Download File

Chemistry Module 1- Lesson 4
Chemistry lesson 4
Study guide - Questions
In this lesson we enter the world of equations and moles. This allows us to calculate number of
atoms and mass of reactants and products. The concept of the mole allows us to actually
calculate quantities of reactant and products from our balanced chemical equations. This branch
of chemistry is called Stochiometry.
Your weekly program.
Develop your own method by all means but this will get you going.








Read the notes attached here. All questions and the final assessment are based on what is
in the notes.
Read the chapter sections in the reading section ( the text has much greater detail than
what you are expected to know) especially look at diagrams and figures
At this stage not much will make sense but that is OK
With your text as reference and this study sheet go through the Power point presentation.
Make notes where needed.
MANDATORY watch all of the embedded video links (you will need internet access to do
this)
Optional - Listen to the audio file of a live lesson; be aware that there will be long pauses
with not much going on at certain points.
Refine your notes / mind maps on the key concepts outlined in these weekly study sheets.
Check your understanding
Now go through your homework questions and answer those
Your study resources







Textbook
Audio files of the actual lessons
Weekly study sheet
Power Point slides
YouTube links and YouTube as a general resource
College forum site
Molecule kit
Chemistry Module 1- Lesson 4
Textbook reference Bettleheim Edition 9
Please read these sections and note take
Chapter 4 sections 4.1 to 4.4
Key Concepts to understand




Balancing chemical equations
Formula and molecular weight
The mole concept
Basic Stochiometry
o Moles to grams and vice versa
o Molar ratios in equations
o Limiting reagents
Use the links in the power point presentation.
Good sources for chemistry videos are
Bozeman science
http://www.bozemanscience.com/
Crash course chemistry – YouTube
Tyler de Witt socratic.org/
Chemistry Module 1- Lesson 4
Study Notes
In order to tackle Stochiometry problems we need to master some foundational concepts first.
These are balancing equations, the molar concept and finding molecular weights.
Chemical equations
Chemical equations do not come already balanced. This must be done before the equation can be
used in any chemically meaningful way.
A balanced equation has equal numbers of each type of atom on each side of the equation.
The Law of Conservation of Mass (lesson 2) is the rationale for balancing a chemical equation.
The essence of this law is - "Matter is neither created nor destroyed."
Therefore, we must finish our chemical reaction with as many atoms of each element as when we
started.
Let’s step through a balancing exercise.
Problem #1: Balance the following equation:
H2 + O2 ---> H2O
It is an unbalanced equation.
This means that there are UNEQUAL numbers of at least one atom on each side of the arrow.
In the example equation, there are two atoms of hydrogen on each side, BUT there are two atoms
of oxygen on the left side and only one on the right side.
Remember this: A balanced equation MUST have EQUAL numbers of EACH type of atom on
BOTH sides of the arrow.
An equation is balanced by changing coefficients in a somewhat trial-and-error fashion. It is
important to note that only the coefficients can be changed, NEVER a subscript.
The coefficient times the subscript gives the total number of atoms.
Sub examples of finding the number of atoms.
(a) 2 H2 - there are 2 x 2 atoms of hydrogen (a total of 4).
Chemistry Module 1- Lesson 4
(b) 2 H2O - there are 2 x 2 atoms of hydrogen (a total of 4) and 2 x 1 atoms of oxygen (a
total of 2).
(c) 2 (NH4)2S - there are 2 x 1 x 2 atoms of nitrogen (a total of 4), there are 2 x 4 x 2 atoms
of hydrogen (a total of 16), and 2 x 1 atoms of sulphur (a total of 2).
Two things you CANNOT do when balancing an equation.
1) You cannot change a subscript.
For example -You cannot change the oxygen's subscript in water from one to two, as in:
H2 + O2 ---> H2O2
True, this is a balanced equation, but you have changed the substances in it. H2O2 is a
completely different substance from H2O. So, it's not the answer to the question that was
asked.
2) You cannot place a coefficient in the middle of a formula.
The coefficient goes at the beginning of a formula, not in the middle, as in:
H2 + O2 ---> H22O
Water only comes as H2O and you can only use whole formula units of it.
Two more points:
1) Make sure that your final set of coefficients are all whole numbers with the lowest common
factor. For example, this equation is balanced:
4 H2 + 2 O2 ---> 4 H2O
However, all the coefficients have the common factor of two. Divide through to eliminate common
factors like this. This means getting the coefficient to the lowest common number.
