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SCH3U
Grade 11, University Preparation
Chemistry
Lesson 9 – The Mole
SCH3U – Chemistry
Lesson 9
Lesson 9 – The Mole
Learning Objectives:
•
•
•
•
Use appropriate terminology related to quantities in chemical reactions including the
mole, and atomic mass
Solve problems related to quantities in chemical reactions by performing calculations
involving quantities in moles, number of particles, and atomic mass
Explain the law of definite proportions
Describe the relationships between Avagadro’s number, the mole concept, and the
molar mass of any given substance
Introduction
When you read the ingredient list on the back of a package of food, have you ever
noticed how much of each ingredient is contained in a serving? We can compare the
quantity of sugar, fat, or vitamins and minerals between different brands as well. The
quantitative information helps us decide which product to select to suit our needs.
Quantities in chemical formulas offer the same kind of important information about the
composition and properties of compounds in a reaction. Take for example water, H2O(l),
and hydrogen peroxide H2O2(l). Both of these compounds contain hydrogen and oxygen.
The only difference between the two, is that there is one extra oxygen atom in the
hydrogen peroxide. Even though this difference seems very small, it results in
significant differences in the properties of both substances. Water for example is stable,
and can be stored easily for long periods of time. Hydrogen peroxide on the other hand
must be stored in a dark container because it is unstable and starts to decompose. It is
also toxic in high concentrations.
In this lesson, we will learn how to measure and communicate quantities when dealing
with entities as small as atoms, ions, and molecules.
Uncertainty in Measurements
There are two types of quantities that are used in science; exact values, and
measurements. Exact values include defines quantities such as 1m = 100cm and
counted values such as 5 test tubes in a rack. Measurements however, are not exact
because there is some uncertainty or error associated with every measurement.
The precision of measurements depends upon the gradations of the measuring device.
Precision may refer to the exactness of a measurement. It is possible for results to be
precise but not accurate. For example, a measurement of 12.74 cm is more precise
than a measurement of 127.4 cm because the first value was measured to hundredths
of a cm where as the second measurement was only measured to the tenth of a cm.
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SCH3U – Chemistry
Lesson 9
No matter how precise a measurement is, it still may not be accurate. Accuracy refers to
how close a given quantity is to an accepted or expected value.
Significant Digits and Certainty
A measurement can only be as accurate and precise as the instrument that produced it.
A scientist must be able to express the accuracy of a number, not just its numerical
value. We can determine the accuracy of a number by the number of significant figures
it contains.
Rules for determining significant digits
1.
All digits 1-9 inclusive are significant
Example: 129 has 3 significant figures
2.
Zeros sandwiched between significant digits are always significant
3.
Place holder zeros are not significant
Example: 0.0025 has 2 significant figures
4.
Trailing zeros in a number are significant only if the number contains a decimal
point
Example: 100.0 has 4 significant figures
100 has 1 significant figure
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SCH3U – Chemistry
5.
Lesson 9
Zeros following a decimal and a significant figure are significant
Example: 0.000470 has 3 significant figures
0.47000 has 5 significant figures
When multiplying and dividing, limit and round to the least number of significant figures
of any of the factors.
When adding and subtracting, limit and round your answer to the least number of
decimal places in any of the factors.
Practice using significant digits!
Using a calculator add 11.7 cm + 3.29 cm + 0.542 cm
Step 1: Add all values together using your calculator
15. 532 cm is the answer on the calculator
Step 2: Use the rules for significant digits to determine the correct answer
Since we added in our calculation we must limit and round our answer to the least
number of decimal places which would be 1 from 11.7 cm.
Therefore, the answer to the question is 15. 5 cm
Scientific Notation
In courses such as it is sometimes necessary to use huge numbers such as
26,890,000,000,000,000,000 (which happens to be the number of molecules of a gas in
a cubic meter). Using these large (or sometimes, extremely small) numbers can easily
lead to mistakes and the use of tons of paper! Scientific notation takes care of this.
Numbers in scientific notation look like the following examples: 4.16 x 10+b and 4.16 x
10-b. b is always a positive, real number. The 10+b tells us that the decimal point is b
places to the right of where it is shown. The 10-b tells us that the decimal point is b
places to the left of where it is shown.
0.00316 can be changed into 3.16 x 10-4
120, 000, 000, 000 can be changed into 1.2 x 1011
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SCH3U – Chemistry
Lesson 9
Proportions in Compounds
In January 1998, nearly double the average annual total of freezing rain fell in one week
in eastern Ontario and Quebec. This created the worst ice storm in decades. A study
undertaken by the Ministry of Health revealed a huge rise in carbon monoxide
poisonings associated with the ice storm. The most common sources of poisoning were
generators, and barbeques being used indoors in locations such as basements and
garages.
Carbon monoxide CO is similar to CO2 which is found in
our atmosphere and is harmless at low concentrations.
The only difference between the two compounds is one
extra oxygen atom. This difference is what causes the
radical differences in properties for both compounds.
The law of definite proportions states that a specific
compound always contains the same elements in definite
proportions by mass. This means the carbon dioxide will
always contain one atom of carbon and two atoms of
oxygen, while carbon monoxide will contain one atom
each of carbon and oxygen.
Isotopes and Average Atomic Mass
The mass of an atom is expressed in atomic mass units. Atomic mass units are relative
measures defined by the mass of Carbon 12. One atom of carbon 12 is assigned a
mass of 12µ. (1µ = 1/12 mass C12) The masses of all other atoms are defined by their
relationship to carbon 12!!!
Example: Oxygen has a mass that is 133% of the mass of C12
133%
× 12.000 μ = 16.0 μ
100%
Atoms that contain the same number of protons but a different number of neutrons are
called isotopes. Most elements are made up of two or more isotopes. Because all of
the atoms in a given element do not have the same number of neutrons, they do not
have the same mass.
Example: Magnesium is 79% Mg24, 10% Mg25, and 11% Mg26
The relative amount in which each isotope is present in an element is called the isotopic
abundance.
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SCH3U – Chemistry
Lesson 9
The average atomic mass of an element is the average of the masses of all the
element’s isotopes. It takes into account the abundance of each isotope within the
element.
