1. Molar Quantities: Balancing Equations
Week 4
Molar Quantities:
Balancing Equations
The subject of Unit Week 4 is the use of quantitative methods for measuring finite amounts of chemical compounds and mixing them together in order to make other compounds or liquid solutions.

2. each side of the equation.
Balancing Equations
CH4 + O2 = CO H2O
Count the atoms on
each side of the equation.
Reactants: 1 C ; 4 H ; 2 O
Products: 1 C ; 2 H ; (2 + 1) = 3 O
Let’s begin by balancing the equation for a simple chemical reaction.
In order to do this, we must count the total number individual atoms contained within reactants and products.
Since this is an algebraic equation, the goal is to make sure that the equation is balanced: in other words, that there is the same number of atoms on each side of the equation.
Note that in the reactants, there is 1 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms.
In the products, there is 1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms.
Note also that 2 of those oxygen atoms are found in the CO2, while 1 oxygen atom comes from the water H20 molecule.

3. Counting H atoms CH4 + O2 = CO2 + 2 H2O Reactants: 1 C ; 4 H ; 2 O
Products: 1 C ; 2 H ; (2 + 1) = 3 O
Double the H in the products:
CH4 + O2 = CO H2O
Products: 1 C ; 4 H ; (2 + 2) = 4 O
Now since there are twice as many H atoms in the reactants than in the products, it is necessary that I double the number of H atoms in the products.
This is most readily accomplished by doubling the number of water molecules in the products.
This is done by placing a coefficient of 2 in front of the H20.

4. Counting O atoms CH4 + 2 O2 = CO2 + 2 H2O Reactants: 1 C ; 4 H ; 2 O
Products: 1 C ; 4 H ; (2 + 2) = 4 O
Double the O in the reactants:
CH O2 = CO H2O
Reactants: 1 C ; 4 H ; 4 O
Products: 1 C ; 4 H ; 4 O
Now since there are twice as many O atoms in the products than in the reactants, it is necessary to double the number of O atoms in the products.
This is done by doubling the number of oxygen O2 molecules in the reactants.
This is done by placing a coefficient of 2 in front of the O2.
Now the equation is balanced.

5. Molar Coefficients CH4 + 2 O2 = CO2 + 2 H2O for this reaction are:
So what is the equation actually saying ?
These molar coefficients are indicating that chemical compounds always react and combine in specific whole number ratios.
The ratios for this particular reaction are 1 : 2 : 1 : 2.

6. Molar coefficients are used for calculations.
1 CH O2 = 1 CO H2O
1 mole of methane gas (CH4) combines with
2 moles of oxygen gas (O2) to produce
1 mole of carbon dioxide (CO2) gas and
2 moles of water vapor (H2O).
In other words, 1 molecule of methane reacts with 2 molecules of oxygen in order to produce 1 molecule of carbon dioxide and 2 molecule of water vapor.
So we need twice as many molecules of oxygen in relation to the number of methane molecules -- because each methane molecule is going to consume 2 molecules of oxygen gas during the reaction.
Similarly, the reaction will produce twice as many water molecules as it will carbon dioxide molecules.
It could just as easily be said that 1 mole of methane reacts with 2 moles of oxygen in order to produce 1 mole of carbon dioxide and 2 mole of water vapor. Similarly, the reaction will produce twice as many moles of water as it will moles of CO2.

7. Mole Ratios CH4 + 2 O2 = CO2 + 2 H2O
What if we had only 0.5 moles of CH4 ???
How much CO2 would be produced ??
CH O2 = CO H2O
(0. 5 moles CH4) x [ 1 mole CO2 / 1 mole CH4 ]
= 0.5 moles CO2
* Note: Moles CH4 cancel out.
Mole ratios are used like conversion factors in order to calculate chemical quantities as follows. Given a known quantity of one reactant or product, mole ratios are used in order calculate the corresponding quantity of any other reactant or product.
For example, if there is only 0.5 mole of CH4, and we wish to determine the corresponding quantity of CO2 which will be produced during the reaction, make a ratio of molar coefficients to use as a conversion factor.
Since we will start with (# moles CH4), then (# moles CH4) will need to be in the ratio’s denominator, so that it will cancel out. Similarly, (# moles CO2) will need to be in the ratio’s numerator, so that it will be included in the answer.
The ratio of 1:1 tells us that for each mole of CH4 consumed, 1 mole of CO2 will be produced. Similarly, for each 0.5 mole of CH4 consumed, 0.5 mole of CO2 will be produced.