2) NO fractions allowed in the final answer, only whole numbers.
Chemistry Module 1- Lesson 4
The final balanced equation is as follows
Some more balanced equation examples
Example 1 shows the numerical logic in the process
Chemistry Module 1- Lesson 4
Example 2 the combustion of methane
Example 3 respiration of glucose for energy
Example 4 Balance the following equation:
Na + H2O ---> NaOH + H2
You will notice that Na and the O are already balanced. So, we look only at the H.
As H must come in twos on the left-hand side, we must have an even number of hydrogen on the
right-hand side. We do this:
Na + H2O ---> 2NaOH + H2
We then balance the H on the left like this:
Na + 2H2O ---> 2NaOH + H2
Notice that the first placing of a 2 messed up the balance of the Na and the O. In addition, notice
that the second placing of a 2 corrected the imbalance in the oxygen.
The last step is to balance the Na: 2Na + 2H2O ---> 2NaOH + H2
Chemistry Module 1- Lesson 4
The mole
In chemistry the mole is a fundamental unit in the SI system, and it is used to measure the amount
of substance. In Latin mole means a "massive heap" of material. It is convenient to think of a
chemical mole as such.
A mole is the same number concept as a dozen eggs
A dozen always means 12 of anything and a mole always means
many atoms or molecules.
This means six multiplied by 10 twenty three times consecutively. Notice the exponential notation.
It is commonly known as Avogadro’s number.
Amedeo Avogadro (1776–1856) was an Italian chemist who in working with gases postulated this
mole concept.
Avogadro's number is equal to 602,214,199,000,000,000,000,000 or more simply, 6.02214199 ×
10 23.
This is a large number! This scale of number allows us to manipulate minute atoms or molecules
by number or mass accurately.
Formula or Molecular weight
This is the weight in grams of an ionic (formula) or covalent compound (molecular). To be able to
calculate this we need to understand molar mass.
First, it is a mass, so it has units of mass, commonly the gram. Second, it concerns the mole
(Avogadro's number). Whether you're dealing with elements or compounds, the molar mass of a
species is the mass in grams of one mole (6.022 x 1023) of that species: one mole of atoms, one
mole of molecules, or one mole of formula units.
To find the molar mass, add the atomic masses of all of the atoms in the molecule. To find the
atomic mass for each element, use the mass given in the Periodic Table. Multiply the subscript
(number of atoms) times the atomic mass of that element and add the masses of all of the
elements.
Chemistry Module 1- Lesson 4
Example 1:
Sodium chloride (NaCl) has a formula weight of 58.443 (atomic weight of Na + atomic weight of Cl)
and a molar mass of 58.443 g/mol.
Formaldehyde (CH 2 O) has a molecular weight of 30.03 (atomic weight of C + 2 [atomic weight of
H]) + atomic weight of O] and a molar mass of 30.03 g/mol.
Example 2:
Calcium phosphate has the formula Ca3(PO4)2. Calculate the mass of one mole of calcium
phosphate.
3 mole of Ca
2 mole of P
8 mole of O.
1 mole of calcium is
40.08g, so
3 mole is
120.24 g
1 mole of phosphorus is
30.9738g, so
2 mole is
61.9476g
1 mole of oxygen is
15.9994g, so
8 mole is
127.9952g
1 mole of Ca3(PO4)2 is
310.18 g
So why does this work? (Optional section)
Carbon 12 is set as the standard reference for molar mass- one mole of this element was
defined as having Avogadro’s number of atoms.
By international agreement, the atomic mass (atomic weight) is the mass of an atom
expressed in atomic mass unit (amu). One atomic mass unit (amu) is defined as a mass
exactly equal to one-twelfth (1/12th) of one carbon-12 atom. This means that the mass of
one carbon-12 is assigned as exactly 12 amu and 12 grams.
The mass of other atoms are determined based on their relative weight. For example, the
hydrogen atom weighs only 8.40 percent of carbon-12 atom. Then the mass of hydrogen
atom is calculated as follows.
Atomic mass of hydrogen = mass of one atom carbon-12 x 8.40%
= 1.008 amu
Chemistry Module 1- Lesson 4
Basic Stochiometry
Now let’s use the knowledge we have covered to work out mass and moles from an equation.
Once we have a balanced equation then the elements and compounds in the equation have a
numerical relationship. This is called a mole ratio. Please note it is a ratio of moles not weight.
Definition: A mole ratio is ratio between the amounts in moles of any two compounds involved in a
chemical reaction. Mole ratios are used as conversion factors between products and reactants in
many chemistry problems.