6
Since the atomic unit is based on carbon-12, why does the periodic
table show a value of 12.01u!?!
C
The isotopes are not present in equal amounts! Carbon is 98.9%
C12, 1.1% C13 and 1X10-10% C14. The average atomic mass is close
to 12 (12.01) because carbon-12 is the most abundant.
12.01
Average Atomic Mass (µ)
The Mole
Chemists can use the average atomic mass to describe the average mass of an atom in
a large sample. Since atoms are very tiny, measurable amounts of substance contain
extremely large numbers of atoms. Chemists must choose a unit that is more
convenient than counting endless numbers of atoms.
The mole is often referred to as the chemist’s dozen. Just as a dozen represents the
number 12, a mole also represents a number. One mole (1 mol) of a substance
contains 6.02 X 1023 particles of a substance. This value is called the Avogadro
constant. Its symbol is NA.
The mole (symbol n) (unit mol) is the quantity that chemists use to measure atoms. It is
defined as the amount of substance that contains as many particles (atoms, molecules,
or formula units) as exactly 12g of carbon12.
1 mol of carbon contains 6.02 X 1023 atoms of C.
1 mol of sodium chloride contains 6.02 X 1023 formula units of NaCl.
1 mol of hydrofluoric acid contains 6.02 X 1023 molecules of HF.
Molar Mass
The Avogadro constant is the factor that converts the relative mass of individual atoms
or molecules to mole quantities expressed in grams.
One mole of an element has a mass expressed in grams numerically equivalent to the
element’s average atomic mass. The mass of one mole of a substance is called the
molar mass (symbol M). Molar mass is expressed in g/mol. The molar mass can be
determined by looking at the periodic table.
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SCH3U – Chemistry
Example:
Lesson 9
Zinc has an average atomic mass of 65.39µ
One mole of Zinc has a mass of 65.39g
Therefore, Zinc has a molar mass of 65.39 g/mol
While you can find the molar mass of an element just by looking at the periodic table,
you need to do some calculations to find the molar mass of a compound.
Practice finding the molar mass of a compound!!
What is the molar mass (the mass of one mole) of sodium hydroxide?
Step 1: Convert the compound into its formula
NaOH
Step 2: Use the periodic table to determine the masses of each element
Na: 22.99 g/mol
O:16.00 g/mol
H: 1.01 g/mol
Step 3: Add the masses together, and check significant digits
The answer is 40.00 g/mol, since the lowest number of decimal places is 2.
Support Questions
1.
Calculate the mass of a molecule of sugar C12H22O11.
2.
What is the molar mass of octane, C8H18, a component of gasoline?
Calculations Involving the Mole
The mole is used to help us “count” atoms and molecules. The relationship
between moles, number of particles and the Avogadro constant is:
N = n X NA
Where:
N = number of particles (atoms, ions, molecules)
n = number of moles
NA = Avogadro constant
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SCH3U – Chemistry
Lesson 9
Practice using N = n x NA!!
A sample contains 2.15 mol of hydrogen peroxide, H2O2 . How many molecules of
H2O2 are in this sample?
Step 1: Determine which formula to use, and if it needs to be re-arranged
N=?
N = n x NA
Step 2: Substitute all the given information into the equation
N = n × NA
N = ( 2.15 mol ) × ( 6.02 × 1023 )
n = 2.15 mol
= 1.2943 × 1024 molecules
NA = 6.02 × 1023
Step 3: Put the final answer into significant digits
Since we were multiplying, we must limit and round to the least number of significant
figures. This would be 3 significant figures due to the 2.15 given in the question.
Therefore the number of molecules in 2.15 mol of hydrogen peroxide is 1.29 x 1024.
Chemists can convert moles of a substance to mass of a substance
using the molar mass. The mathematical relationship is:
m=nXM
Where:
m = mass in grams
n = number of moles
M = molar mass in g/mol
Practice using m = n x M!!
A dilute solution of hydrogen peroxide, H2O2, in water is used as a disinfectant.
For a particular experiment, 3.50 moles of hydrogen peroxide are required. What
mass of hydrogen peroxide is this?
Step 1: Determine which formula to use, and if it needs to be re-arranged
m=?
m=nxM
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SCH3U – Chemistry
Lesson 9
Step 2: Substitute all the given information into the equation
m = n×N
n = 3.50 mol
M = (use the periodice table)
m = ( 3.50 mol ) × ( 34.02 g/mol )
= 2 (1.01) + 2 (16.00 )
= 119.07 g
= 34.02 g/mol
Step 3: Put the final answer into significant digits
Since we were multiplying, we must limit and round to the least number of significant
figures. This would be 3 significant figures due to the 3.50 given in the question.
Therefore the mass of hydrogen peroxide is 119 g.
Now that you have learned how the number of particles, number of moles, and mass of
a substance are related, you can convert from one value to another!
This type of calculation is usually a two-step process:
# Particles
6.02 X 1023
Moles (mol)
Molar
Mass
Mass (g)
Another Practice Problem!!
What is the mass of 3.45 X 1024 molecules of copper (II) sulphate?
Step 1: Determine which formula to use, and if it needs to be re-arranged
m=?
but since we have number of molecules in the question, we must
also use the formula N = n x NA
m=nxM
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SCH3U – Chemistry
Lesson 9
Step 2: Substitute all the given information into the equations
m = n×N
m=?
n=?
N = n × NA
M = CuSO4
n=?
N = 3.45 × 1024
= 63.55 + 32.06 + 4 (16.00 )
NA = 6.02 × 1023
= 159.61 g/mol
Step 3: Complete the equation that has the most information first
n=
N = n × NA
N
NA
3.45 × 1024
6.02 × 1023
= 5.730897 mol
=
Step 4: Complete the second equation to determine the final answer, don’t forget to
check significant digits
m = ( 5.730897 mol ) × (159.61 g/mol )
m = n ×M
= 914.708 g
= 915 g
Therefore, there are 915 g of copper(II)sulphate.
Support Questions
3.
Calculate the molar mass of ethanol (C2H5OH).
4.
