Transcript Chapters 10 and 11 Notes

Chemical Quantities and
Chemical Reactions
Chemical Quantities
Measuring Matter
• Often measure the amount of something
by one of three different methods:
–count
–mass
–volume
Measuring Matter
• By count:
- Example: 1 dozen apples = 12 apples
• By mass:
- Example: 1 dozen apples = 2.0 kg apples
• By volume:
- Example: 1 dozen apples = 0.20 bushel
apples
Measuring Matter
• Can convert among units:
• Example:
1 dozen apples
12 apples
1 dozen apples
0.20 bushel apples
2.0 kg apples
1 dozen apples
Finding Mass from a Count
• What is the mass of 90 average-sized
apples if 1 dozen of the apples has a mass
of 2.0 kg?
• # apples dozens apples mass apples
• Conversion factors:
1 dozen apples
12 apples
2.0 kg apples
1 dozen apples
Finding Mass from a Count
90 apples X 1 dozen apples X 2.0 kg apples
12 apples
1 dozen apples
= 15 kg apples
What is a Mole?
• mole - (mol) 6.02  1023 representative
particles of a substance
- SI unit for measuring the amount of a
substance
-The number 6.02  1023 is called
Avogadro’s number (Italian scientist)
- helped clarify difference between
atoms and molecules
What is a Mole?
• Mole of any substance contains
Avogadro’s number of representative
particles, or 6.02  1023 representative
particles
• Representative particle - refers to
species present in a substance
- usually atoms, molecules, or formula
units
What is a Mole?
Converting Number of Atoms to
Moles
• Magnesium is a light metal used in the
manufacture of aircraft, automobile
wheels, tools, and garden furniture. How
many moles of magnesium is 1.25 X 1023
atoms of magnesium?
• Conversion factors:
1 mol Mg
6.02 X 10 23 atoms Mg
Converting Number of Atoms to
Moles
1.25 X 10 23 atoms Mg X
1 mol Mg
6.02 X 1023 atoms Mg
= 2.08 X 10-1 mol Mg = 0.208 mol Mg
What is a Mole?
•Converting Moles to Number of Particles
Converting Moles to Number of
Atoms
• Propane is a gas used for cooking and
heating. How many atoms are in 2.12 mol
of propane (C3H8)?
• Conversion factors:
6.02 X 1023 molecules C3H8
1 mol C3H8
11 atoms
1 molecule C3H8
Converting Moles to Number of
Atoms
• Example Continued:
2.12 mol C3H8 X 6.02 X 1023 molecules C3H8
1 mol C3H8
X
11 atoms
1 molecule C3H8
= 1.4039 X 1025 atoms = 1.40 X 1025 atoms
The Mass of a Mole of an
Element
• Atomic mass of an element expressed in
grams is the mass of a mole of the element
• Molar mass- mass of a mole of an
element
- Examples:
molar mass of Carbon = 12.0 g
molar mass of Hydrogen = 1.0 g
The Mass of a Mole of an Element
The Mass of a Mole of an
Element
• One molar mass of carbon, sulfur,
mercury, and iron are shown
The Mass of a Mole of an Element
• You must know formula of compound to
calculate the mass of a mole of an
element
• To calculate the molar mass of a
compound:
- find the number of grams of each
element in one mole of the compound
- add the masses of the elements in the
compound
Finding the Molar Mass of a
Compound
• The decomposition of hydrogen peroxide
(H2O2) provides sufficient energy to launch
a rocket. What is the molar mass of
hydrogen peroxide?
• 1 mole H = 1.0 g H
• 1 mol O = 16.0 g O
Finding the Molar Mass of a
Compound
• Example Continued:
2 mol H X 1.0 g H = 2.0 g H
1 mol H
2 mol O X 16.0 g O = 32.0 g O
1 mol O
Molar mass of H2O2 = 2.0 g + 32.0 g = 34.0 g
The Mole-Mass Relationship
• It’s very important to use the formula of the
substance when calculating molar mass
• Use molar mass of an element or compound
to convert between the mass of a substance
and the moles of a substance
Converting Moles to Mass
• The aluminum satellite dishes are resistant
to corrosion because aluminum reacts with
oxygen in the air to form a coating of
aluminum oxide (Al2O3). This tough,
resistant coating prevents any further
corrosion. What is the mass of 9.45 mol of
aluminum oxide?
• Moles = 9.45 mol Al2O3
• 1 mol Al2O3 = 102.0 g Al2O3
Converting Moles to Mass
• Example Continued:
mass = 9.45 mol Al2O3 X 102.0 g Al2O3
1 mol Al2O3
= 964 g Al2O3
Converting Mass to Moles
• When iron is exposed to air, it corrodes to
form red-brown rust. Rust is iron(III) oxide
(Fe2O3). How many moles of iron(III)
oxide are contained in 92.2 g of pure
Fe2O3?
• Mass = 92.2 g Fe2O3
• 1 mol Fe2O3 = 159.6 g Fe2O3
Converting Mass to Moles
• Example Continued:
Moles = 92.2 g Fe2O3 X 1 mol Fe2O3
159.6 g Fe2O3
= 0.578 mol Fe2O3
Mole-Volume Relationship
• Avogadro’s hypothesis – equal volumes
of gases at same temperature and
pressure contain equal numbers of
particles
Mole–Volume Relationship
• Volume of a gas varies with temperature
and pressure
- Example: balloon filled with helium
shrinks when outside on a cold day
- The volume of a gas is usually
measured at a standard temperature
and pressure
•Standard temperature and pressure
(STP) - means a temperature of 0°C and a
pressure of 101.3 kPa, or 1 atmosphere
(atm)
Mole-Volume Relationship
• At STP, 1 mol or, 6.02  1023 representative
particles, of any gas occupies a volume of
22.4 L
• Molar volume – quantity of 22.4 L
Calculating the Volume of Gas at
STP
• Sulfur dioxide (SO2) is a gas produced by
burning coal. It is an air pollutant and one
of the causes of acid rain. Determine the
volume, in liters of 0.60 mol SO2 gas at
STP.
• Moles = 0.60 mol SO2
• 1 mol SO2 = 22.4 L SO2
• Conversion factor:
22.4 L SO2
1 mol SO2
Calculating the Volume of a Gas at
STP
• Example Continued:
Volume = 0.60 mol SO2 X 22.4 L SO2
1 mol SO2
= 13 L SO2
The Mole–Volume Relationship
•Calculating Molar Mass from Density:
Calculating Molar Mass of a Gas at
STP
• The density of a gaseous compound
containing carbon and oxygen is found to
be 1.964 g/L at STP. What is the molar
mass of the compound?
• Density = 1.964 g/L
• 1 mol (gas at STP) = 22.4 L
Molar mass = 1.964 g X 22.4 L
1L
1 mol
= 44.0 g/mol
The Mole Road Map
The Mole Road Map
The Mole Road Map
The Mole Road Map
The Percent Composition of a
Compound
• Percent composition- percent by mass of
each element in the compound
• Percent by mass of an element in a
compound is the number of grams of the
element divided by the mass in grams of the
compound, multiplied by 100%
Percent Composition
Calculating Percent Composition
from Mass Data
• When a 13.60-g sample of a compound
containing only magnesium and oxygen is
decomposed, 5.40 g of oxygen is
obtained. What is the percent composition
of this compound?
• Mass = 13.60 g
• Mass of oxygen = 5.40 g
• Mass of magnesium = 13.60g – 5.40 g =
8.20 g Mg
Calculating Percent Composition
from Mass Data
• Example Continued:
• % Mg = 8.20 g X 100% = 60.3%
13.60 g
• % O = 5.40 g X 100% = 39.7%
13.60 g
• 60.3% + 39.7% = 100%
The Percent Composition of a
Compound
•Percent Composition from the Chemical
Formula
Calculating Percent Composition
from a Formula
• Propane (C3H8), the fuel commonly used
in gas grills, is one of the compounds
obtained from petroleum. Calculate the
percent composition of propane.
• Mass of C in 1 mol C3H8 = 36.0g
• Mass of H in 1 mole C3H8 = 8.0g
• Molar mass of C3H8 = 44.0g/mol
Calculating Percent Composition
from a Formula
• Example Continued:
• % C = 36.0 g X 100% = 81.8%
44.0g
• % H = 8.0g X 100% = 18%
44.0g
Add up to 100% when expressed as 2
significant figures
The Percent Composition of a
Compound
•Percent Composition as a Conversion
Factor
–You can use percent composition to
calculate the number of grams of any
element in a specific mass of a compound.
The Percent Composition of a
Compound
•Propane (C3H8) is 81.8% carbon and 18%
hydrogen. You can calculate the mass of
carbon and the mass of hydrogen in an 82.0
g sample of C3H8.
Empirical Formulas
• Empirical formula- gives the lowest
whole-number ratio of the atoms of the
elements in a compound
–The empirical formula of a compound
shows the smallest whole-number ratio of
the atoms in the compound
Determining Empirical Formula
• A compound is analyzed and found to
contain 25.9% nitrogen and 74.1%
oxygen. What is the empirical formula of
the compound?
