Summer Assignment



But there is still work to do…

Informational Websites

Below you will find some very helpful website that I advise you to work on in your spare time. These sites are interactive and will provide you with some tutorial support during the summer.

1. AP Chemistry Wiki Pages (This one will be very helpful. All of the class notes and instructional videos will be found here.)



2. Dimensional Analysis

Tutorial: 

3.  Ionic Nomenclature

Tutorial: 

4. Electron Configuration

Tutorial: 

5. Balancing Chemical

Equations:  

6. Redox Reaction Tutorials:

  

Scientific Notation, Significant Figures, and the

Factor-Label Method of Solving Problems

Scientific Notation

Scientific notation (Chang, p. 21) is a type of exponential notation in which only one digit is kept to the left of the decimal point. Example: 8.4050 x 10-8.

Significant Figures

It is reasonable that a calculated result can be no more precise than the least precise piece of information that went into the calculation. Thus it is common practice to write numbers in scientific notation with only the last place containing any uncertainty. When we do this we are keeping only the “significant figures” (Chang, pgs 23-24).

To determine the number of significant figures in a number, you read the number from left to right and count all digits starting with the first non-zero digit. Do not count the exponential part.

Thus the number 0.002050 contains 4 significant figures and is written in scientific notation as 2.050 x 10-3. The trailing zeros in a non-decimal number such as 1200 may or may not be significant: the number may be written as 1.2 x 102, 1.20 x 102 or 1.200 x 102 depending on whether it has 2, 3, or 4 significant figures.

Significant Figures in Derived Quantities

When doing calculations, you should use all the digits allowed by your calculator in all intermediate steps. Then in the final step, round off your answer to the appropriate number of “significant figures” such that only the last decimal place contains any uncertainty. You do this by following the rules:

• When adding or subtracting, first express all numbers with the same exponent. Then the number of decimal places in the answer should be equal to the number of decimal places in the number with the fewest decimal places.

• In multiplication or division, the number of significant figures in the answer should be the same as that in the factor with the fewest significant figures.

When using these rules, assume that exact numbers have an infinite number of significant figures (or decimal places). For example, there are exactly 12 inches in one foot.

Solving Problems Using Dimensional Analysis: The Factor-Label Method

Units may be used as a guide in solving problems. First decide what units you need for your answer. Then determine what units you are given in the problem, and what conversion factors will take you from the given units to the desired units. If the units cancel out properly, chances are that you are doing the right thing! The basic set up is

[pic]

Conversion factors are added until the new units are the same as the units desired. Each conversion factor has a denominator equivalent to the numerator but in different units.

Assignment # 2

1. Carry out the following mathematical operations and express your answer in scientific notation using the proper number of significant figures.

(a) (4.28 x 10-4) + (3.564 x 10-2)

(b) (0.00950) x (8.501 x 107) [pic] 3.1425 x 10-11

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2. Use the factor-label method to solve the following problem. Show your work, and give your answer in scientific notation using the proper number of significant figures.

The calorie (1 cal = 4.184 J) is a unit of energy. The burning of a sample of gasoline produces 4.0 x 102 kJ of heat. Convert this energy to calories. (103 J = 1 kJ.)

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3. Use the factor-label method to solve the following problem. Show your work, and give your answer in scientific notation using the proper number of significant figures.

The distance from the sun to the earth is 93 million miles. How many minutes does it take light from the sun to reach earth?

Useful information: 1 km = 0.6214 mile, c = speed of light = 3.00 x 108 m/s)

Atomic Mass, Moles, and the Periodic Table

Atomic Mass and Molar Mass

Isotopic masses cannot be obtained by summing the masses of the elementary particles (neutrons, protons, and electrons) from which the isotope is formed. This process would give masses slightly too large, since mass is lost when the neutrons and protons come together to form the nucleus.

Atomic masses (also called atomic weights) are thus assigned relative to the mass of a particular carbon isotope, [pic], which is assigned the mass of 12 amu exactly. Likewise 1 mole of [pic] has a mass of exactly 12 g. Atomic masses and molar masses of other isotopes are calculated based on their mass relative to that of Carbon-12.

Masses of “average” atoms are found by summing isotopic masses, weighting each isotopic mass by its abundance (see Chang, p. 76). Thus one “average” C atom has a mass of 12.01 amu, and the mass of 1 mole of “average” carbon atoms has a mass of 12.01 g. These average masses are what are given on the periodic chart.

What is a Mole?

Since atoms and molecules are so tiny, it is convenient to talk about a large number of them at a time. The chemical counting unit is known as the mole. A mole is defined as the amount of substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the [pic] isotope. It has been found experimentally that

[pic]

This value is known as Avogadro’s number. Just like 1 dozen of anything always contains 12 items, 1 mole of anything always contains 6.022 x 1023 items.

Molecular Masses and Compound Masses

Molecular masses are found by summing atomic masses (see Chang, pp. 81-82). They are often called molecular weights. Thus the mass of 1 mole of water, H2O, would be 2 x (molar mass of H) plus 1x (molar mass of O) or [(2 x 1.008 g) + (1 x 16.00 g)] = 18.02 g.

Ionic compounds such as NaCl do not contain molecules. Their formulas give the relative numbers of each kind of atom in the sample. What we mean by the molar mass (or the molecular weight) of an ionic compound is really the formula weight. The formula weight is the sum of the atomic masses in the formula.

Percent Composition of Compounds

The percent composition by mass is the percent by mass of each element in a compound. If there are n moles of an element per mole of compound, the percent by mass of the element is calculated using the equation,

[pic]

The sum of the % compositions of all elements in a compound is 100%.

Assignment # 3

Exercises

1. The atomic mass scale gives masses in atomic mass units (amu) relative to the mass of carbon-12.

(a) What is the mass of one 12C atom in atomic mass units (amu)? ____________

(b) What is the mass of an average C atom in atomic mass units (amu)? ____________

(c) What is the mass of an average Cl atom in amu? ____________

(d) What is the mass of an average Br atom in amu? ____________

2. The molar mass scale gives masses in grams (g) relative to the mass of 12C.

(a) What is the mass in grams of 1 mole (mol) of 12C? ___________

(b) What is the mass in grams of 1 mole (mol) of carbon? ___________

(c) What is the mass in grams of 1 mole (mol) of Cl? ___________

(d) What is the mass in grams of 1 mole (mol) of Na? ___________

3. How many 12C atoms are present in a mole of 12C ?

4. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H10O.

(a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol.

(b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g?

Assignment # 4

Naming Inorganic Compounds

To name a compound you must first decide whether the substance is an ionic or molecular compound. Ionic compounds are easily recognized since they usually contain both metallic and non-metallic elements. The most common exception to this rule are ionic compounds containing the ammonium ion, NH4+, such as (NH4)2CO3 or NH4Br which contain no metal ions. Molecular compounds typically contain only non-metallic atoms (and metalloids).

Conventions for naming ionic compounds are given. To successfully follow the rules, however, you must be first learn the names of common ions. Names of ionic compounds do not give the number of each type of ion in the formula: the chemist is supposed to be able to figure that out from his/her knowledge of ion charges and the requirement that salts be neutral (and thus have a sum of zero for the ion charges in the formula).

Binary compounds of the non-metals are named following the guidelines given in Chang on pp. 62-64. Note that when naming these molecular compounds, the number of atoms of a given type is commonly indicated with a prefix (di-, tri-, tetra, etc.).

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Exercises

1. Complete the following chart of corresponding ion names and formulas.

|Cation Name |Formula | |Anion Name |Formula |

|(1) potassium ion | | |(11) nitrate ion | |

|(2) |Fe3+ | |(12) |H2PO4- |

|(3) ammonium ion | | |(13) hydrogen carbonate (or | |

| | | |bicarbonate) ion | |

|(4) |Ba2+ | |(14) |MnO4- |

|(5) silver ion | | |(15) perchlorate ion | |

|(6) |Cu2+ | |(16) |S2- |

|(7) zinc ion | | |(17) acetate ion | |

|(8) |Co2+ | |(18) dichromate ion | |

|(9) hydrogen ion | | |(19) |CO32- |

|(10) chromium(III) ion | | |(20) sulfite ion | |

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2. Complete the following chart of corresponding compound names and formulas. Circle the names of all non-ionic (i.e., molecular) compounds.

|Compound Name |Formula | |Compound Name |Formula |

|(1) silver nitrate | | |(11) sodium hydrogen phosphate | |

|(2) |Ni(CH3CO2)2 | |(12) |SO3 |

|(3) ammonium sulfate | | |(13) potassium permanganate | |

|(4) |P2O5 | |(14) |Al2S3 |

|(5) sodium oxide | | |(15) cobalt(III) sulfate | |

|(6) |NH4NO3 | |(16) |Ag2CrO4 |

|(7) nitrogen trichloride | | |(17) |SrF2 |

|(8) |NaHCO3 | |(18) sulfur hexafluoride | |

|(9) iron(II) acetate | | |(19) |NH3 |

|(10) carbon tetrachloride | | |(20) |LiClO3 |

Assignment # 5

Empirical and Molecular Formulas

The empirical formula of a compound gives the simplest whole number ratio of different types of atoms in the compound. All salt formulas are empirical formulas. On the other hand, the molecular formula of a compound may or may not be the same as its empirical formula. For example, the molecular formula of butane is C4H10 while its empirical formula is C2H5. The molecular formula gives the true number of each kind of atom in a molecule.