What is the mass of 0.90 mol of sodium?
5.
How many moles are there in 18.9 g of carbon?
6.
What is the mass of 5.34 X 1023 molecules of sodium hydrogen carbonate?
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SCH3U – Chemistry
Lesson 9
Key Question #9
1.
A recipe for sweet and sour sauce calls for:
500 g water
200 g sugar (C12H22O11)
25 g vinegar (HC2H3O2)
15 g citric acid (C6H8O7)
5 g salt (NaCl)
Convert the recipe into amounts in moles. (10 marks)
2.
A thyroid condition called goiter can be treated by increasing iodine in the diet.
Iodized salt contains calcium iodate, Ca(IO3)2 which is added to table salt.
(5 marks)
a. How many atoms of Iodine are in 1.00 x 10-2 mole of calcium iodate?
b. What is the mass of calcium iodate that contains that any atoms of iodine?
3.
Suppose that there is a prestigious award given by the Academy of Science each
year to the most significant scientific concept. Write a paper nominating the mole
concept for this award, citing the mole’s role, and importance in the application of
chemical reactions in society, industry, and the environment. (10 marks)
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SCH3U
Grade 11, University Preparation
Chemistry
Lesson 10 – Chemical Formulae
SCH3U – Chemistry
Lesson 10
Lesson 10 – Chemical Formulae
Learning Objectives:
•
•
•
•
Conduct an inquiry to calculate the percent composition of a compound
Determine the empirical and molecular formulae of various chemical compounds,
given molar masses and percent composition data.
Explain the relationship between the empirical and molecular formula of a chemical
compound.
Analyse processes in the home, the workplace, and the environmental sector that
involve the use of chemical quantities and calculations
Introduction
Using molar mass values from a periodic table, we can calculate the mass of reactants
and products in chemical reactions. This however can only be done if we already know
the chemical formula.
When new substances are produced, we need to determine their chemical formula. In
order to do this we need to find out the chemical composition of the compound. This
means we need to find out what elements it is made up of, and the quantities of each
element.
In this lesson we will determine the composition by mass of a substance and then
convert the mass amounts to percentages, to give us the percent composition. We can
then use atomic mass and molar mass to determine the correct chemical formula.
Percent Composition
Percent composition is the percentage by mass of each element in a compound.
Percent Composition = mass of element in compound x 100
(% comp.)
total mass of compound
Let’s look at the percentage composition of water. Water is formed when hydrogen is
allowed to react with oxygen. This reaction gives off large amounts of energy. The
results of an experiment revealed that 2.5g of hydrogen, when completely reacted,
produced 22.5g of water. What is the percentage composition of water by mass?
Since 2.5g of hydrogen combines with oxygen to produce 22.5g of water, that means
the mass of oxygen present initially must be (22.5 - 2.5 = 20.0g) due to the law of
conservation of mass.
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SCH3U – Chemistry
2.5g
× 100
22.5g
= 11.1%
%H =
Lesson 10
20.0g
× 100
22.5g
= 88.9%
%O =
Practice calculating Percent Composition!!
Sodium fluoride is added to toothpaste and to water supplies to help reduce tooth
decay. When a 3.65g sample of the compound was analyzed, it was found to
consist of 2.00g of sodium and 1.65g of fluorine. Calculate the percentage by
mass of each element in the compound.
Step 1: Use the equation for % composition, to determine the % for each element in the
compound.
2.00
× 100
3.65
= 54.8%
%Na =
1.65
× 100
3.65
= 45.2%
%F =
The %Na in this reaction is 54.8%, and the %F in this reaction is 45.2%.
Support Questions
1.
65.4g of zinc reacts completely with 32.1g of sulphur to produce zinc sulphide.
a. What mass of zinc sulphide would be formed?
b. Calculate the percent composition of zinc sulphide.
c. What mass of sulphur is present in 43.1g of zinc sulphide?
2.
Calculate the percent composition of mercury hydroxide if a 1.173g sample is
made up of 1.003g mercury, 0.16g oxygen, and the remainder hydrogen.
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SCH3U – Chemistry
Lesson 10
Lab: What Makes Popcorn Pop?
Purpose
Each kernel of corn is made up of not only corn, but
also water. This water is what allows popcorn to
pop. The purpose of this lab is to determine the %
composition of popcorn.
Materials
Popping corn
Popcorn Popper
Balance or scale
Procedure
•
•
•
•
•
Measure the mass of some unpopped popping corn
Pop the popping corn
Allow the popcorn to cool, and then measure the mass again.
Assume any difference in masses is caused by the loss of water
from the kernels
Calculate the percentage of water in the sample of popcorn
Observations
Create an observation table that includes all quantitative and
qualitative observations that you made
Analysis
What is the % composition of popping corn?
Empirical Formulas
If we are given an unknown substance and we want to find its chemical formula we
need to identify the elements that are in the compound, as well as the number of each
of the elements. We start by determining the % composition by mass, and then we
convert that mass into moles which gives us the subscripts associated with each
element. A formula that is derived this way is known as the empirical formula. The
empirical formula of a compound (also known as the simplest formula) shows the
smallest whole number ratio of the elements in the compound.
To determine the empirical formula of a compound we will use the percent composition
calculation and the mole concept.
Example 1
Analysis of a compound of sulphur and oxygen, produced during the burning of sulphurcontaining fuels, showed that it contained 50.0% sulphur and 50.0% oxygen by mass.
What is the empirical formula of the compound?
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SCH3U – Chemistry
Lesson 10
Step 1: Assume 100g sample size.
S = 50.0 g
O = 50.0 g
Step 2: Calculate the number of moles of each element. n =
S
50.0 g
32.06 g/mol
= 1.5596 mol
n=
m
M
O
50.0 g
16.00 g/mol
= 3.125 mol
n=
Step 3: Divide by the smallest # of moles.
S
O
1.5596 mol
1.5596 mol
=1
3.125 mol
1.5596 mol
= 2.00
=
=
Step 4: State the empirical formula
The mole ratio for S and O is 1:2, making the chemical formula SO2
Summary: Determining Empirical Formulas
1.
Find the mass of each element in 100 g of the compound, using percent
composition.