• Percent of nitrogen = 25.9% N
• Percent of oxygen = 74.1% O
Determining Empirical Formula
• Since percent means parts per 100, you
can assume that 100.0 g of the compound
contains 25.9 g N and 74.1 g O.
•
•
25.9 g N X 1 mol N = 1.85 mol N
14.0 g N
74.1 g O X 1 mol O = 4.63 mol O
16.0 g O
Determining Empirical Formula
• Divide each molar quantity by the smaller
number of moles:
1.85 mol N = 1 mol N
1.85
4.63 mol O = 2.50 mol O
1.85
Determining Empirical Formula
• Can’t have fraction : N1O2.5
• Obtain lowest whole-number ratio by
multiplying each part of the ratio by the
smallest whole number that will convert
both subscripts into whole numbers
1 mol N X 2 = 2 mol N
2.5 mol O X 2 = 5 mol O
Empirical Formula = N2O5
Empirical Formulas
• Ethyne (C2H2) is a gas used
in welder’s torches. Styrene
(C8H8) is used in making
polystyrene
•These two compounds of
carbon have the same
empirical formula (CH) but
different molecular formulas
Molecular Formulas
• Molecular formula of a compound is either
the same as its experimentally determined
empirical formula, or it is a simple wholenumber multiple of its empirical formula
- must know compound’s molar mass to
determine molecular formulas
Molecular Formulas
Methanal, ethanoic
acid, and glucose all
have the same
empirical formula—
CH2O.
Molecular Formulas
Finding Molecular Formula
• Calculate the molecular formula of a
compound whose molar mass is 60.0
g/mol and empirical formula is CH4N.
• Empirical formula = CH4N
• Molar mass = 60.0 g/mol
• Empirical formula mass = 30.0 g/mol
• 60.0/30.0 = 2 (multiply empirical by 2)
• Molecular formula = C2H8N2
Chemical Reactions
Writing Chemical Equations
•Word Equations
–To write a word equation:
• write the names of the reactants to
the left of the arrow separated by plus
signs
• write the names of the products to
the right of the arrow, also separated
by plus signs
•Reactant + Reactant  Product + Product
Writing Chemical Equations
•Methane + Oxygen  Carbon dioxide + Water
Writing Chemical Equations
•iron + oxygen  iron(III) oxide
Writing Chemical Equations
•Hydrogen Peroxide  Water and Oxygen
Writing Chemical Equations
•Chemical Equations
– Chemical equation- is a representation
of a chemical reaction
• the formulas of the reactants (on the left)
are connected by an arrow with the
formulas of the products (on the right)
Writing Chemical Equations
• Skeleton equation- a chemical equation
that does not indicate the relative amounts
of the reactants and products
•Here is the equation for rusting:
Fe + O2  Fe2O3
Writing Chemical Equations
• You can indicate the physical states of
substances by using symbols in a
chemical reaction:
- use (s) for a solid
- use (l) for a liquid
- use (g) for a gas
- use (aq) for aqueous or a substance
dissolved in water
Fe(s) + O2(g)
Fe2O3(s)
Writing Chemical Equations
Writing Chemical Equations
• Catalyst- is a substance that speeds up
the reaction but is not used up in the
reaction
•Without Catalyst
With Catalyst
Writing Chemical Equations
• A catalyst is neither a reactant or product
- its formula is written above the arrow in a
chemical equation
Example:
H2O2(aq)
MnO2
H2O(l) + O2(g)
Writing a Skeleton Equation
• Hydrochloric acid and solid sodium
hydrogen carbonate are reacted together.
The products formed are aqueous sodium
chloride, water, and carbon dioxide gas.
Write a skeleton equation for this reaction.
• Solid sodium hydrogen carbonate:
NaHCO3(s)
• Hydrochloric acid: HCl(aq)
• Water: H2O(l)
• Carbon dioxide gas: CO2(g)
Writing a Skeleton Equation
• Example Continued:
NaHCO3(s) + HCl(aq)
NaCl(aq) + H2O(l)
+ CO2(g)
Balancing Chemical Equations
•This is a balanced equation for making a
bicycle:
• Coefficients—small whole numbers that
are placed in front of the formulas in an
equation in order to balance it
Balancing Chemical Equations
•A chemical reaction is also described by a
balanced equation
• Balanced equation- each side of the
equation has the same number of atoms of
each element and mass is conserved
Balancing Chemical Equations
•To write a balanced chemical equation:
- write the skeleton equation
- use coefficients to balance the equation
so that it obeys the law of conservation of
mass
Writing a Balanced Chemical
Equation
• Hydrogen and oxygen react to form water.
The reaction releases enough energy to
launch a rocket. Write a balanced
equation for the reaction.