Example 1:
2 H2(g) + O2(g) → 2 H2O(g)
The mole ratio between O2 and H2O is 1:2. For every 1 mole of O2 used, 2 moles of H2O are
formed.
The mole ratio between H2 and H2O is 1:1. For every two moles of H2 used, 2 moles of H2O is
formed.
Using our skill we can also work out which reactants will be used up first in a particular process –
this is called finding the limiting reagent.
The key skills in Stochiometry problems are
o Balance the equation
o Be able to determine the mole ratios in the equation
o Be able to convert moles to grams and vice versa
Let’s get to work on real problem
The reaction of powdered aluminium and iron(II)oxide
Here is the balanced equation for the reaction
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l)
Chemistry Module 1- Lesson 4
This reaction produces so much heat the iron that forms is molten. Because of this, railroads use
the reaction to provide molten steel to weld steel rails together when laying track.
Note the brackets tell us the state of the chemical species (s) means solid. In the products section
the (l) means the Iron is liquid hence the heat. In the title of the question (ii) against the Iron means
it has an overall charge of 2- for the ion.
Here is the question
Suppose that in one batch of reactants 4.20mol Al was mixed with 1.75mol Fe2O3. Which
reactant, if any, was the limiting reactant?
Calculate the mass of iron (in grams) that can be formed from this mixture of reactants.
How do we approach this question – firstly determine what has been asked of you. Note that there
are two parts to the question!
Part 1 - Which reactant is the limiting reactant – this means which reactant runs out first in
the process.
Part 2 – Calculate the mass in grams of iron that can be formed with this mixture of
reactants
Notice that we are given information in moles and asked to give the answer in grams. This
means moles need to be converted to grams. Also we will need to find out the molar ration
between Fe2O3 and Al.
With all chemistry problems and most scientific problems you arrive at the answer by using
logical steps rather than guessing. It’s important to set out your work neatly, step wise so
that you can check your calculations at the end and also record your reasoning process.
In an assignment the reasoning and calculations hold as much weight as the actual correct
answer.
In your text book dimensional analysis is used for these problems but I will outline an easier
method for you to use.
First step – balance the equation, this has already been done.
Done
Chemistry Module 1- Lesson 4
Second step - work out the molar ration between the two compounds mentioned. In this
question it is Fe2O3 and Al
The molar ration is 2:1 Al to Fe2O3
Third step - use the molar ratio to work out which reactant is not fully used up. The other
species is then the limiting reagent. This will give us the answer to the first part of the
question.
Taking the 1.75 moles of Fe2O3 as one part and multiplying by two gives 3.5 moles of Al
needed to react with it. We are given 4.2 moles of Al so there is some left over.
The limiting reactant is the Fe2O3 as it is completely used up.
Fourth step – Find out the molar ratio between the limiting reagent and the product 2Fe(l)
Looking at the balanced equation we see the ratio is 1:2 Fe2O3 to 2Fe
Fifth step - Find the amount of moles of 2Fe formed by the reaction of Fe2O3.
1.75 moles of Fe2O3 multiplied by 2 gives 3.5 moles of 2Fe using the molar ratio.
Sixth step - Now convert the moles into grams for the final part of the answer.
3.5 moles of Fe with formula weight of (55.8) 111.6 grams = 3.5 x 55.8 = 195.3 grams.
The given mixture of reactants produces 195.3 grams of Iron Fe.
Note: the answer is presented as a concluding statement with the number of the answer
and the scientific units in this case
Chemistry Module 1- Lesson 4
Home work
Question 1: Write out the molar ratios for these questions
Example A: N2 + 3H2 ---> 2NH3
Write the molar ratios for N2 to H2 and NH3 to H2.
Example B: 2SO2 + O2 ---> 2SO3
Write the molar ratios for O2 to SO3 and O2 to SO2.
Example C: PCl3 + Cl2 ---> PCl5
Write the molar ratios for PCl3 to Cl2 and PCl3 to PCl5.
Question 2: Determine the mass in grams of the following:
1.35 moles Fe
Question 3: Calculate the number of moles of this compound:
21.5 g CaCO3
Question 4: Using the following equation answer three questions about the reaction.
N2 + H2 ---> NH3
Question A – Balance the equation
Question B – What is the molar ratio between N2 and NH3
Question C – If you have 2.00 mol of N2 reacting with sufficient H2, how many moles of NH3 will be
produced?
Chemistry Module 1- Lesson 4