2.
Find the amount in moles of each element by converting the mass in 100 g to
moles, using the molar mass of the elements
3.
Find the whole number ratio of atoms in 100 g to determine the empirical
formula. Reduce the ratio to lowest terms.
Support Questions
3.
Police were called in to investigate the death of a diplomat in Paris, France.
There was no witness and the only clue was a very small “foreign object” stuck in
the back of the victim’s head. Analysis revealed the following information:
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SCH3U – Chemistry
Lesson 10
potassium 2.7g, uranium 16.9g, nitrogen 0.970g, and boron 0.750 g. What is the
EF of the “foreign object”?
4.
Recently, a number of shoppers fell ill. Their illness was not identified until the
forensic lab in Toronto analyzed a bit of the left toenail of each victim. Analysis
revealed the following: sulphur 10.1%, oxygen 5.05%, uranium 75.1%, and the
remainder was phosphorus. What is the simplest formula for the cause?
Molecular Formulas
Different molecules may have the same percentage composition, but contain different
numbers of atoms in a molecule. For example, ethyne, which is used in welders torches
and benzene used as a solvent, both have the same percentage composition by mass,
so they also have the same empirical formula CH. A molecular formula is needed to tell
us the actual number and kind of atoms in a molecule of the substance.
The molecular formula for ethyne is C2H2, and the molecular formula for benzene is
C6H6. The molecular formula of a compound represents the actual number of atoms in
each molecule.
Example 1
The empirical formula of a compound of carbon and hydrogen was found by experiment
to be CH2. Using a mass spectrometer, the molar mass was found to be 42 g/mol.
What is the molecular formula of the compound?
Step 1: Calculate the molar mass of the empirical formula
CH2 = 12.01 + 2(1.01)
= 14.03 g/mol
Step 2: Determine the ratio by dividing the molecular mass by empirical formula mass.
Molecular Mass
Empirical Formula Mass
42 g/mol
=
14.03 g/mol
= 2.99 or 3
Ratio =
Step 3: State the molecular formula.
CH2 x 3
Therefore, the molecular formula for this compound is C3H6
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SCH3U – Chemistry
Lesson 10
Summary: Determining Empirical Formulas
From the empirical formula and measured molar mass;
1.
2.
Calculate the molar mass of the empirical formula
Compare the measured molar mass of the substance with the molar mass
derived from the empirical formula and increase subscripts in the empirical
formula by the multiple needed to make the two molar masses equal
From percentage composition and measured molar mass;
1.
2.
3.
Find the mass of each element in one mole of the compound by multiplying the
percentage by the molar mass of the compound.
Use the molar mass of the element to convert the mass of the element to amount
in moles.
The mole ratio of the elements in the compound provides the subscripts in the
molecular formula.
Table of Molecular Formula Vs. Empirical Formula
Name of Compound
Molecular Formula
Empirical Formula
Magnesium Oxide
Mg2O2
MgO
Calcium Carbonate
CaCO3
CaCO3
Glucose
C6H12O6
CH2O
Support Questions
5.
Caffeine is a vasoconstrictor, meaning that it causes contraction of blood
vessels. Migraine headaches are attributed to dilation of the blood vessels in the
head, therefore, caffeine has a potential for treatment of this condition. The
percentage composition of caffeine is: C-49.48%, H-5.15%, O-16.49%, and N28.87%. The molar mass of caffeine is 194 g/mol. What is the molecular
formula of caffeine?
6.
The gypsy moth is an insect that has caused heavy losses in the forestry
industry. The larvae and eggs are carried long distances on cars and other
vehicles. It is believed that it is the larvae or caterpillar which causes the
damage by eating the leaves. Research shows that GYPTOL is responsible for
the attraction of gypsy moths. While difficult to synthesize, the compound
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SCH3U – Chemistry
Lesson 10
contains 72.48% carbon, 11.41% hydrogen, and 16.11% oxygen and has a molar
mass of 298 g/mol. What is its molecular formula?
Key Question #10
1.
The cheese making process began thousands of years ago, but today it is an
industrial process that uses technology. Enzymes act as chemical catalysts that
change the mild fats and proteins into cheese. Moisture in the final product is
controlled by various methods of salting, moulding, and pressing. Using the
internet research the ripening process in cheese making, and report in one page
your findings about how the percentage of moisture and milk fats in different
chesses is related to their characteristic shape, texture and flavour. Be sure to
include at least 3 different types of cheeses in your explanation. (10 marks)
2.
Choose a vitamin that you will research. Some options include; vitamin A, B3,
B6, B9, B12, C, D, and vitamin E. Create a poster advertising the importance of
your vitamin. Be creative! In addition to the importance of your vitamin, be sure to
include its chemical formula, molar mass, and % composition. You must show
how you calculated molar mass and % composition on the back of your poster.
(15 marks)
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SCH3U
Grade 11, University Preparation
Chemistry
Lesson 11 – Quantitative Analysis
SCH3U – Chemistry
Lesson 11
Lesson 11 – Quantitative Analysis
Learning Objectives:
•
•
•
Use appropriate terminology related to quantities in chemical reactions such as
stoichiometry
Calculate the corresponding mass, or quantity in moles or molecules, for any given
reactant or product in a balanced chemical equation as well as for any other reactant
or product in the chemical reaction
Explain the quantitative relationships expressed in a balanced chemical equation
using appropriate units of measure (moles, grams, atoms, ions, molecules)
Introduction
One of the most useful applications of chemistry is the synthesis of new substances that
benefit society. New medicines are developed every year, and new materials for
clothing, sports equipment and cleaning products are advertised daily.
This is the reason why the discovery of new and better ways to make these existing
products is so important. New procedures always require an understanding of both
qualitative and quantitative aspects of chemical reactions. Companies are particularly
interested in the costs related to new chemical processes.
If a company is interested in making a product, they are going to make sure they
combine reactants in a way that will use them both up. The will be wasting money if they
put excess reactant into the reaction. In this lesson we will learn how to determine the
ratios in which reactions proceed.