• Determine the correct formulas for all of
the reactants and products
– Hydrogen = H2
– Oxygen = O2
– Water = H2O
Writing a Balanced Chemical
Equation
• Write the skeleton equation:
H2(g) + O2(g)
H2O(l)
• Determine the number of atoms of each
element between the reactants and
products:
- reactants = 2 hydrogen, 2 oxygen
- products = 2 hydrogen, 1 oxygen
Writing a Balanced Chemical
Equation
• Balance the elements one at a time
• If no coefficient is written, it is assumed to
be 1
H2(g) + O2(g)
2H2O(l)
2H2(g) + O2(g)
2H2O(l)
Writing a Balanced Chemical
Equation
• Practice Problem:
• Balance the following:
Cu(s) + AgNO3(aq)
Cu(NO3)2(aq) + Ag(s)
Answer:
Cu(s) + 2AgNO3(aq)
2Ag(s)
Cu(NO3)2(aq) +
Classifying Reactions
•The five general types of reaction are:
-combination
-decomposition
-single-replacement
-double-replacement
-combustion
Classifying Reactions
• Combination reaction- is a chemical
change in which two or more substances
react to form a single new substance
•Example:
2Mg(s) + O2(g)
2MgO(s)
Writing Combination Reactions
• Copper and sulfur are reactants in a
combination reaction. Complete the
equation for the reaction:
Cu(s) + S(s)
?
Cu(s) + S(s)
CuS(s)
Classifying Reactions
• Decomposition reaction- is a chemical
change in which a single compound breaks
down into two or more simpler products
Decomposition Reaction
• Example:
2HgO(s)
2Hg(l) + O2(g)
Practice: Decomposition of water using
electrolysis
H2O(l) electricity ?
electricity
H2O(l)
H2(g) + O2(g)
electricity
2H2O(l)
2H2(g) + O2(g) (balanced)
Classifying Reactions
• Single-replacement reaction- is a
chemical change in which one element
replaces a second element in a compound
Single-Replacement Reaction
• Example:
2K(s) + 2H2O(l)
2KOH(aq) + H2(g)
Practice: Write a balanced equation for the
following:
Zn(s) + H2SO4(aq)
Zn(s) + H2SO4(aq)
?
ZnSO4(aq) + H2(g)
Classifying Reactions
• Activity series of
metals lists metals in
order of decreasing
reactivity
Classifying Reactions
• Double-replacement reaction- is a
chemical change involving an exchange of
positive ions between two compounds
Double-replacement reaction
• Example:
Na2S(aq) + Cd(NO3)2(aq)
CdS(s) +
2NaNO3(aq)
Practice: Write the equation for the following
double-replacement reaction:
• CaBr2(aq) + AgNO3(aq)
?
• CaBr2(aq) + AgNO3(aq)
AgBr(s) +
Ca(NO3)2(aq)
• CaBr2(aq) + 2AgNO3(aq)
2AgBr(s) +
Ca(NO3)2(aq)
Classifying Reactions
• Combustion reaction- is a chemical
change in which an element or a compound
reacts with oxygen, often producing energy
in the form of heat and light
Combustion Reactions
• Example: Combustion of gasoline
2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(l)
Practice: write the combustion of benzene
C6H6(l)
• C6H6(l) + O2(g)
• 2C6H6(l) + 15O2(g)
CO2(g) + H2O(g)
12CO2(g) + 6H2O(g)
Predicting the Products of a
Chemical Reaction
•The number of elements and/or
compounds reacting is a good indicator of
possible reaction type and thus possible
products
Net Ionic Equations
• Complete ionic equation is an equation
that shows dissolved ionic compounds as
dissociated free ions
Net Ionic Equations
• Spectator ion - ion that appears on both
sides of an equation and is not directly
involved in the reaction
• Net ionic equation- is an equation for a
reaction in solution that shows only those
particles that are directly involved in the
chemical change
- balanced with respect to both mass
and charge
Net Ionic Equations
•Sodium ions and nitrate ions are not
changed during the chemical reaction of
silver nitrate and sodium chloride so the net
ionic equation is:
Writing Net Ionic Equations
• Aqueous solutions of iron(III) chloride and
potassium hydroxide are mixed. A
precipitate of iron(III) hydroxide forms.
Identify the spectator ions and write a
balanced net ionic equation.
Fe3+(aq) + 3Cl-(aq) + 3K+(aq) + 3OH-(aq)
Fe(OH)3(s) + 3K+(aq) + 3Cl-(aq)
Spectators = K+ and Cl• Fe3+(aq) + 3OH-(aq)
Fe(OH)3(s)
Predicting the Formation of a
Precipitate
•You can predict the
formation of a
precipitate by using
the general rules for
solubility of ionic
compounds