8. Mole Ratios What if we had only 0.5 moles of CH4 ???
How much H20 would be produced ??
CH O2 = CO H2O
(0. 5 moles CH4) x [ 2 mole H2O / 1 mole CH4 ]
= 1.0 moles H2O
*Note: Moles CH4 cancel out.
So what if we wanted to know how much H20 will be produced from 0.5 moles of CH4 ? Again, since we begin with (# moles CH4), then (# moles CH4) will need to be in the ratio’s denominator, so that it will cancel out.
Similarly, (# moles H2O) will need to be in the ratio’s numerator, so that it will be included in the answer. The ratio of 2:1 tells us that for each mole of CH4 consumed, 2 mole of H2O will be produced (or twice as much water as methane).
Thus, for each 0.5 mole of CH4 consumed, 1.0 moles of CO2 will be produced. Similarly, if we had 10 moles of CH4, then 20 moles of water would be produced.

9. The Mole Because atoms are so small, the normal units
of mass --  the gram and the kilogram are much too large to be convenient.
One mole of something consists of x units of that substance.
One mole of carbon contains
6.02 x atoms of carbon.
One mole of water contains
6.02 x molecules of water.
Note that in the measurement of chemical quantities, we are not using a scale of mass. This is due to the fact that chemical compounds always react and combine in specific whole number ratios – molecule by molecule -- not gram by gram. In order to introduce the concept of molar mass, we need to understand the definition of the mole.
Because atoms and molecules are so small, the normal units of mass --  the gram and the kilogram -- are much too large to be convenient. The mole is simply a shorthand way of referring to a very large number. The concept is the same as that of a dozen, which represents the number 12. But in this case, there are far more atoms in one gram of substance than a dozen, or 12. In fact, there are something on the order of 10^22 power or 10^23 power. Thus, we use Avogadro’s number, which is equal to 6.0 x 10^23 power, in order to define how many particles (either atoms or molecules) there are in 1 mole of substance.

10. Molar Mass
In this context, the molar mass is defined as the amount of mass in 1 mole of that substance. The molar masses of the elements are given in the Periodic Table of the Elements. The numbers are located just below the chemical symbol for each particular element. A few of the elements and their respective molar masses are listed in the table shown in this slide.

11. Molar Mass of Compounds
Molar Mass (or molecular weight ) of a
molecule or compound is obtained by summing the molar masses
of the atoms of the compound.
Molar masses of the elements
are located below the element’s symbol
in the Periodic Table of the Elements.
The molar mass (or molecular weight) of a molecule or compound is obtained by summing (or adding together) the molar masses of each of the individual atoms of the compound.
Again, the molar masses of the elements are given in the Periodic Table of the Elements. The numbers are located just below the chemical symbol for each particular element.

12. 1 mole of carbon C atoms and
Molar Mass: CH4
1 mole of CH4 molecules contains :
1 mole of carbon C atoms and
4 moles of hydrogen H atoms.
Mass of 1 mol of methane:
Sum of the masses
C atoms and H atoms.
1 mol C = 1 x g  =  g
4 mol H = 4 x g  =  g
                 1 mol CH4   =  g
For example, the molar mass of methane, CH4, is determined adding together the molar masses of 1 carbon atom and 4 hydrogen atoms.
The carbon atom has a molar mass of 12 grams, and the hydrogen atom has a molar mass of 1 atom. So the total molar mass of the CH4 methane molecule is 16 grams.
In other words, there are 16 grams in 1 mole of methane gas.
Another way of saying this is that 6 x 10^23 molecules of methane CH4 weighs in at 16 grams.

13. Mole-Mass Calculations
What if I do not know how many moles of
reactant I have ? What if, instead, I have:
200 g of CH4 = ???? moles CH4
In a mole-mass calculation, we have the
added burden of converting mass to moles.
For this, we need the molar masses of
the reactants and/or products.
Now: What if we do not know how many moles of reactant are present in the reaction?
What if, instead, we know that there are 200g of CH4 present?
After all, in the laboratory, we do not have scales which measure chemical compounds in moles. In the chemical laboratory, we measure chemical quantities in grams. The problem is that they react mole by mole – not gram by gram.
Thus, in a typical mole-mass calculation, we have the added burden of converting mass to moles.
For this, we need the molar masses of the reactants and/or products.

