
compound indicated that it is 92. 23% C
and 7.82% H. The mass spectrum
indicated the molar mass is 78 g/mol.
What is the molecular formula of this
unknown compound?
Limiting Reactants
During photosynthesis a reaction mixture of
carbon dioxide and water is converted to a
molecule of glucose.
Ham and Cheese as “limiting” reagents…
Limiting Reagents
• The limiting reactant is completely
consumed in the chemical reaction.
 The amount of product formed
depends on the amount of the limiting
reagent available.
Example
10.0 g of methane (CH4) is burned in 20.0
g of oxygen (O2) to produce carbon
dioxide (CO2) and water (H2O).
a. What is the limiting reactant?
b. How many grams of water will be
produced?
Percent Yield
• Theoretical Yield: the calculated
amount of product formed from a given
amount of reactant(s)
• Actual Yield: the measured amount of
product formed
Percent Yield =
Actual Yield
x 100%
Theoretical Yield
Percent Yield Problem
Aluminum burns in bromine liquid
producing aluminum bromide. In a
certain experiment, 6.0 g of aluminum
was reacted with an excess of bromine
to yield 50.3 g aluminum bromide.
Calculate the theoretical and percent
yields for this experiment.
The Empirical* molecular formula can
be directly calculated from percent
composition data
*based on experiment(s)
ChemTour: Avogadro’s Number
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This ChemTour provides a step-by-step explanation of how
to use Avogadro’s number to convert from moles to
molecules or atoms and vice versa.
ChemTour: Balancing Equations
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This ChemTour reviews the procedure for balancing the
mass of reactants on the left side of the equation against
the mass of products on the right side.
ChemTour: Carbon Cycle
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This ChemTour shows how carbon is cycled and stored in
Earth’s environment.
ChemTour: Percent Composition
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PC | Mac
This ChemTour provides a step-by-step explanation for
finding the percent composition of elements in a substance
from the molecular formula and for finding the empirical
formula from the percent composition data.
ChemTour: Limiting Reactant
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PC | Mac
Learn to identify the limiting reactant for a reaction and to
determine the maximum yield of products.
A balloon is filled with 0.2 mole of hydrogen (H2) gas
and 0.1 mole of oxygen (O2) gas. If the mixture in the
balloon is ignited, how many moles of water will
form?
A) 0.1 mole
Function of Water: Moles
B) 0.2 mole
C) 0.3 mole
Consider the following arguments for each answer and
vote again:
A. Since each water molecule (H2O) contains 1 oxygen
atom, 0.1 mole of oxygen should form 0.1 mole of
water.
B. One mole of H2O has 2 moles of hydrogen atoms and 1
mole of oxygen atoms, so 0.4 mole of hydrogen atoms
and 0.2 mole of oxygen atoms will form 0.2 mole of
H2O.
C. In a chemical reaction the number of moles must
always be conserved, so 0.3 mole of water (0.1 + 0.2)
will form.
Function of Water: Moles
A 1-liter balloon is filled with hydrogen (H2) and
oxygen (O2) such that the molar ratio of
hydrogen to oxygen is 2:1.
If the mixture in the balloon is ignited, approximately
how many grams of water will form? Note: A gas is
about 1000 times less dense than its corresponding
liquid.
A) ~1 milligram
Function of Water: Masses
B) ~1 gram
C) ~1 kilogram
Consider the following arguments for each answer and
vote again:
A. Gases are very light, so even if the reaction goes to
completion, only a milligram or so of water will form.
B. A 1-liter gas mixture of hydrogen and oxygen will
react to form ~1 liter of water vapor, which has a mass
of ~1 gram.
C. The balloon contains 1 liter of gas, so the reaction will
form ~1 liter of liquid water, which has a mass of ~1
kilogram.
Function of Water: Masses
A piece of elemental phosphorus is combusted in an
atmosphere of excess oxygen until it is completely
converted to an oxide of phosphorus, PxOY.
Elemental analysis shows the product to contain
43.6% phosphorus and 56.4% oxygen by mass. What
is the empirical formula of the phosphorus oxide?
A) PO
Combustion of Phosphorus
B) P2O3
C) P2O5
Consider the following arguments for each answer and
vote again:
A. Atoms of phosphorus and oxygen react one by one to
form a compound whose empirical formula is therefore
PO.
B. Since the product is ~40% phosphorus and ~60%
oxygen, the ratio of phosphorus to oxygen is ~2:3,
suggesting that the formula is P2O3.