Quantitative Analysis
When a driver is asked to breathe into a breathalyzer, the instrument analyzes the
breath sample and measures the quantity of alcohol in the exhaled air. At a swimming
pool, a lifeguard may take a sample of water to analyze the quantity of acid, or amount
of chemicals needed to disinfect the water. In both these examples, the quantity of a
substance or sample is measured. This type of analysis is known as quantitative
analysis. Quantitative analysis uses numerical values to
describe a compound.
There are several different techniques used in
quantitative analysis. If the substance we are testing is in
solution, we can allow it to react until it forms a
precipitate. We can then collect and measure the
quantity of the precipitate and use our knowledge of
chemical reactions to calculate the quantity of the
substance in the original solution.
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A Breathalyzer Device
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SCH3U – Chemistry
Lesson 11
If we proceed with this type of strategy we need to make sure that the quantity is
completely reacted and that none of it remains in solution. In order to do this, we need
to ensure that enough reactant is added to take part in the reaction. This way the
reaction will occur until all of the reactant has been used up.
Balancing: A Refresher
In the unit there are many calculations that depend on a correctly balanced equation. If
you are having difficulty with naming compounds, or balancing, refer back to Unit 1, and
Unit 2 for extra practice. The following rules summarize how to balance an equation by
looking at the skeleton equation (unbalanced equation). Use the following rules to
balance this equation.
Na3PO4(aq) + CaCl2(aq) Æ Ca3(PO4)2(s) + NaCl(aq)
1.
Write out the chemical formula for each reactant and product including the state
of matter for each one.
2.
Try balancing any atom that is not in a polyatomic ion and is not oxygen or
hydrogen.
2 Na3PO4(aq) + 3 CaCl2(aq) Æ Ca3(PO4)2(s) + 6 NaCl(aq)
3.
If possible balance polyatomic ions as a group
4.
Balance the remaining atoms and molecules, taking into account the oxygen and
hydrogen atoms and water; leave until last any elements such as H2 or O2.
5.
Check the final reaction equation to ensure that there is the same number of
each type of atom before and after the reaction.
Stoichiometry
The quantitative relationship among reactants and products is called stoichiometry. The
term stoichiometry is derived from two Greek words: stoicheion (meaning "element")
and metron (meaning "measure"). On this subject, you often are required to calculate
quantities of reactants or products.
Stoichiometry calculations are based on the fact that atoms are conserved. They cannot
be destroyed or created. Numbers and kinds of atoms before and after the reactions are
always the same. This is the basic law of nature. Consider the following reaction:
C3H8 + 5O2 Æ 3CO2 + 4H2O
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SCH3U – Chemistry
Lesson 11
This reaction can be interpreted in terms of molecules: 1 molecule of propane, C3H8,
and 5 molecules of oxygen create 3 molecules of carbon dioxide and 4 molecules of
water.
This however is not very practical. Chemical reactions usually involve billions of atoms
or molecules.
Therefore, we consider reactions in terms of the mole instead. Recall that the mole is
6.02 X 1023 particles. The reaction above is then interpreted as follows: 1 Mole of
propane and 5 Moles of oxygen create 3 Moles of carbon dioxide and 4 Moles of water
Examining the ratio of moles in a reaction provides us with
useful pieces of information.
Example 1
The combustion of methanol (an alcohol which is added to
gasoline to enhance its combustibility) occurs according to the
following equation:
2CH3OH + 3O2 Æ 2CO2 + 4H2O
If 3.5 moles of methanol are burned in excess oxygen, how
many moles of carbon dioxide and water are produced?
Step 1: Make sure the equation is balanced (if not, then
balance it!!!)
2CH3OH + 3O2 Æ 2CO2 + 4H2O
Step 2: Write down the mole ratio from the balanced equation
Because there is excess (extra) oxygen in the reaction we do not need to include any
ratios that contain O2
2CH3OH : 2CO2 This ratio states that 2 moles of methanol produces 2 moles of carbon
dioxide
2CH3OH : 4H2O This ratio states that 2 moles of methanol produces 4 moles of water
Step 3: Determine the new ratios
2CH3OH 2CO2
:
3.5CH3OH ?CO2
The top is the original ratio from Step 2
The bottom is the new ratio from the question
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2CH3OH 4H2O
:
3.5CH3OH ?H2O
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SCH3U – Chemistry
Lesson 11
Step 4: Solve for the new ratio using algebra
?CO2 =
( 3.5CH3OH)( 2CO2 )
2CH3OH
?H2O =
= 3.5
( 3.5CH3OH)( 4H2O )
2CH3OH
=7
This means that 3.5 moles of CO2
are produced when 3.5 moles of
CH3OH are used.
This means that 7 moles of H2O
are produced when 3.5 moles of
CH3OH are used.
Using our knowledge of stoichiometry, paired with the mass formula for moles we can
determine the mass of any reactant or product in a balanced chemical equation.
If we know the mass of one of the reactants or products, then we can calculate the
mass of another using the mole ratio of a balanced equation.
Example 2
Consider the following reaction: Fe2O3 + 3CO Æ 2Fe + 3CO2
What mass of iron (Fe) can be formed from 425 g of iron ore (Fe2O3)?
Step 1: Write a balanced chemical equation
Fe2O3 +
3CO Æ 2Fe + 3CO2
Step 2: Convert the measured mass into an amount in moles
m
M
425g (of Fe 2O3 )
=
2 ( 55.85 ) + 3 (16.0 )
n=
Convert mass of Fe2O3 to moles of Fe2O3
= 2.66124 mol
Step 3: Use the mole ratio in the balanced chemical equation to predict the number of
moles of the desired substance.
Fe 2O3
2Fe
:
2.66 (mole Fe 2O3 ) ?Fe
?Fe =
This is the original ratio
This is the new ratio from Step 2
2.66 (mole Fe 2O3 ) ( 2Fe )
Fe 2O3
= 5.32 mol of Fe
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SCH3U – Chemistry
Lesson 11
Step 4: Convert moles of Fe to mass of Fe using molar mass
m = ( 5.32 mol ) x ( 55.85 g/mol )
= 297.122g
m=nxM
= 297g
Therefore the mass of Fe produced is 297 g.
