Chapter 3 Formulas, Equations and Moles

Balancing Chemical Equations

What does mother nature lets you change and what doesn’t she?

H2 + O2 → H2O

2H2 + O2 → 2 H2O

Coefficients in black (bold) are changeable (numbers appearing before a substance)

Subscripts in bold red are not changeable: Changing the subscript changes the chemical that is formed or used and you can’t do that!

Example:

The combustion of butane:

CH3CH2CH2CH3 + O2 → CO2 + H2O

Remember: conservation of mass

Equivalent symbols: → (means “goes to”) can be replaced or is equivalent to a mathematical “ = ” sign

CH3CH2CH2CH3 + O2 = CO2 + H2O

Application of the conservation of mass for carbon and hydrogen results in:

CH3CH2CH2CH3 + O2 = 4CO2 + 5H2O

Balancing the oxygens:

CH3CH2CH2CH3 + 13/2O2 = 4CO2 + 5H2O or

2CH3CH2CH2CH3 + 13O2 = 8CO2 + 10H2O

Notice that we made an assumption here, that we had as much oxygen as we needed. In reality, this may not be the case. Therefore you also need to know the limiting reagent.

Limiting reagent: the reagent present in the least amount based on the stoichiometry of the reaction.

The stoichiometry of the reaction is the ratio of reagents that are needed for the balanced reaction:

2CH3CH2CH2CH3 + 13O2 = 8CO2 + 10H2O

Sometimes by changing the coefficients in front of the reactants, you can change the chemistry, but not control it except superficially. Coefficients are changed by changing the relative amounts of reagents.

In the case of the reaction we just looked at, if O2 is the limiting reaction, the product composition may change:

In a more limited amount of air (O2)

CH3CH2CH2CH3 + O2 = CO + H2O

CH3CH2CH2CH3 + O2 = 4CO + 5H2O

CH3CH2CH2CH3 + 9/2O2 = 4CO + 5H2O

2CH3CH2CH2CH3 + 9O2 = 8CO + 10H2O

Less than the required amount of oxygen will lead to incomplete combustion and a mixture of CO and CO2, carbon soot, etc.

C + H2O = CO + H2 water gas

CO + H2 + O2 = CO2 + H2O + heat

water gas.mov

K + H2O → KOH + H2

K + H2O → KOH + 1/2H2

2K + 2H2O → 2KOH + H2

Na + H2O → NaOH + H2

2Na + 2H2O → 2NaOH + H2

Li + H2O → LiOH + H2

2Li + 2H2O → 2LiOH + H2

movie MF Ch 6 07

Fermentation of glucose by yeast to yield ethyl alcohol

1. C6H12O6 → CO2 + CH3CH2OH

C6H12O6 → 2CO2 + 2CH3CH2OH

Fermentation of sucrose

2. C12H22O11 → CO2 + CH3CH2OH C12H22O11 → 4CO2 + 4CH3CH2OH ?

What else is present when we ferment grape juice or grains?

H2O

C12H22O11 + H2O → 4CO2 + 4CH3CH2OH

Atomic weight = sum of the number of protons and neutrons

Molecular weight = sum of the number of protons and neutrons that make up the molecule

Gram atomic weight = weight in grams of 6.02x 1023 atoms

Gram molecular weight = weight in grams of 6.02x 1023 molecules or formula units (used for for ionic solids, for example: NaCl)

A wonder drug?The classic white willow tree, Salix alba, provides more than just shade and shelter for natures animals.

Salicin is the key ingredient that is isolated from the tree and can be converted into one of our most reliable and heavily used drugs, aspirin!

10.1

White willow tree

It is also used as an analgesic-pain reliever, and as an anti-inflammatory agent.

Aspirin is used as an antipyretic-fever reduction.

salicyn

C

CC

C

CC CO2H

OHH

H

H

H

C

CC

C

CC CH2

OHH

H

H

H

OC

CC

C

CO

H

HO

OH

HOH

HH

H

What is the gram molecular weight of aspirin = C9H8O4 ?

1 molecule has a mass of 9*(mass of C)+8*(mass of H) + 4*(mass of O)

Mass of C atom*6.02x1023 = 12 g/mol

Mass of H atom*6.02x1023 = 1 g/mol

Mass of O atom*6.02x1023 = 16 g/mol

Total = 9*12+8+4*16 = 180 g /mol

Aspirin

C

CC

C

CC CO2H

OCOCH3

H

H

H

H

Aspirin is made by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3) according to the following reaction:

C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2HHow many grams of acetic anhydride (C4H6O3) are needed to react with 4.5 g of salicylic acid (C7H6O3) and how many grams of aspirin would be formed? gMW aspirin (C9H8O4)= 180 9*12+8*1+4*16 = 180 g/molgMW acetic anhydride (C4H6O3) = 4*12+6+3*16 = 102 g/molgMW salicylic acid (C7H6O3) = 7*12+6+3*16 = 138 g/molgMW acetic acid (CH3CO2H) = 2*12+4+2*16 = 60 g/mol

Note that 138 + 102 = 240; 180 + 60 = 240 ; mass is balanced

One last question! Is the reaction balanced?

