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Chapter 3
Chemical Reactions and Reaction Stoichiometry
Atomic mass calculation:
Ex: Carbon occur as a mixtures of three isotopes in
nature. 98.89% 12C, 1.11% 13C, 0.01% 14C . Calculate
its atomic mass.
Atomic Mass of Carbon
=(.9889x12)+(.0111X13)+(.0001x14) (
=12.01 amu
2
Average Atomic Mass Practice
An element occurs in nature as a mixture of two
isotopes such as 7.50% with an atomic mass of
6.01521 grams and 92.50% with a mass of
7.016003 grams. Calculate the average atomic
mass of the element.
• The mole is a unit that allows us to count the
number of atoms or molecules in a substance
• The mole is also related to mass
Avogadro’s Number( Mole)
Avogadro’s number is the number of atoms in
12.01 grams of carbon.12.01 g sample of carbon
contains 6.02 x 1023 carbon atoms.
Mole (mol) is a unit of measure for an amount of a
chemical substance.
Mole is Avogadro’s number of particles.
1 mol = Avogadro’s number = 6.02 x 1023 particles
Mole Calculation Steps
1 mol = 6.02 x 1023 particles(molecules)
1 mol = GFM( gram formula mass)
First we need to find the conversion factor that
allows us to convert between mass and moles
Molecules
Moles (mol)
Mass (g)
The conversion factor is called:
gram formula mass (GFM) or molecular mass
grams
The unit for gram formula mass is:
mol
Samples of One Mole Quantities
=
6.02 x 1023 C atoms
1 mole Al =
6.02 x 1023 Al atoms
1 mole S
6.02 x 1023 S atoms
1 mole C
=
1 mole H2O =
6.02 x 1023 H2O molecules
1 mole CCl4 =
6.02 x 1023 CCl4 molecules
Calculating Molar Mass/ molecular
mass/ gram formula mass(GFM)
The molar mass of a substance is the sum of the
molar masses of each element.
Ex: What is the gfm of iron(II) nitrate?
The sum of the atomic masses is as follows:
55.85+ 2(14.01) + 6(16.00 )= 179.87 amu
The molar mass for Fe(NO3)2 = 179.87 g/mol
1 mol Fe(NO3)2 = 179.87g
Mole calculation Steps
1. Write down given and find
2. Choose correct conversion factor:
– “Avogadro’s #” or “GFM”( gram formula mass)
3. Multiply given by correct conversion factor
Mole Calculations
• How many potassium atoms are in 3.6740 mol
of K?
3.6740 mol K x
6.02 x 1023 atoms
1 mol K
= 2.2117 x 1024 atoms
Mole Calculations
Avogadro’s number is used to convert
particles of a substance to moles.
Ex:-How many moles of CO2 are in
2.50 x 1024 molecules CO2?
2.50 x 1024 molecules CO2 x 1 mole CO2
6.02 x 1023 molecules CO2
= 4.15 mole CO2
Using Avogadro’s Number
Avogadro’s number is used to convert
moles of a substance to particles.
How many Cu atoms are in 0.50 mole Cu?
0.50 mole Cu x 6.02 x 1023 Cu atoms
1 mole Cu
= 3.0 x 1023 Cu atoms
Moles Practice
How many atoms are in 5.00 moles of aluminum
nitrate?
Moles  Molecules  Atoms
Moles Practice
How many moles of oxygen gas (O2) are in
7.5 x 1024 molecules?
Percent Composition
Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound
Ex: What is the % of iron in iron(II) nitrate?
% Fe = 55.85x100/179.87 = 31.05 %
16
Mass Percent Practice
What is the percentage of sodium in sodium
sulfate?
Empirical Formula(EF)
• The empirical formula of a compound is the
simplest whole number ratio of atoms of each
element in a molecule.
• The molecular formula of benzene is C6H6.
EF= CH.
• The molecular formula of octane is C8H18.
EF= C4H9.
Ex:
1) Which of these is NOT empirical formula:
•
•
•
•
NH
CH3O2
N6H9O3
CH8O12
2) Which of these is an empirical formula?
• N2H5O7
• N2H4O8
• C2H4
• C2O4
Empirical Formula Calculations
What is the empirical formula for a compound
that is 60.3% Mg and 39.7% O?
Element
Mass (g) or
Mass %
g or %_____
atomic mass
mol
Simple Ratio
(divide by
smallest #)
Mg
60.3
60.3/24.31
2.480
1
O
39.7
39.7/16
2.481
1
Answer: EF = MgO
Empirical Formula Practice
Aspirin contains 60.0% carbon, 4.48% hydrogen,
and 35.5% oxygen. What is its empirical
formula?
Molecular Formula (MF)
The actual molecular formula is some multiple
of the empirical formula, ( EF)n.
Molecular Formulas
Remember, these give the actual number
of atoms, not just the ratio.
1. Determine the empirical formula.
2. Find the empirical formula mass
3. Divide the molecular mass by empirical
formula mass & multiply your empirical
formula by this number
Molecular Formula Calculations
Ex: The empirical formula is SO2 and the molecular mass of
the compound was experimentally determined to be
128.128g/mol. What is the molecular formula?
EF mass =
S x 1 = 32.066 x 1 = 32.066
O x 2 = 15.999 x 2 = 31.998
EF mass = 64.064
n= MF mass/ EF mass = 128.128/64.064 = 2
Molecular Formula =( SO2 )2 = S2O4
Molecular Formula Practice
A compound has 49.3% carbon, 6.90% hydrogen,
and 43.8% oxygen. The molar mass of the
compound is 146 grams/mol. Determine the
empirical and molecular formulas.
Balancing chemical equations
Potassium iodide reacts with lead nitrate to
yield lead iodide and potassium nitrate.
2 KI + Pb(NO3)2  PbI2 +2 K(NO3)
Balancing Practice
1) Methane (CH4) reacts with oxygen to form
carbon dioxide and water.
2) Nitrogen gas reacts with hydrogen gas to
form nitrogen trihydride.
Stochiometry
2 NO(g) + O2(g) → 2 NO2(g)
In a balanced chemical equation “2 moles of NO
gas react with 1 mole of O2 gas to produce 2
moles of NO2 gas.”The coefficients indicate the
mole ratio, of reactants and products.
Stochiometry Practice
15.0 grams of copper (II) chloride react with 20.0 grams of sodium
nitrate to form copper(II) nitrate and sodium chloride.
a) How much sodium chloride can be formed?
b) What is the limiting reactant for the reaction?
c) If 11.3 g of sodium chloride are formed in the reaction, what is
the percentage yield of the reaction?
Stochiometry Practice
Sodium reacts with water to form sodium
hydroxide and hydrogen gas. How many moles
of hydrogen gas are produced from 5.4 grams
of sodium?
Stochiometry Practice
Copper (II) chloride reacts with aluminum to
form aluminum chloride and copper. How
many grams of aluminum would it react to
form 145 grams of copper?
Percent Yield
When you perform a laboratory experiment,
the amount of product collected is the
actual yield. The amount of product
calculated from a limiting reactant problem
is the theoretical yield.
The percent yield is the amount of the actual
yield compared to the theoretical yield.
actual yield
x 100 % = percent yield
theoretical yield
Percent Yield & Limiting Reactant Practice
Ex: 54.0 grams of silver chloride combine with
32.0 grams of magnesium nitrate to produce
silver nitrate.
a) How many grams of magnesium chloride will
be produced?
b) Which reactant is the limiting reactant?
c) After the reaction, you only obtained 14.35 g
of magnesium chloride. What is percent
yield?