Empirical formulas may be easily determined from experimental data.

• Usually you must first determine how many grams of each type of atom are in the compound. If percent composition data is given, assume that you have 100.0 g of the compound; then the number of grams of each element is equal to the percentage for that element.

• The next task is convert the grams of each element to moles of the element. Be sure to keep at least three significant figures in your answers.

• The final step is to write the molar amounts of each element as subscripts in the formula. Then divide all molar subscripts by the smallest value in the set. At this point, the subscripts may all be very close to whole numbers; if so, you are finished. If one (or more) of the subscripts is not close to a whole number, multiply all molar subscripts by the simple factor which makes all subscripts whole numbers.

Once the empirical formula is determined, the molecular formula is easily found if the molar mass (molecular weight) of the molecule is also known. You first calculate the molar mass of the empirical formula. Then you divide the molar mass of the molecule by the molar mass of the empirical formula. The division should give a simple whole number. That number is the factor by which all subscripts in the empirical formula must be multiplied to obtain the molecular formula.

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Exercises

1. The molecular formula of the antifreeze ethylene glycol is C2H6O2. What is the empirical formula?

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2. A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound?

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3. Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One of them is a blue solid that comes apart, reversibly, in the gas phase. It contains 36.84% N. What is the empirical formula of this oxide?

______________________________________________________________________________

4. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine. What is the empirical formula of the indium compound?

Types of Chemical Reactions

One skill that chemists learn over time is that of writing and balancing equations. The first task is deciding what type of reaction is taking place. In this chapter we study three types:

Precipitation Reactions: In these reactions two soluble salts usually react to form to an insoluble salt (the precipitate!) and a soluble salt. The cations of the reacting salts exchange anions. See Chang, Table 4.2, p. 119 for solubility guidelines.

Acid-Base Reactions: Most commonly an acid of the type HX or H2X reacts with a basic hydroxide to form a salt plus water. Alternatively, the acid may react with ammonia (NH3) to form an ammonium salt (but no water). These are proton transfer reactions in which H+ (the proton) is transferred from the acid to the base.

Oxidation-Reduction Reactions: These are reactions in which one type of atom increases in oxidation number (is oxidized) and another type of atom decreases in oxidation number (is reduced). A large number of oxidation-reduction (redox) reactions contain one or more reactants or products, which are pure elements.

Note that hydroxides can react with acids in acid-base reactions, and also with other salts in precipitation reactions.

Writing Balanced Ionic Equations

The first step in writing a balanced equation is predicting the products of the reaction as discussed above. Then the steps below are completed in sequence:

Balance the Molecular Equation: In the “molecular” equation, nothing is broken up into ions. Salt formulas are written so that the cation charges exactly balance out the anion charges so that the salt is neutral. Then the equation is balanced for atoms.

Balance the Total Ionic Equation: The first step in writing an ionic equation is to decide what species should be broken up into ions. The rules below should help!

|Break up into Ions |Do NOT break up! Leave “as is”! |

|Strong Acids. HCl, HBr, HI, HNO3, HClO4, and H2SO4 are the most common |Weak Acids. Nearly all acids are weak. |

|examples; assume other acids are weak. |Weak Bases. Nearly all bases are weak. |

|Strong Bases. NaOH, KOH, or Ba(OH)2 are the most common examples; assume |Insoluble Salts. Most salts are insoluble. |

|other bases are weak. |Non-electrolytes or Weak Electrolytes. Examples include H2O, |

|Soluble Salts. Salts of the alkali metals, salts containing the NH4+ ion, the|gases, pure elements, hydrocarbons, and alcohols. |

|NO3- ion, and other salts as specified in Chang, Table 4.2, p. 119. | |

Balance the Net Ionic Equation: Identify all spectator ions: these are ions that are identical on both sides of the balanced total ionic equation. Remove the spectator ions from the equation. What remains is the net ionic equation. Finally, simplify the stoichiometric coefficients if all of them are divisible by a common factor.

