CHEM 1405 Concept Review: Chemical Quantities & Reactions

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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 Al₂(SO₄)₃:                            

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

 $Mass\:Percent\:of\:Element\:X=\frac{mass\:of\:element\:X\:in\:1\:mole\:of\:a\:compound}{molar\:mass\:of\:the\:compound}\times100\%$

 

Example: What is the mass percent of oxygen in Al₂(SO₄)₃?

 $Mass\:\%\:of\:O\:in\:Al_2\left(SO_4\right)_3=\frac{192.00\:g\:O}{342.17\:g\:Al_2\left(SO_4\right)_3}\times100\%=56.112\%$

 

 

Writing and Balancing Chemical Reaction Equations

To write an equation from a written description of a reaction, follow the following steps:

1.  Write the correct chemical formula for each substance indicated.

2.  Place the reactants on the left side of the equation with a plus sign between reactants if more than one is present.

3.  Draw an arrow pointing right (from reactants to products).

4.  Place the products on the right side of the equation with a plus sign between products if more than one is present.

 

            Balancing equations is simply applying the Law of Conservation of Mass.  The number of atoms of each type on the reactant (left) side MUST equal the number of atoms of each type on the product (right) side.  For a complete guide on balancing equations, check out the guide "How to Balance Any Equation,” available online at the TCC Learning Commons. 

            In addition to indicating how many molecules of each reactant will produce how many molecules of each product, the coefficients of a balanced equation indicate the mole ratios of between reactants and products. 

Example:  The equation "$2H_2\left(g\right)+O_2\left(g\right)\longrightarrow2H_2O\left(l\right)$" tells us that 2 moles of H₂ react with 1 mole of O₂ to form 2 moles of H₂O.

 

 

Stoichiometry

Stoichiometry is 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.  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?

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

IMPORTANT NOTE: Notice that the only way to go from one substance to another is a mole to mole ratio.

 

 

Limiting Reagents and Theoretical Yield

Limiting reactant: The reactant that is completely used up in a reaction.  The limiting reactant, since it runs out first, is the one that determines how much product can be formed.

Excess Reactant:  Any reactant that has some amount left over at the end of a reaction.

Theoretical Yield: The amount of product that could be formed under perfect laboratory conditions.  The theoretical yield is determined by and MUST be calculated using the limiting reagent.

 

Steps to Determine the Limiting Reactant, the Theoretical Yield, and the Excess Reactant(s) Left Over

Example:  Identify the limiting reagent if 30.0 g of sulfuric acid reacts with 25.0 g of aluminum hydroxide.

 

Step 1: Write the balanced chemical equation for the reaction, predicting products as needed.

$3H_2SO_4\left(aq\right)+2Al\left(OH\right)_3\left(aq\right)\longrightarrow Al_2SO_4\left(aq\right)+6H_2O\left(l\right)$

Step 2: Convert each reactant to the SAME product.  The reactant that produces less product is the one that runs out first, making it the LIMITING REAGENT.  The lower amount of product is the THEORETICAL YIELD.

Step 3: Convert the amount limiting reagent the problem starts with to the excess reagent needed to react with it.

Step 4: Subtract the amount of excess reagent needed to react with the limiting reagent from the amount of excess reagent originally present.

 

 

Percent Yield 

$\%\:Yield=\frac{Actual\:Yield\:\left(must\:be\:given\:in\:a\:problem\:or\:taken\:from\:experimental\:data\right)}{Theoretical\:Yield\:\left(Calculated\:from\:the\:limiting\:reagent\right)}\times100\%$

Example:  What is the percent yield for the above problem if 9.87g H₂O are actually produced in the reaction?

$\frac{9.87\:g\:H_2O}{11.0\:g\:H_2O}\times100\%=89.7\%$

 

 

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: