Announcements
HOUR EXAM 1
July 18 6-7:30PM
Chapter 3
--Want me to do recitation again?
--Skip Combustion Analysis & Isomers (p.82-83 in
Principles of Chemistry Text)
See me if you donʼt understand!
Mass Relationships,
Stoichiometry and Chemical
Formulas
Chapter 3: 6, 10, 12, 14, 16, 18, 20, 23, 26, 28, 30, 32, 33, 38, 40, 41, 43,
45, 49, 51, 65, 66, 69, 71, 73, 75, 83, 85, 89, 93, 95, 3.119 (Principles of
Chemistry)
Chapter 3: 3.7, 3.11, 3.13, 3.15, 3.17, 3.19, 3.26, 3.29, 3.34,3.36, 3.38,
3.42, 3.45, 3.52, 3.56, 3.62, 3.64, 3.66, 3.72, 3.74, 3.82, 3.88, 3.93, 3.95,
3.97, 3.99, 3.101,3.114, 3.118, 3.119 (4th Chemistry Molecular Nature of
Matter)
Learning Objectives
1. Understand
mass.
Another stoichiometry example from Silberberg
relative atomic masses, average isotopic
2. Connect
the dots between amu & grams & the periodic
table via the mole.
3. Compute
a molecular & molar mass of a substance
from a formula.
factor label method to convert between
grams<=>moles<=>molecules
Copper metal is obtained from copper(I) sulfide
containing ores in multistep-extractive process. After
grinding the ore into fine rocks, the first step is to
heat it strongly with oxygen gas to form powdered
cuprous oxide and gaseous sulfur dioxide.
(0) Write a balanced chemical equation for this process
4. Using
(a) How many moles of molecular oxygen are required to fully
roast 10.0 mol of copper(I) sulfide?
5. %
(b) How many grams of sulfur dioxide are formed when 10.0
mol of copper(I) sulfide is roasted?
6.
Mass To Empirical Formula and Vis-versa
Balancing equations and mastering stoichiometry
7. Limiting
Reagent, Yields, Solution Stoichiometry
(0) Write a balanced chemical equation for this process
Cu2S(s) + O2(g)
2Cu2S(s) + 3O2(g)
Cu2O(s) + SO2(g)
unbalanced
2Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of
copper(I) sulfide?
3 mol O2 = 15.0 mol O
mol O2 = ? = 10.0 mol Cu2S x
2
2 mol Cu2S
(c) How many kilograms of oxygen are required to form 2.86
kg of copper(I) oxide?
2Cu2S(s) + 3O2(g)
(c) How many kilograms of oxygen are required to form 2.86 kg
of copper(I) oxide?
kg O2 = 2.86 kg Cu2O x
103g Cu2O 1 mol Cu2O
x 3mol O2 x
x
kg Cu2O
143.10g Cu2O 2mol Cu2O
kg O2 = 0.959 =
(b) How many grams of sulfur dioxide are formed when 10.0 mol
of copper(I) sulfide is roasted?
g SO2 = 10.0 mol Cu2S x
2mol SO2 x 64.07g SO2
= 641g SO2
2mol Cu2S 1 mol SO2
2Cu2O(s) + 2SO2(g)
1 kg O2
32.00g O2
x
1000 g O2
1 mol O2
Suppose two reactants---neither in excess. Which
reactant limits how much can be produced? (chemists
call it the limiting reagent....a must know for Chem 7
students).
Is H2 or O2 the limiting reagent?
Do You Understand Limiting Reagents?
Phosphorus trichloride is a commercially important
compound used in the manufacture of pesticides. It
is made by the direct combination of phosphorus P4
and gaseous chlorine. Suppose 323 g of chlorine is
combined with 125 g P4. Determine the amount of
phosphorous trichloride that can be produced when
these reactants are combined.
NOTE YOU HAVE TWO REACTANTS AND YOU
HAVE TO DETERMINE WHICH ONE LIMITS THE
REACTION!
P4 (s) + Cl2 (g) ! PCl3 (l)
Step 1. Convert words to formulas
P4 (s) + 6Cl2 (g) ! 4PCl3 (l) Step 2. Balance
Step 3. Determine reactant that
produces the least product.
Do You Understand Limiting Reagents?