C. Although the mass ratio is 4.36/5.64, the molar ratio is
1.4/3.5=2/5, so the empirical formula can only be P2O5
Combustion of Phosphorus
A 1-liter bulb is filled with pure
oxygen. Elemental phosphorus is
added to the bulb in increments of 2
milligrams, corresponding to the blue
points in the plots below.
Which plot correctly shows the accumulated amount
of P4O10 produced (y-axis) as a function of the mass
of phosphorus added (x-axis)?
A)
B)
Product Yield of Phosphorus Oxide
C)
Consider the following arguments for each answer and
vote again:
A. Until sufficient phosphorus is added to achieve the
correct molar ratio of phosphorus to oxygen (2:5), no
reaction will occur, after which P4O10 will form as
more phosphorus is added.
B. The amount of P4O10 product will accumulate as
phosphorus is added until all oxygen is used up, at
which point no more P4O10 can be produced.
C. As long as oxygen is present, the amount of P4O10 will
be constant. When the oxygen is depleted, the amount
of P4O10 will steadily decrease.
Product Yield of Phosphorus Oxide
A sample of oxygen (O2) gas is
isotopically enriched such that
50% of the atoms are 16O and
50% of the atoms are 18O.
What is the correct mass spectrum for this gas?
A)
Mass Spectrum of Oxygen
B)
C)
Consider the following arguments for each answer
and vote again:
A. Three isotopic species, 16O-16O, 16O-18O, and 18O18O, exist with equal probabilities, so there should be
three lines of equal intensity in the mass spectrum.
B. Three isotopic species, 16O-16O, 16O-18O , and 18O18O, exist, and 16O-18O has twice the probability
(16O-18O, 18O-16O) so there should be a 1:2:1
intensity pattern.
C. Two isotopic species, 16O-16O and 18O-18O, exist with
equal probabilities because molecules can only be
formed from the same isotopes, so 16O-18O cannot
exist.
Mass Spectrum of Oxygen
Consider the oxidation of 1 gram of a
metal bromide, MBr, in excess
MnO2/H2SO4
4 MBr + MnO2 + 2 H2SO4 →
2 M2SO4 + MnBr2 + 2 H2O + Br2
Which metal bromide will yield the largest quantity of Br2?
A) Potassium
Oxidation of Bromides
B) Rubidium
C) Cesium
Consider the following arguments for each answer
and vote again:
A:
Potassium has the lowest molar mass and so 1.0
gram of KBr will have the most moles and produce the
most Br2.
B:
Rubidium's molar mass is closer to bromine's than
either of the other two metals, so RbBr will produce the
most Br2.
C:
Cesium has the highest molar mass and so 1.0
gram of CsBr has the most metal and will produce the
most Br2.
Oxidation of Bromides
A compound contains hydrogen,
oxygen, carbon, and sulfur in the
mass ratios 1:2:3:4.
What is the empirical formula of the compound?
A) H8OC2S
B) HO16C12S32
Empirical Formula of H:O:C:S
C) HO2C3S4
Consider the following arguments for each answer and
vote again:
A. Although the mass ratios are 1:2:3:4, the molar ratios
are 1:0.125:0.25:0.125, so the empirical formula is
H8OC2S.
B. Although the mass ratios are 1:2:3:4, the ratios of the
relative atomic masses are 1:16:12:32,
so the
molecular formula should be HO16C12S32.
C. Since the molecule is 10% hydrogen, 20% oxygen,
30% carbon and 40% sulfur the empirical formula is
obviously HO2C3S4.
Empirical Formula of H:O:C:S
One mole of an unknown hydrocarbon is combusted
with an excess of oxygen to produce 88 grams of
carbon dioxide (44 g/mol) and 36 grams of water
vapor (18 g/mol).
What is the molecular formula of the hydrocarbon?
A) C2H2
B) CH2
Molecular Formula of a Hydrocarbon
C) C2H4
Consider the following arguments for each answer
and vote again:
A. Two moles of CO2 and 2 moles of H2O are formed
from 1 mole of CxHY, so the molecular formula must
be C2H2.
B. The products contain 2 moles of carbon atoms and 4
moles of hydrogen atoms, which is a ratio of 1:2, so
the molecular formula is simply CH2.
C. The products coming from 1 mole of CxHY molecules
contain 2 moles of carbon atoms and 4 moles of
hydrogen atoms, so the molecular formula is C2H4.
Molecular Formula of a Hydrocarbon