Support Questions
1.
What mass of calcium will produce 10g of calcium oxide?
Ca + O2 Æ CaO
2.
Find the mass of Fe2O3 formed from 125g of FeS.
FeS + O2 Æ Fe2O3 + SO2
3.
Sodium hydroxide reacts with carbonic acid to produce water and sodium
carbonate. What mass of sodium carbonate will be produced from 60g of sodium
hydroxide?
Agricultural Applications
We often use excess reagents (extra reactants) in our everyday lives, where little may
be sufficient, but more may be better. When we wash our laundry for example, we may
use a little extra detergent to make sure our clothes are completely rid of dirt. When we
nourish our crops, we may use a little extra fertilizer to make sure that we reap the
largest harvest possible. But there is always the possibility of harm when we use a little
extra. In the case of fertilizers, it seems that a little excess can accumulate into a large
excess, with accompanying consequences.
Fertilizers supply nitrogen to the soil. It is needed for plant growth, and nitrates are the
main source of nitrogen for plants. Since nitrates are in short supply in most soils,
farmers and gardeners add synthetic chemical fertilizers to their soil. Of the large
quantities of fertilizers used by both home owners and farmers, the unused excess
amounts are washed from the soil and roadways by rain, and irrigation into ground
water, rivers, and lakes.
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SCH3U – Chemistry
Lesson 11
In areas of intensive agriculture the fertilizer that is not taken up by crops, is harmful to
local ecosystems. How much damage could a little fertilizer do? Once in a lake, the
nitrates from fertilizers do what they are supposed to do, stimulate plant growth. In
areas where the nitrate concentration is high there is rapid growth of plants and algae
on the surface of the water. When the algae dies, the decomposing bacteria consume
large amounts of dissolved oxygen from the water. This deprives fish and other
organisms of needed oxygen. Entire habitats have been destroyed by the nutrient run
off that forms sludge.
Key Question #11
1.
Take a Stand: Controlling the Use of Fertilizers
Background: To address the problem of what to do about nutrient run off, groups of
researchers have proposed some form of government initiative to regulate the amount
of fertilizers used by farmers and homeowners. Many groups have special interests in
the proposal:
•
•
•
•
City officials; concerned about cost and safety of water supply
Environmentalists; concerned about the effect on ecosystems
Tourism Industry (boating, fishing); concerned about the loss of income resulting
from contamination of resources
Farmers; concerned about calculating amounts of fertilizer needed, and the
instability of the weather that causes run off, also their need to maximize crop yields
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SCH3U – Chemistry
Lesson 11
Research: Use library and internet resources to collect information about the issue at
hand, then select a role (from above) and support your position using your research.
Evaluation: In your area, a group of concerned citizens wants the government to
regulate the amount of fertilizers used. The mayor has organized a panel to discuss this
and has asked for a representative from everyone with an opinion on the issue to be
present. All of the information will then be included in a non bias report which will be
sent off to the Ministry of Agriculture. You are a member of this panel, you can be a
farmer, city official, environmentalist, high school student, a fisher, a restaurant owner,
or even a parent with a small child. Your task is to select a role, and a position. Your
position is to be reflected in a one page paper. Be sure to be convincing. Your opinion
counts! It can make a difference! (10 marks)
2.
How many molecules are contained in a 51g sample of water? Show all your
work, and don’t forget to use significant digits in your final answer! (5 marks)
3.
Ammonia reacts with oxygen gas to produce water and nitrogen gas. What mass
of oxygen gas must be used to react with 19g of ammonia? Show all your work
and don’t forget to use significant digits in your final answer! (5 marks)
4.
What mass of copper II hydroxide precipitate is produced by the double
displacement reaction in a solution of 2.67 g of potassium hydroxide with excess
aqueous copper II nitrate? (5 marks)
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SCH3U
Grade 11, University Preparation
Chemistry
Lesson 12 – The Yield of a Chemical Reaction
SCH3U – Chemistry
Lesson 12
Lesson 12 – The Yield of a Chemical Reaction
Learning Objectives:
•
•
•
•
Assess on the basis of research, the importance of quantitative accuracy in industrial
chemical processes and the potential impact on the environment if the quantitative
accuracy is not observed.
Use appropriate terminology related to quantities in chemistry including percentage
yield and limiting reagent
Solve problems related to quantities in chemical reactions by performing calculations
involving percentage yield and limiting reagents
Conduct an inquiry to determine the actual yield, theoretical yield, and percentage
yield of the products of a chemical reaction
Introduction
When we predict the mass of reactants and products, we base our calculations on the
balanced equation for the chemical reaction. This calculation however is not often the
same as the actual amount that is used or produced in a reaction. The calculation
shows how much of the reactant of product should be used or made.
Often the amount of product we obtain is less that what we calculated. Most often this
loss is due to experimental errors such as losing product during the transfer of solutions,
filtering of precipitates, and splattering during heating. These losses can be reduced by
improving technical skills.
In this lesson we will compare a calculated yield to an actual yield.
Limiting and Excess Reagents
When a reaction occurs, we need to ensure that enough of the reactant is present for it
to occur. Usually, more than required, or an excess of one of the reactants is used. This
makes sure that the reaction will continue until all of the other reactant is used up. The
reactant that gets completely used up is the limiting factor in the reaction.
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SCH3U – Chemistry
Lesson 12
Lab: Limiting Reagents
Background
When sodium hydrogen carbonate (baking soda) reacts with an acid
such acetic acid (vinegar), carbon dioxide is produced. In this activity
you will compare amounts of gas produced when a fixed amount of
baking soda is mixed with increasing amounts of vinegar.
Materials
100 mL baking soda (5 tbsp)
300 mL vinegar
3 small Ziploc bags
Measuring cup
Measuring spoon
Procedure
•
•
•
•
•
•
•
•
Evaluation
Using the measuring spoon, carefully place 15mL or 3 tsp of
baking soda into the bottom corner of one of the baggies. Fold the
corner upward and tape it in place so that the solid does not come
into contact with anything else.
Carefully pour 30mL of vinegar into the other bottom corner of the
bag, make sure it does not come into contact with the solid.