124.0 g of solid Al metal is reacted with 601.0 g of
iron(III) oxide to produce iron metal and aluminum
oxide. Calculate the mass of aluminum oxide
formed.
mol PCl3 = 323 g Cl2 ×
1 mol Cl2
4 mol PCl3
×
= 3.04mol PCl3
70.91 g Cl2
6 mol Cl2
1. Write a balanced equation for all problems
mol PCl3 = 125 g P4 ×
1 mol P4
4 mol PCl3
×
= 4.04mol PCl3
123.88 g P4
1 mol P4
2. Two reactant masses and “no excess” = limiting reagent.
Step 4. Cl2 is the limiting reagent and
determines the amount of PCl3
g PCl3 = 125 g P4 ×
1 mol P4
4 mol PCl3
137.32 g PCl3
×
×
= 417.5 g PCl3
123.88 g P4
6 molCl2
1 mol PCl3
124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
g Al
124 g Al x
Al2O3 + 2Fe
mol Al2SO3 produced
mol Al
1 mol Al
27.0 g Al
g Fe2O3
601 g Fe2O3 x
x
1 mol Al2O3
2 mol Al
mol Fe2O3
1 mol Fe2O3
160. g Fe2O3
x
x
101.96 g Al2O3
= 234 g Al2O3
1 mol Al2O3
mol Al2SO3 produced
1 mol Al2O3
101.96 g Al2O3
x
= 383 g Al2O3
1 mol Fe2O3
1 mol Al2O3
Al make the least and is the limiting reagent!
234 g Al2O3 can be produced
3. Work in moles (grams => moles => equation stoichiometry)
4. Determine maximum theoretical amount of product for both
reactants. The limiting reagent is the one that produces the
least.
Try it At Home:Answer is in your book!
Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4),
ignite on contact to form nitrogen gas and water
vapor. How many grams of nitrogen gas form when
1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are
mixed?
Try it At Home:
Hydrazine(N2H4) and dinitrogen tetraoxide(N2O4),
ignite on contact to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when 1.00 x 102
g of N2H4 and 2.00 x 102 g of N2O4 are mixed?
2 N2H4(l) + N2O4(l)
BALANCE!
mol N2H4
1.00 x 102g N2H4
2.00 x 102 g N2O4 X
3 mol N2
2 mol N2H4
1 mol N2O4
92.02 g N2O4
= 4.68 mol N2
= 2.17 mol N2O4
3 mol N2
mol N2 = 2.17 mol N2O4 X
1 mol N2O4
A + B
N2H4 is the limiting reactant
because it produces less
product, N2, than does N2O4.
C
(main product)
(reactants)
D
(side products)
= 6.51 mol N2
4.68mol N2
28.02g N2
mol N2
= 131g N2
The % Yield of a chemical reaction is the ratio of
product mass obtained in the lab over the
theoretical (i.e. calculated) X 100.
from lab data
% Yield =
Real reactions have side-reactions and reduce the
amount of product obtained vs the “theoretical
amount”.
= 3.12mol N2H4
32.05g N2H4
mol N2 = ? = 3.12 mol N2H4 X
3 N2(g) + 4 H2O(l)
When we do chemical reactions calculations in the
class, they are “ideal and yield 100% product. This is
the theoretical value”.
Actual Amount
x 100
Theoretical Amount
from limiting
Learning Check: Calculating Percent Yield
Silicon carbide (SiC) is made by allowing silicon
dioxide to react with powdered carbon (C) under high
temperatures. Carbon monoxide is formed as a byproduct. Suppose 100.0 kg of silicon dioxide is
processed in the lab and 51.4 kg of SiC is recovered
What is the percent yield of SiC from this process?
reagent calculation
Actual Amount amount of product obtained
from a actual reaction in the laboratory. It’s given in a word
problem.
Theoretical Amount is the amount of product that would
result if all the limiting reagent reacted.
Learning Check: Calculating Percent Yield
SiO2(s) + C(s)
SiO2(s) + 3C(s)
SiC(s) + CO(g) 1. Converts words to formulas
SiC(s) + 2CO(g)
2. Balance
3. We are given one reactant, we must assume excess of the other (C)
mol SiO2 = 102 kg SiO2 ×
1 mol SiO2
1 mol SiC
103 g SiO2
×
×
= 1664 mol SiC
1 kg SiO2
60.09 g SiO2
1 mol SiO2
kg SiC = 1664 mol SiCl ×
40.10 g SiC
1 kg SiC
×
= 66.73 kg SiC
1 mol SiC
1000 g SiC = 66.73 kg SiC
4. We get the actual from experiment and theoretical from calculation and
plug them into the yield equation (units of mass should be the same)
% Yield =
kg Actual
51.4 kg
× 100 =
× 100 =
= 77.0%
77.0%
kg Theoretical
66.73 kg
Notice how the word excess is missing
and the amount of the other reactant.
Must assume other reactant is excess!