If all the ions are spectator ions so that nothing is left for your net ionic equation, no reaction has taken place!

Net Ionic Equations Fact IQ Sheet

Net Ionic Equation: A balanced chemical equation in which only the ions/compounds involved in the reaction (rxn) are shown. The ions that are not involved in the rxn are called the spectator ions.

How To Write Balanced Net Ionic Equations

1) Balance the equation.

2) Break all the aqueous (aq)* compounds up into ions, but do not break up polyatomic ions. The charges on these ions are the same as the ones you use when naming the compounds. (e.g. Na2SO4(aq) → 2Na+(aq) + SO4-2(aq)) Do not break up solids (s) liquids (l) or gases (g). The atoms that make up the compounds in these phases will not break up into ions, (e.g. H2O(l), CO2(g), NaI(s)). NOTE: If you are not given the products, you will need to be able to use the solubility rules to predict the products. Some professors require you learn these rules, while other professors provide them on exams.

* Aqueous means dissolved in water; ionic compounds (metal/non-metal) dissolve in water by breaking up into ions. This has to do with the fact that water is very polar (has a positive and a negative end) and ions have charges. Polar compounds like to mix with other polar or charged compounds, not with non-polar compounds. Example: Oil and water don't mix. This is because water is polar and oil is non-polar.

3) Cancel the ions that are the same on both sides of the equation.

4) Write down what is left.

Example: Write the net ionic equation for the following equation, which, when balanced, might be called the "whole formula" or overall reaction equation: FeCl3(aq) + Na2CO3(aq) → Fe2(CO3)3(s) + NaCl(aq)

1) Balance to get the overall reaction equation: 2 FeCl3(aq) + 3 Na2CO3(aq) → Fe2(CO3)3(s) + 6 NaCl(aq)

2) Complete ionic eq.: 2 Fe3+(aq)+ 6 Cl-(aq)+ 6 Na+(aq) + 3 CO32-(aq) → Fe2(CO3)3(s)+ 6 Cl-(aq)+ 6 Na+(aq)

3) Cancel spectator ions: 2 Fe3+(aq)+ 6 Cl-(aq)+ 6 Na+(aq) + 3 CO32-(aq) →Fe2(CO3)3(s)+ 6 Cl-(aq)+ 6 Na+(aq)

4) Net ionic Equation: 2 Fe3+(aq) + 3 CO32-(aq) → Fe2(CO3)3(s)

Assignment # 6

Exercises

For each of the following reactions, complete the chart. Be sure to balance all of your equations.

1. Mg(OH)2(s) + HCl(aq)

|(a) |Reaction type: |Formulas of Products Formed: |

|(b) |Molecular Equation | |

|(c) |Total Ionic Equation | |

|(d) |Net Ionic Equation | |

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2. AgNO3(aq) + K2Cr2O7(aq)

|(a) |Reaction type: |Formulas of Products Formed: |

|(b) |Molecular Equation | |

|(c) |Total Ionic Equation | |

|(d) |Net Ionic Equation | |

______________________________________________________________________________

3. NH3(aq) + HC2H3O2(aq)

(or CH3COOH)

|(a) |Reaction type: |Formulas of Products Formed: |

|(b) |Molecular Equation | |

|(c) |Total Ionic Equation | |

|(d) |Net Ionic Equation | |

______________________________________________________________________________

4. NaOH(aq) + H2SO4(aq)

|(a) |Reaction type: |Formulas of Products Formed: |

|(b) |Molecular Equation | |

|(c) |Total Ionic Equation | |

|(d) |Net Ionic Equation | |

______________________________________________________________________________

5. H2S(aq) + Ba(OH)2(aq)

|(a) |Reaction type: |Formulas of Products Formed: |

|(b) |Molecular Equation | |

|(c) |Total Ionic Equation | |

|(d) |Net Ionic Equation | |

Chemical Stoichiometry Problems

Calculating the yield of a chemical reaction is a process at the heart of chemistry. While there are many ways a problem can be phrased, in all cases the stoichiometric coefficients in the balanced reaction are used to determine the mole ratios between reactants and products. Thus the first step is usually calculating the moles of each species available. If an amount is given in grams, the molar mass is used as a conversion factor to change grams to moles.

Limiting Reagent Problems

In some problems, amounts of more than one species are given. In that case your first task is to determine which species is the limiting reagent. Just as you can make only 1 bicycle from 2 wheels and 4 handlebars (with 3 handlebars left over), and only 2 bicycles from 8 wheels and 2 handlebars (with 4 wheels left over), in chemical reactions some species are limiting while others may be present in excess.

In the case of a bicycle, we need [pic]. We obtain analogous information about the relative amounts of species that react from the stoichiometric coefficients in a balanced chemical equation. For example, in Exercise (2) below the equation

CO(g) + 2 H2(g) → CH3OH (l)

tells us we need [pic]. If we have more than 2 moles of H2 for each mole of CO, CO will be the limiting reagent and the excess H2 will not react. Conversely, if we have more than 1 mole of CO for every 2 moles of H2, H2 will be the limiting reagent and the excess CO(g) will be left over. In each case, the yield of CH3OH is determined by the moles of limiting reagent available.

Calculating the Theoretical Yield

The theoretical (maximum possible) yield is based on the amount of limiting reagent available. The yield is calculated in steps:

• Calculate moles of all reactants available. If amounts are given in grams, convert grams to moles using the molar mass of each reactant as your conversion factor: [pic].

• NOTE: Skip this step if you have already identified the limiting reagent. To determine which reagent is limiting, use the mole ratio obtained from the balanced equation for the reaction to find the moles of reactant B needed to react with the available moles of reactant A. If the moles of B available are less than the moles of B needed, reactant B is the limiting reagent and reactant A is in excess. Conversely, if the moles of B available are more than the moles of B needed, A is the limiting reagent and B is in excess.

• Calculate the moles of product based on the moles of limiting reagent available; use the stoichiometric ratio of [pic] as the conversion factor.

• If you are asked for the yield in grams, convert the yield in moles to a yield in grams using the molar mass as your conversion factor: [pic]

Percent Yield

Most reactions do not go to completion, and so the actual yield is less than the percent yield.

Assignment # 7

Exercises: Use your own paper if you want more space.

1. Ammonia is produced by the reaction

3 H2(g) + N2(g) → 2 NH3(g)

(a) If N2(g) is present in excess and 55.6 g of H2(g) reacts, what is the theoretical yield of NH3(g)?

(b) What is the percent yield if the actual yield of the reaction is 159 g of NH3(g)?

___________________________________________________________________________

2. Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction

CO(g) + 2 H2(g) → CH3OH (l)

A mixture of 1.20 g H2(g) and 7.45 g CO(g) are allowed to react.

(a) Which reagent is the limiting reagent?

(b) What is the yield of CH3OH? [Assume theoretical yield in g is what is wanted here.]

(c) How much of the reagent present in excess is left over?

(d) Suppose the actual yield is 7.52 g of CH3OH. What is the % yield?

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are reactions in which oxidation numbers change. Oxidation numbers are either real charges or formal charges which help chemists keep track of electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device.

Oxidation cannot occur without reduction.

In a redox reaction, the substance oxidized contains atoms which increase in oxidation number. Oxidation is associated with electron loss (helpful mnemonic: LEO = Loss of Electrons, Oxidation).

The substance reduced contains atoms which decrease in oxidation number during the reaction. Reduction is associated with electron gain (helpful mnemonic: GER = Gain of Electrons, Reduction).

An oxidizing agent is a substance which oxidizes something else: it itself is reduced! Also, a reducing agent is a substance that reduces another reactant: it itself is oxidized.

A disproportionation reaction is a reaction in which the same element is both oxidized and reduced.

How to Assign Oxidation Numbers: The Fundamental Rules

The oxidation number of any pure element is zero. Thus the oxidation number of H in H2 is zero.

The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of Cl in the Cl- ion is -1, that for Mg in the Mg+2 ion is +2, and that for oxygen in O2- ion is -2.

The sum of the oxidation numbers in a compound is zero if neutral, or equal to the charge if an ion.

The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in compounds is +2. The oxidation number of F is -1 in all its compounds.

The oxidation number of H is +1 in most compounds. Exceptions are H2 (where H = 0) and the ionic hydrides, such as NaH (where H = -1).

The oxidation number of oxygen (O) is -2 in most compounds. Exceptions are O2 (where O = 0) and peroxides, such as H2O2 or Na2O2, where O = -1.

For other elements, you can usually use the sum rule above to solve for the unknown oxidation number.

Examples:

NO(g) has O = -2, so N = +2.

NO2 (g) has two oxygen atoms and each has O = -2. Thus N + 2(-2) = 0, so N = +4.

