CHEM 1406 Concept Review: Chemical Quantities & Reactions

Back to Chemistry 1406 Concept Reviews

The Mole

            The mole is a central concept in chemistry.  Moles tell us the amount of something.  Specifically, 1 mole of an element contains 6.022*10²³ atoms of that element.  1 mole of a compound contains 6.022*10²³ molecules of that compound.  The number 6.022*10²³ is known as Avogadro’s number.  1 mole of a substance also has a mass equal to the formula mass number of grams.  (Example: 1 mole of Cu = 63.55 g Cu)

 

Molar Mass of a Compound:  The molar mass of a compound is the mass of 1 mole of that compound.  Molar mass is calculated by taking the sum of all the molar masses of each atom in the compound.  See the following example on how to calculate molar mass:

Example:  Find the molar mass of $Al_2SO_4\left(aq\right)$:     Al: 2 x 26.98 =       53.96

                                                                                  S: 3 x 32.07 =       96.21

                                                                                 O: 12 x 16.00 =   192.00

                                                                                                               342.17 g/mol

 

In addition, the coefficients of a balanced equation can tell you the mole ratio of one substance to another. 

Example:  The equation $2H_2\left(g\right)+O_2\left(g\right)\:\longrightarrow\:2H_2O\left(l\right)$  tells us that 2 moles of H₂ react with 1 mole of O₂ to form 2 moles of H₂O.

 

All of these relationships allow us to convert between grams and moles of different substances.  The use of calculations that use molar mass and mole ratios in balanced chemical reactions to calculate different amounts of reactants or products is referred to as Stoichiometry.  A summary and map of these relationships and how they can be used to convert from one quantity to another is below:

 

 

 

Example Stoichiometry Calculations

Example equation:       $3H_2SO_4\left(aq\right)+2Al\left(OH\right)_3\:\longrightarrow\:Al_2\left(SO_4\right)_3\left(aq\right)+6H_2O\left(l\right)$

 

Moles to Moles:  How many moles of water can be produced from 5.40 moles of H₂SO₄?

$5.40\:mol\:H_2SO_4\times\left(\frac{6\:mol\:H_2O}{3\:mol\:H_2SO_4}\right)=10.8\:mol\:H_2O$

 

Moles to Mass:  How many grams of water can be produced from 5.4 moles of H₂SO₄?

$5.40\:mol\:H_2SO_4\times\left(\frac{6\:mol\:H_2O}{3\:mol\:H_2SO_4}\right)\times\left(\frac{18.02\:g\:H_2O}{1\:mol\:H_2O}\right)=195\:g\:H_2O$

 

Mass to Moles:  How many moles of water can be produced from the reaction of 25.0 g Al(OH)₃?

$25.0g\:Al\left(OH\right)_3\times\left(\frac{1\:mol\:Al\left(OH\right)_3}{78.01\:g\:Al\left(OH\right)_3}\right)\times\left(\frac{6\:mol\:H_2O}{2\:mol\:Al\left(OH\right)_3}\right)=0.961\:mol\:H_2O$

 

Mass to Mass:  How many grams of water can be produced from the reaction of 25.0 g Al(OH)₃?

$25.0\:g\:Al\left(OH\right)_3\times\left(\frac{1\:mol\:Al\left(OH\right)_3}{78.01\:g\:Al\left(OH\right)_3}\right)\times\left(\frac{6\:mol\:H_2O}{2\:mol\:Al\left(OH\right)_3}\right)\times\left(\frac{18.02\:g\:H_2O}{1\:mol\:H_2O}\right)=17.3\:g\:H_2O$

 

 

 

Types of Reactions

Synthesis (Combination): Two or more reactants (often elements) combine to form a single product. 

General Form:   $A+B\:\longrightarrow\:C$                                      Example:   $2Na\left(s\right)+Cl_2\left(g\right)\:\longrightarrow\:2NaCl\left(s\right)$

 

Decomposition:  A single reactant breaks down into multiple products.

General Form:   $C\:\longrightarrow\:A+B$                                      Example:   $CaCO_3\left(s\right)\:\longrightarrow\:CaO\left(s\right)+CO_2\left(g\right)$

 

Combustion:  Rapid reactions that produce a flame.  Generally, they will usually involve a hydrocarbon (C_xH_y) reacting with oxygen (O₂) to produce water (H₂O) and carbon dioxide (CO₂).

General Form:   $C_xH_y+O_2\:\longrightarrow\:CO_2+H_2O$                Example:   $C_3H_8\left(g\right)+5O_2\left(g\right)\:\longrightarrow\:3CO_2\left(g\right)+4H_2O\left(l\right)$

 

Single Replacement:  A type of oxidation-reduction reaction where one element displaces another. 

General Form:    $A+BC\:\longrightarrow\:AC+B$             Example:   $2Al\left(s\right)+3CuCl_2\left(aq\right)\:\longrightarrow\:2AlCl_3\left(aq\right)+3Cu\left(s\right)$

 

Double Replacement:  A type of reaction where two compounds exchange ions to form two new compounds.  This will be discussed further in Chapter 4.

General Form:   $AB+CD\:\longrightarrow\:AD+CB$           Example:   $K_2CO_3\left(aq\right)+MgI_2\left(aq\right)\:\longrightarrow\:2KI\left(aq\right)+MgCO_3\left(s\right)$

 

 

Oxidation Reduction Reactions

Oxidation-Reduction (Redox) Reaction:  Any reaction that involves the exchange of electrons between atoms.

Single replacement reactions and Combustion reactions are always oxidation-reduction reactions.  Synthesis and Decomposition reactions are usually redox reactions. 

Oxidation: An increase in oxidation number (This is caused by a substance losing electrons in a reaction.)  In biological systems, oxidation can be described as either a gain of oxygen or a loss of hydrogen in a molecule.

Reduction: A decrease in oxidation number (This is caused by a substance gaining electrons in a reaction.) In biological systems, reduction can be described as either a gain of hydrogen or a loss of oxygen in a molecule.

 

Simplified Rules for Determining Oxidation Number

1.  Hydrogen always has an oxidation number of +1 when bonded to non-metals and -1 when bonded to metals.

2.  Fluorine has an oxidation number of -1 in ALL compounds.

3.  Oxygen nearly always has an oxidation number of -2. The only major exception is peroxides, in which the oxidation number is -1.

4.  The oxidation number of any element in its elemental state is 0. Examples:  O as O₂(g), S in S₈(s), Al(s) by itself, etc.

5.  In any ionic compound, the oxidation number of any monatomic ion (single-atom ion) is the same as its charge.

6.  The sum of the oxidation numbers of all atoms within a compound must add up to the overall charge of the compound.

Example: