CHEM 1411 Concept Reviews: Chemical Quantities & Aqueous Reactions

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Mole Conversion Map

 

Stoichiometry:  The field of study that examines the quantities of substances consumed and produced in chemical reactions.

 

 

Limiting Reagents and Theoretical Yield

When two substances react, there will nearly always be extra of one or the other substance left over unreacted.  This reactant is called the excess reactant.  The limiting reactant is the one that runs out first and is therefore completely used up in the reaction.  As such, the limiting reactant is the one that determines how much product can be formed.  Converting the amount of limiting reactant to the amount of product expected to be produced gives you a quantity known as the theoretical yield.  To determine both the limiting reactant AND the theoretical yield, simply convert each amount of reactant to the same product.  Whichever reactant gives you a LOWER amount of product is your limiting reactant, and the amount of product calculated from the limiting reactant will be your theoretical yield, which can be used for other calculations, including percent yield.

 

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

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

$30.0\:g\:H_2SO_4\times\left(\frac{1\:mol\:H_2SO_4}{98.08\:g\:H_2SO_4}\right)\times\left(\frac{6\:mol\:H_2O}{3\:mol\:H_2SO_4}\right)\times\left(\frac{18.02\:g\:H_2O}{1\:mol\:H_2O}\right)=11.0\:g\:H_2O$

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

Since 25.0 g Al(OH)₃ can produce 17.3 g H₂O but 30.0 g H₂SO₄ can only produce 11.0 g H₂O, the H₂SO₄ reagent will run out first.  Therefore, 30.0 g H₂SO₄ is the LIMITING REAGENT, and the 11.0 g H₂O that could be produced is the THEORETICAL YIELD.

 

 

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?

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

 

 

 

Solutions and Electrolytes

Solution:  A homogeneous mixture of 2 or more substances.  A solution is composed of a solvent and solute(s).

Solute:  The dissolved substance in a solution.

Solvent:  The dissolving media of a solution.  The solvent breaks down and assimilates solutes into itself.  The most common solutions have water as a solvent.  Solutions with water as a solvent are called aqueous.

Electrolytes:  Substances that form ions when dissolved in solution.  Electrolytes can be weak or strong.

                Strong Electrolytes:          Substances that completely separate into their component ions when dissolved.

                                                                (All soluble ionic compounds and strong acids are strong electrolytes.)

                Weak Electrolytes:            Substances that exist in solution mostly as neutral molecules, with only a small

                                                                fraction separating into ions.  (These are mostly weak acids and bases.)

Nonelectrolytes:  Substances that do NOT form ions when dissolved in solution. These are usually molecular.

 

 

 

Metathesis Reactions

Metathesis (Exchange) Reactions:  Reactions that involve the exchange or “changing out” of certain ions for others.  These reactions are also known as “double replacement” reaction.   Positive ions switch places and get new “partners” or negative ions.  The new products are formed based on the charges of the ions being put together.
General Form:   $AB+CD\longrightarrow AD+CB$ 

 

For a metathesis reaction to occur, there must be some net change to the system (i.e. the removal of ions from the system).  There are four conditions under which a metathesis reaction will be “driven forward”:

1. The formation of a gas. (gas evolution reaction)                 Example:  $2HCl\left(aq\right)+Na_2S\left(aq\right)\longrightarrow2NaCl\left(aq\right)+H_2S\left(g\right)$
2. The formation of a liquid. (acid-base neutralization rxn)   Example:  $HCl\left(aq\right)+NaOH\left(aq\right)\longrightarrow H_2O\left(l\right)+NaCl\left(aq\right)$
3. The formation of a solid. (precipitation reaction)          Example:  $K_2CO_3\left(aq\right)+MgI_2\left(aq\right)\longrightarrow2KI\left(aq\right)+MgCO_3\left(s\right)$
4. The formation of a weak electrolyte.                              Example:  $NaF\left(aq\right)+HBr\left(aq\right)\longrightarrow HF\left(aq\right)+NaBr\left(aq\right)$

 

 

 

How to Write a Net Ionic Equation

1. Write and balance the molecular equation (predicting products if not given).

                Example:  $K_2CO_3+MgI_2\longrightarrow2KI+MgCO_3$

 

2. Determine the phases of each substance in the reaction. For ionic substances, use the solubility rules (p. 121 in the textbook) to determine if the substance is soluble (aq) or insoluble (s).

                Example:  $K_2CO_3\left(aq\right)+MgI_2\left(aq\right)\longrightarrow2KI\left(aq\right)+MgCO_3\left(s\right)$

 

3. Write the complete ionic equation by breaking up any aqueous strong electrolytes (see above definition) into separated ions. The phases of all separated ions should be aqueous. 
   Note: Remember to include the charges of the ions and do NOT break apart polyatomic ions.

                Example:  $2K^+\left(aq\right)+CO_3^{\:2-}\left(aq\right)+Mg^{2+}\left(aq\right)+2I^-\left(aq\right)\longrightarrow2K^+\left(aq\right)+2I^-\left(aq\right)+MgCO_3\left(s\right)$

 

4. Cancel out any ions that are present in the same form on both the product side and the reactant side of the equation. These ions are called spectator ions since they are not involved in the reaction.  Rewrite what’s left.

                Example:  $Mg^{2+}\left(aq\right)+CO_3^{\:2-}\left(aq\right)\longrightarrow MgCO_3\left(s\right)$

 

 

 

Oxidation-Reduction (Redox) Reactions

Oxidation: An increase in oxidation number (This is caused by a substance losing electrons in a reaction.)

Reduction: A decrease in oxidation number (This is caused by a substance gaining electrons in a reaction.)

 

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.

 

 

 

Molarity and Dilution

When substances are dissolved in solution, it is helpful to know the concentration or how much of a substance is dissolved per unit of volume.  While there are several different ways to measure concentration, one of the most useful for chemistry is Molarity, which can be calculated using the following equation:

$Molarity\:of\:A=\frac{moles\:of\:A}{liters\:of\:solution}$   or in abbreviated form...     $M=\frac{mol}{L}$

 

The above relationship can be used to derive the Dilution Equation shown below:

$M_{concentrated}\times V_{concentrated}=M_{diluted}\times V_{diluted}$