14. Mole-Mass Calculations
The steps for performing a successful mole-mass calculation are as follows.
First we make sure the equation for the reaction is properly balanced with the appropriate molar coefficients.
Then, using the masses that we are given, and using the proper molar masses for each component, we convert the masses of reactants or products to moles.
Next, we use the molar coefficients from the balanced equation in order to set up the necessary mole ratios.
We then use these mole ratios in order to calculate the number of moles of the desired reactant or product.
Finally, if we wish to report the final answer in grams, then we convert our answers from moles to grams by using the proper molar masses.

15. Mass CH4 – Moles CH4 What if I have: 200 g of CH4 ?
How many moles of CH4 is that ?
Molar mass of CH4 = g / mol
( 200 g ) x ( 1 mol / 16.0 g )
= mol CH4
*Note: Inversion of the molar mass
Note that in converting from mass to moles, it is necessary to invert the molar mass. Since we will be starting with the mass in grams, the molecular weight in grams must in the denominator, so that these terms cancel. With moles of substance in the numerator, the answer will be in moles.
For example, 200 grams of methane has a molar mass of 16.0 grams per mole. But if we multiply those two terms together, the units will not cancel properly. But if we invert the molar mass to mole per grams (1 mole per 16 grams) then the grams will cancel out, and we are left with 12.5 moles of CH4.

16. Moles CH4 – Moles CO2 will be produced by 12.5 moles of CH4 ??
How many moles of CO2
will be produced
by 12.5 moles of CH4 ??
CH O2 = CO H2O
(12.5 moles CH4) x [ 1 mole CO2 / 1 mole CH4 ]
= moles CO2
Then if we want to know how many moles of CO2 will be produced by those 12.5 moles of CH4, we use the mole ratio 1 : 1 for CO2 : CH4 to determine that 12.5 moles of CO2 will be produced.

17. Moles CO2 – Mass CO2
How many grams in 12.5 moles of CO2 ? Molar mass of CO2: Mass of 1 mol C = 1 x 12.0 g = 12.0 g Mass of 2 mol O = 2 x 16.0 g = 32.0 g Mass of 1 mol CO2 = 44.0 g ( 12.5 mol ) x ( 44.0 g / mol ) = 550 g CO2
And if we wish to report the answer in grams, then we need to determine the molar mass of CO2. 1 carbon atom weighs 12 grams per mole, and each oxygen atom weighs 16 grams per mole. So each mole of CO2 weighs in at ( ) = 44 grams. Then 12.5 moles of CO2 weighs in at (12.5 x 44) = 550 grams.
In summary, 200 grams of methane gas could theoretically produce 550 grams of CO2 gas. Of course, this result assumes that there is sufficient oxygen present in the reaction area.

18. Heats of Reaction CH4 + 2 O2 = CO2 + 2 H2O
Don’t forget about the most important part
of any fuel combustion reaction such as :
CH O2 = CO H2O
It produces heat !!
It is well worth noting here that one of the main reaction product of any reaction such as this – the combustion of a hydrocarbon fuel in the presence of oxygen – will be heat.
In other words, a certain amount of heat is liberated for each mole of reactant consumed and each mole of product produced.

19. Exothermic Reaction
Reactions like this one which liberate heat are called exothermic reactions, as evidenced in this plot of energy versus progress of reaction.
One observation that can be made from this plot is that the amount of potential energy stored within the reactants is higher than that of the products. That same thing would be true for any hydrocarbon fossil fuel. The products contain less energy, and the difference between the 2 lies in the heat or thermal energy which is released during the reaction.

20. Exothermic Reaction
The other factor evidenced by this plot is the hill or bump on top which represents the activations energy. Reactions such as this do not occur spontaneously, but instead they require a boost or kick-start of some sort in order get the reaction going. This is typically comes in the form of heat. But once the reaction starts, it is much like lighting a fire in the woods. It will continue to its completion until all of the wood fuel has turned to ash.
Notice that the quantity on the plot defined as the heat of reaction is equal to the difference between the heat content of the reactants and the heat content of the products.

21. Endothermic Reaction
Notice the difference in appearance in the plot of on endothermic reaction in which heat is absorbed. In this case, the heat of reaction is still defined as the difference in heat content between reactants and products. But in this case, it has a different sign.
In an exothermic reaction where heat is liberated, the heat of reaction will be negative.
In an endothermic reaction where heat is absorbed, the heat of reaction is positive.

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