Aspirin is made by reacting salicyclic acid (C7H6O3) with acetic anhydride (C4H6O3) according to the following reaction:

C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H

138 g/mol 102g/mol = 180g/mol 60g/mol

mass of reactants = mass of products

How many grams of acetic anhydride are needed to react with 4.5 g of salicylic acid and how many grams of aspirin would be formed?

4.5g/138g/mol = 0.0326 mol salicylic acid

Since the coefficients in front of both salicyclic acid, acetic anhydride and aspirin are 1, 0.0326 mol of aspirin should be theoretically formed;

this corresponds to x/180g/mol = 0.0326 mol or x = 5.87 g

5.87 is called the theoretical yield of salicylic acid.

Suppose you actually isolated 3.0 g when you ran this reaction, the actual yield would be 3.0/5.87x100=51%

How many grams of acetic anhydride are needed to react with 4.5 g of salicylic acid?

C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H138 g/mol 102g/mol 180g/mol 60g/mol

salicylic acid + acetic anhydride → aspirin + acetic acid

4.5g/138g/mol = 0.0326 mol C7H6O3

0.0326 mol C4H6O3 are also needed or 0.0326 mol x 102g/mol =3.33 g

Usually an excess of one reagent is used; the other reagent is called the limiting reagent

Glucose: C6H12O6

What is the % composition of glucose (the % C, H and O by mass)?

Suppose we had a mole of glucose, what is the molecular weight?

C = 6*12 = 72g/mol

H = 12*1 = 12g/mol

O = 16*6 = 96 g/mol

Total = 180 g/mol

C = 72/180 = 0.40 40%

H = 12/180 = 0.067 6.7%

O = 96/180 = 0.533 53.3%

Lets examine a series of reactions to make alum: KAl(SO4)2

.12H2OWhat is Alum?

The most common form, potassium aluminum sulfate, or potash alum, is one form that has been used in food processing. Another, sodium aluminum sulfate, is an ingredient in commercially produced baking powder. (Have you never noticed the faint metallic taste in baking powder? It comes from the alum.) The potassium-based alum has been used to produce crisp cucumber and watermelon-rind pickles as well as maraschino cherries, where the aluminum ions strengthen the fruits' cell-wall pectins.

Alum is approved by the U.S. Food and Drug Administration as a food additive, but in large quantities — well, an ounce or more — it is toxic to humans.

Alum's antibacterial properties contribute to its traditional use as an underarm deodorant. It has been used for this purpose. Today, potassium alum is sold commercially for this purpose as a "deodorant crystal," often in a protective plastic case.

Alum in powder or crystal form, or in styptic pencils, is sometimes applied to cuts to prevent or treat infection.

Powdered alum is commonly cited as a home remedy for canker sores.

Lets examine the following series of reactions to make alum

Al + KOH + H2O → KAl(OH)4 + H2 ↑

Let’s start by balancing the reaction:

Al and K are balanced; How about O and H? 2O vs 4O and 3H vs 6H

If we change the coefficient of KOH, the K is not balanced; changing the coefficient on H2O doesn’t effect anything else

Al + KOH + 3H2O → KAl(OH)4 + 1.5H2 ↑

2Al + 2KOH + 6H2O → 2 KAl(OH)4 + 3H2 ↑

Now suppose we add dilute sulfuric acid to this compound

KAl(OH)4 + H2SO4 → KAl(SO4)2.12H2O

KAl(OH)4 + 2H2SO4 → KAl(SO4)2.4H2O

Remember Alum is: KAl(SO4)2. 12H2O

KAl(OH)4 + 2H2SO4 +8H2O → KAl(SO4)2.12H2O

2Al + 2KOH + 6H2O → 2 KAl(OH)4 + 3H2 ↑ step 1

2 KAl(OH)4 + 4H2SO4 + 16H2O → 2KAl(SO4)2.12H2O step 2

_______________________________________________

2Al + 2KOH + 6H2O → 2 KAl(OH)4 + 3H2 ↑

KAl(OH)4 + 2H2SO4 +8H2O → KAl(SO4)2.12H2O

Overall the reaction is:

2Al +2KOH +22 H2O + 4H2SO4 = 2KAl(SO4)2.12H2O +3H2

This equations represents the overall stoichiometry of the reaction. It was generated by a process of mass balance, so that we can use this reaction tocalculate the overall yield of the reaction. Suppose we started with 0.5 g of Al and used an excess of all the remaining reagents. What would be the theoretical yield of alum (KAl(SO4)2

.12H2O)?

2Al +2KOH +22 H2O + 4H2SO4 = 2KAl(SO4)2.12H2O +3H2

Al = 27 g /molKAl(SO4)2

.12H2O = 474 g/molH2SO4 = 98g/mol

0.5g/27g/mol = 0.0185 mol AlHow many mol of Al is required to produce 2 mol of alum?

2 mol Al produces 2 mol of alumHow many mol of alum is produced from 0.0185 mol Al?

0.0185 mol alum; What is the theoretical yield?474g/mol x 0.0185 mol = 8.77 g alum