1. Consider the combustion of propane: C3H8 + O2  CO2 + H2O (unbalanced)
QUESTION:
Consider the combustion of propane: C3H8 + O2  CO2 + H2O (unbalanced)
Using the smallest set of whole numbers, the sum of the coefficients in the balanced chemical equation is A. 7, B. 9, C. 12, D. 13
In a chemical equation, number of atoms must be the same on both sides of the arrow, for each element.
Consider the combustion of propane: C3H8 + O2  CO2 + H2O
Using the smallest set of whole numbers, the sum of the coefficients in the balanced chemical equation for this reaction is A. 7, B. 9, C. 12, D. 13
PAUSE CLICK In a chemical equation, the number of atoms must be the same on both sides of the arrow, for each element.
CLICK Let’s count the atoms in the unbalanced equation.
THere are three elements shown in the equation... carbon, oxygen, and hydrogen
Carbon atoms
HIGHLIGHT C in cell
are found in propane on the left
HIGHLIGHT C3H8
There are three carbon atoms in C3H8.
CIRCLE 3 in C3H8 and 3 in table
On the right, carbon is found in carbon dioxide,
HIGHLIGHT CO2
There is one carbon atom in CO2.
CALLOUT “subscript implied = 1”
CIRCLE 1 in table.
Hydrogen atoms
HIGHLIGHT H in table
are found in propane on the reactant side.. HIGHLIGHT C3H8
and there are 8 hydrogen atoms in propane
CIRCLE 8 in C3H8
CIRCLE 8 in table
On the product side, hydrogen atoms are found in water..
HIGHLIGHT H2O
and there are two hydrogen atoms in H2O
CIRCLE 2 in H2O
CIRCLE 2 in table
For oxygen, we find 2 atoms on the reactant side...
CIRCLE 2 in O2 and 2 in table
and three atoms on the right
CIRCLE 3 in table
Two oxygen atoms from CO2 CIRCLE 2 in CO2
and one from H2O
CALLOUT “implied subscript = 1”
We can see that none of the three elements are balanced.
CLICK CONTINUED ON NEXT SLIDE
ELEMENT
LEFT
RIGHT C 3 1 H 8 2 O

2. Here’s how we balance the equation.
C3H O  CO H2O O2
CO2
H2O
CO2 O2
H2O O2
H2O O2
C3H O  CO H2O
Here’s how we balance the equation.
Let’s start with carbon.
Since we have 3 carbons on the left and only one on the right, we need two more on the right.
The way to do this is to put more carbon dioxide on the right
CLICK CLICK
So, now we have one...
HIGHLIGHT first CO2
two...
HIGHLIGHT second CO2
three...
HIGHLIGHT third CO2.
CLICK At this point carbon is balanced.
As a result of adding more CO2’s on the product side, we now have more oxygen atoms on the right.
CLICK We now have 7 oxygen atoms on the right. Six in the three CO2’s...
Two...
HIGHLIGHT 2 in first CO2
Four...
HIGHLIGHT 2 in second CO2
Six...
HIGHLIGHT 2 in third CO2
and one in water
HIGHLIGHT O in H2O
At this point carbon is balanced, but hydrogen and oxygen are still not balanced.
Let’s try to balance hydrogen next.
HIGHLIGHT Hydrogen row in table
We have 8 hydrogen atoms on the left and only two on the right.
We need to put more hydrogens on the product side. To do this we add more water molecules
CLICK CLICK CLICK So, now, we have two...
HIGHLIGHT 2 in first H2O
four...
HIGHLIGHT 2 in second H2O
six...
HIGHLIGHT 2 in third H2O
eight...
HIGHLIGHT 2 in fourth H2O
CLICK Now, hydrogen is balanced.
Again, we have affected the total number of oxygens,.. Since adding water not only adds hydrogen but also oxygen.
Let’s recount the oxygen atoms on the right.
We now have 10.
CLICK Two...
seven...
HIGHLIGHT O in first H2O
HIGHLIGHT O in second H2O
nine...
HIGHLIGHT O in third H2O
ten...
HIGHLIGHT O in fourth H2O
At this time, carbon is balanced.... we have three on each side....
HIGHLIGHT
Hydrogen is also balanced... we have 8 on each side...
HIGHLIGHT 8
But the oxygen is not balanced... we only have 2 on the left
HIGHLIGHT 2
and 10 on the right
HIGHLIGHT 10
This means we need to put more oxygen on the left.
We do this by adding more O2’s to the reactant side.
CLICK CLICK CLICK CLICK Now, we have
2...
HIGHLIGHT 2 in first O2
4...
HIGHLIGHT 2 in second O2
6...
HIGHLIGHT 2 in third O2
8...
HIGHLIGHT 2 in fourth O2
and 10...
HIGHLIGHT 2 in fifth O2
So the balanced equation has one C3H8...
BOX C3H8
...five O2’s
BOX the five O2’s
...three CO2’s
BOX the three CO2’s
...and four H2O’s
BOX the four H2O’s
We can summarize everything by writing the chemical equation like this....
CLICK We put a coefficient of 5 in front of O2...
... a coefficient of 3 in front of CO2...
...and a coefficient of 4 in front of H2O...
We don’t need to put a coefficient in front of C3H8....
CALLOUT “Implied coefficient is 1”
If no coefficient is written, it is implied to be one.
CLICK
CONTINUED ON NEXT PAGE
ELEMENT
LEFT
RIGHT C 3 1 H 8 2 O 3 8 10 7 10

3. QUESTION:
Consider the combustion of propane: C3H8 + O2  CO2 + H2O (unbalanced)
Using the smallest set of whole numbers, the sum of the coefficients in the balanced chemical equation is A. 7, B. 9, C. 12, D. 13
C3H O  CO H2O
Smallest set of whole number coefficients: greatest common factor is 1.
So the sum of the coefficients is 1
CALLOUT “implied 1”
plus 5
HIGHLIGHT
plus 3
HIGHLIGHT plus 4
...or equals 13.
The correct answer is 13.
We know that what we have here is the smallest set of whole number coefficients because the greatest common factor of 1, 5, 3, and 4 is 1.
It’s not possible to divide all of them by a same number to get a set of smaller whole numbers.
CLICK PAUSE END RECORDING

4. Video ID: © 2008, Project VALUE (Video Assessment Library for Undergraduate Education), Department of Physical Sciences Nicholls State University Author: Glenn V. Lo Narrator: Funded by Louisiana Board of Regents Contract No. LA-DL-SELECT-13-07/08