Flatten and press the bag gently to remove as much air as
possible, then seal the bag.
Remove the tape, and shake the bag to allow both reactants to
mix well. Allow the reaction to continue until no more gas is
produced (bubbles stop). Keep the bag sealed for later
observations.
Repeat the procedure with another bag, using the same amount of
baking soda, but increasing the amount of vinegar to 60mL.
Repeat one more time, this time using 120mL of vinegar.
In an observation table record the relative volumes of gas in the
three sealed bags. Is there evidence that the amount of product
formed is related to the amount of reactants used? Is there
evidence that some of the baking soda has not completely reacted
in one or more of the bags?
Open each bag and add 15mL of vinegar to each. Is there
evidence that a shortage of vinegar was limiting the production of
carbon dioxide gas in some bags?
Add 15mL of baking soda to each bag. Is there evidence that a
lack of baking soda was limiting the production of carbon dioxide
in some bags?
Complete an observation table that contains all of your results, do not
forget to include the answers to the questions asked in the procedure.
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SCH3U – Chemistry
Lesson 12
Calculating Limiting and Excess Reagents
When reactants in a chemical reaction are mixed together, the quantities that react are
determined by the mole ratios in a balanced equation for the reaction. It is important to
know the minimum quantities of each reactant needed for a desired quantity of product.
Generally, industrial processes and lab procedures are designed so that one of the
reactants is used as the limiting reagent, and the other reactants are used in slight
excess. Consider the balanced equation:
CH4(g) + 2O2(g) Æ CO2(g) + 2H2O(l)
Methane and oxygen gas react in the ratio 1mol CH4 : 2mol O2
Suppose 1 mole of methane (CH4) and 3 moles of oxygen (O2) are mixed and allowed
to react. Then all of the methane and 2 moles of oxygen are used up, but 1 mole of
oxygen is “left over” (un-reacted). The reactant that is completely used up (methane) is
called the limiting reactant. After this reactant is used up, the reaction will no longer
proceed. The reactant that is left over (oxygen) is called the excess reactant. By
determining the limiting reactant, chemists can predict how much product will be
produced in a chemical reaction.
Example 1
Zinc and sulphur react to form zinc sulphide. If 6.00g of zinc and 4.00g of sulphur are
available for the reaction, determine the limiting reactant and then determine mass of
zinc sulphide produced.
Step 1: Determine the balanced chemical equation
Zn +
S Æ ZnS
Step 2: Calculate the moles of each given reactant. The reactant with the smallest
number of moles is the limiting reagent.
Zn
S
n=
m
M
6.0g
=
65.38g / mol
= 0.0918mol
4.00g
32.06g / mol
= 0.1248mol
n=
Therefore Zn is the limiting reagent.
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SCH3U – Chemistry
Lesson 12
Step 3: Use the mole ratio of the limiting reactant in the balanced chemical equation to
predict the number of moles of the desired substance.
Zn
ZnS
:
0.0918 mol Zn ?ZnS
( 0.0918 mole Zn)(1 ZnS )
?ZnS =
1 Zn
= 0.0918 mol of ZnS
Step 4: Convert moles to mass using molar mass
m = n ×M
= 0.0918 × 97.44
= 8.945g
= 8.95g
Therefore the mass of ZnS produced is 8.95 g.
Support Questions
1.
The reaction between hydrochloric acid and zinc metal produces hydrogen gas
and zinc chloride. If 50g of each hydrochloric acid and zinc are mixed, determine
the amount of zinc chloride that would be produced.
2.
When 75g of sodium sulphite and 80.0g of hydrogen bromide are reacted, the
products formed are sodium bromide and hydrogen sulphite. Calculate the mass
of sodium bromide produced.
The Yield of a Chemical Reaction
You have learned how chemists can use stoichiometry to predict the amount of product
that can be theoretically expected from a chemical reaction. This theoretical yield is not
always the same as the amount of product that is actually obtained from an experiment.
The amount of product that is actually made is called the actual yield. Quite often, it is
experimental error that leads to the actual yield being less than the theoretical
expectation. However, sometimes there may be a competing reaction that is taking
place, or the reaction in question may not go to completion due to limitations on
temperature and pressure.
Percentage Yield =
Actual Yield
Theoretical Yield
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× 100%
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SCH3U – Chemistry
Lesson 12
Example 2
Ammonium nitrate is and an important compound used both as a fertilizer and as an
explosive. It is produced by reacting ammonia with concentrated nitric acid.
NH3(s) + HNO3(aq) Æ NH4NO3(s)
a. What mass of ammonium nitrate can theoretically be produced from the reaction of
375.0g of ammonia with excess nitric acid?
b. If the percentage yield is 88.5%, what mass of ammonium nitrate is actually obtained
from 375.0g of ammonia?
Step 1: Determine the balanced chemical equation, and the number of moles of limiting
reactant.
NH3(s) + HNO3(aq) Æ NH4NO3(s)
NH3(S)
n=
=
m
M
375.0g
17.04g/mol
Since the question tells us that nitric acid is
in excess, we know that ammonia is the
limiting reactant.
= 22.007 mol
Step 2: Use the mole ratio to determine the moles of product.
NH3
NH4NO3
:
22.007 mol NH3 ?NH4NO3
?NH4NO3 =
Original Ratio
New Ratio
( 22.007 mol NH3 )(1 NH4NO3 )
1 NH3
= 22.007 mol of NH4NO3
Step 3: Calculate the mass of product.
m = n ×M
= 22.007 × 80.06
= 1761.88g
= 1761g
Therefore the theoretical mass of NH4NO3 produced is 1761 g.
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SCH3U – Chemistry
Lesson 12
Step 4: Calculate the percentage yield.
Since we are given % yield in the question we are looking for the actual yield.
% yield =
actual yield
× 100%
theoretical yield
actual yield
× 100
1761
( 88.5 )(1761)
actual yield =
100
= 1558 g
88.5 =
Therefore the actual amount of NH4NO3 produced is 1558g.
Support Questions
3.
Can the actual yield ever be greater than the theoretical yield? Explain.
4.
Iron is produced from its ore Fe2O3 by heating with carbon monoxide in a blast
furnace. If the industrial process produced 635 kg of iron from 1000 kg of Fe2O3,
what is the percentage yield of iron in the process? The equation for the reaction
is given below.
Fe2O3 + 3 CO Æ 2 Fe + 3 CO2
Thought Lab: Percentage Yield
Purpose
What is the mass of precipitate formed when 3.43g of barium hydroxide
in solution reacts with an excess of sulphuric acid?
Hypothesis
Calculate the theoretical yield of barium sulphate
Evidence
•
•
•
A white precipitate formed when the barium hydroxide solution was
mixed with the sulphuric acid.
Mass of filter paper = 0.96 g
Mass of filter paper and precipitate = 5.25 g
Analysis
What is the mass of precipitate formed when 3.43g of barium hydroxide
in solution reacts with an excess of sulphuric acid?
Evaluation
What is the percentage yield of this reaction?
Copyright © 2009, Durham Continuing Education
Page 34 of 39
SCH3U – Chemistry
Lesson 12
Chemistry in Technology
One of the most useful applications of chemistry is the synthesis of new substances that
benefit society. New medicines are developed every year, and new materials for
clothing, sports, equipment and cleaning products are advertised daily. This is why they
the discovery of new and better ways of synthesizing existing chemicals is so exciting.
New procedures often follow the development of new technology.
The production of soda ash ( sodium carbonate) which is used as a cleaner commonly
known as washing soda was first extracted from plant ashes, and used to make soap.
As the demand for soap increased, there wasn’t enough sodium carbonate. The French
Academy of Science offered a prize for the development of a method to make sodium
carbonate from common substances. As a result of this, in 1775, the Leblanc process
was developed.
To make sodium carbonate using the Leblanc process, sodium chloride and sulphuric
acid are heated, limestone and coal are added and heated again to a high temperature.
Quantitatively the Leblanc process was inefficient. It required burning a lot of coal,
which is expensive, and generated hydrogen chloride which is a sever air pollutant. One
of the by products was an insoluble reside that had no commercial value. In other
words, a great deal of resources and energy was put into the reaction, and only a small
proportion of the materials were recovered as a useful product.
It wasn’t until 1867, almost a hundred years later that a new process was developed for
the production of sodium carbonate. This new technique – the Solvay process, was one
third the cost of the older Leblanc process. It reacts chalk (calcium carbonate) with salt
(sodium chloride) to produce baking soda, and washing soda (sodium carbonate). All of
the intermediate products are recycled as reactants in further reactions, and the only
one by product of this reaction can be used as road salt.
The Leblanc process was costly, not only in terms of the cost of raw materials, and fuel,
but also in terms of damage to the environment. The newer Solvay process was more
economically, and environmentally friendly. What if the Solvay process had been more
costly? Would it still have replaced the Leblanc process? If becoming environmentally
friendly means making less profit, the loss of profit will likely be passed on to the
consumer in the form of higher prices. Are we willing to pay the cost?
Support Questions
5.
Should we be willing to pay the cost for improved technology? Explain your view
point.
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SCH3U – Chemistry
Lesson 12
Key Question #12
Take A Stand: Should We Be Willing to Pay for Better
Technology?
1.
Question: Who should pay the costs of improved technology?
Research: Choose one industry (pulp and paper, nickel smelting, gasohol, etc) and
research different companies in that industry to see what efforts have been made to
develop environmentally friendly solutions. Your research should include the following
components:
•
•
•
•
the reason for changing technologies
the benefits of the new technology
the costs involved in changing to the new technology
the consumer demand or support for the new technology
Evaluation: Choose one product made by the industry you researched and take a stand
on whether we should be willing to pay the cost for improved technology. (10 marks)
2.
Propane (C3H8) is a gas at room temperature, but it exists as a liquid under
pressure in a propane tank. It reacts with oxygen gas in the air to make carbon
dioxide and water vapour. Write the balanced chemical equation for this
reaction. (2 Marks)
3.
What mass of carbon dioxide gas is expected when 97.5g of propane reacts with
500.0g of oxygen gas? Hint: determine the limiting reactant. (4 Marks) What is
the percent yield if only 250g of carbon dioxide are released? (1 Mark)
4.
Submit your observation table for the Limiting Reagents Lab. Do not forget to
include the answers to the questions within the procedure. (8 marks)
5.
For the Percentage Yield Lab, submit your hypothesis, analysis, and evaluation
sections. (10 marks)
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SCH3U
Grade 11, University Preparation
Chemistry
Support Question Answers
SCH3U – Chemistry
Support Question Answers
Lesson 9
1.
342 g/mol
2.
114.26 g/mol
3.
46.08 g/mol
4.
m=nxM
20.7 g
5.
n = m/M
1.57 mol
6.
n = N/NA
0.887 mol
m=nxM
74.5 g
Lesson 10
Zn + S Æ ZnS
65.4 + 32.1= 97.5 g
Zn = 67%
S = 33%
14.19 g
1.
a.
b.
c.
2.
Hg = 85%
3.
KUNB
4.
SOUP
5.
C8H10O2N4
6.
C18H33O3
O = 14%
H = 1%
0.178 mol
m = nxM
Lesson 11
1.
n = m/M
7g
2.
n = 1.42 mol, 2:1, ratio, n = 0.71 mol, m =113 g
3.
n = 1.5 mol, 1:1 ratio, n = 1.5 mol, m = 124.5g (sig dig 100 g)
Lesson 12
1.
n = 1.37 mol (HCl limiting), 2: 1 ratio, m = 93.35g
2.
n = 0.988 mol (HBr limiting), 2:2 ratio, m = 102 g
3.
Yes it is possible due to experimental errors. If there is contamination of the
product then the mass of the actual product could be greater than the theoretical
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SCH3U – Chemistry
Support Question Answers
yield. Also, if we are drying a precipitate, it is possible that there is still water in
the sample. This would also cause the actual yield to be greater than expected.
4.
n = 6262 mol, 1:2 ratio, n = 12523 mol, m = 699 000g(theoretical)
%yield = 91%
5.
Answers will vary.
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