SO42- has O = -2. Thus S + 4(-2) = -2. Solving the equation gives S = -2 + 8 = +6.

K2Cr2O7 has K = +1 and O = -2. Thus 2(+1) + 2 Cr + 7(-2) = 0; 2 Cr = 12; Cr = +6.

Recognizing Oxidation-Reduction Reactions

Oxidation-reduction reactions are reactions in which one type of atom increases in oxidation number (is oxidized) and another type of atom decreases in oxidation number (is reduced). Thus to show that a reaction is a redox reaction, you need to calculate oxidation numbers for the atoms in the reactants and products, and document that changes are taking place. There are, however, a few useful generalizations.

A large number (but not all!) of oxidation-reduction reactions contain one or more reactants or products which are pure elements. Why is this true? Also, all electrochemical reactions are redox reactions.

Most acid-base reactions and most precipitation reactions are not redox reactions. Why? Give some examples!

How to Balance Equations for Oxidation-Reduction Reactions

How to Balance Redox Reactions Using the Method of Half-Reactions

Oxidation-reduction reactions are often tricky to balance without using a systematic method. We shall use the method of half-reactions which is outlined in detail below.

Method in Acidic (or Neutral) Solution

Suppose you are asked to balance the equation below:

NO2– + MnO4– → NO3– + Mn+2 (in acid solution)

Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to know which is which):

NO2– → NO3–

MnO4– → Mn+2

Next, balance for atoms. First do this for atoms other than O and H. (Both equations above are already balanced for N and Mn, so no change is needed in this example.) Then balance for O atoms by adding H2O to the reaction side deficient in O:

H2O + NO2– → NO3–

MnO4– → Mn+2 + 4 H2O

This leaves H atoms unbalanced. In acidic (or neutral) solution, balance for H atoms by adding H+ to the side deficient in H:

H2O + NO2– → NO3– + 2 H+

8 H+ + MnO4– → Mn+2 + 4 H2O

The next step is to balance for charge. To do this, add electrons (e-) to the more positive side:

H2O + NO2- → NO3- + 2 H+ + 2 e-

5 e- + 8 H+ + MnO4- → Mn+2 + 4 H2O

Now you need to multiply the equations by appropriate factors so that the number of electrons lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the reduction half-reaction (GER):

5 x [ H2O + NO2- → NO3- + 2 H+ + 2 e- ]

2 x [ 5 e- + 8 H+ + MnO4- → Mn+2 + 4 H2O ]

Then, sum the above equations to obtain

5H2O + 5NO2- + 10 e- + 16H+ + 2MnO4- → 5NO3- + 10H+ + 10 e- + 2Mn+2 + 8H2O

Finally, simplify by subtracting out species that are identical on both sides. Our final balanced redox equation is

5 NO2- + 6 H+ + 2 MnO4- → 5 NO3- + 2 Mn+2 + 3 H2O

Check this equation to confirm that it is balanced for atoms and balanced for charge.

Method in Basic Solution

Suppose you are asked to balance the equation below:

I- + MnO4- → I2 + MnO2 (in basic solution)

Begin by writing the unbalanced oxidation and reduction half-reactions (you do not need to know which is which):

I- → I2

MnO4- → MnO2

Next, balance for atoms. First do this for atoms other than O and H:

2 I- → I2

MnO4- → MnO2

Then balance for O atoms by adding H2O to the reaction side deficient in O:

2 I- → I2

MnO4- → MnO2 + 2 H2O

This leaves H atoms unbalanced. In basic solution (just as in acidic or neutral solution) first balance for H atoms by adding H+ to the side deficient in H:

4 H+ + MnO4- → MnO2 + 2 H2O

In basic solution, follow this step by neutralizing the H+; do this by adding an equivalent amount of OH- to both sides of the equation.

4 OH- + 4 H+ + MnO4- → MnO2 + 2 H2O + 4 OH-

Then form water on the side which has both H+ and OH- (recall that H+ + OH- ( H2O): in this case we form 4 H2O on the left:

4 H2O + MnO4- → MnO2 + 2 H2O + 4 OH-

Next simplify the water by subtracting 2 H20 from both sides. The half-reactions are now:

2 I– → I2

2 H2O + MnO4– → MnO2 + 4 OH–

At this point the equations should be balanced for atoms. The next step is to balance for charge. To do this, add electrons (e–) to the more positive side:

2 I– → I2 + 2 e–

3 e– + 2 H2O + MnO4– → MnO2 + 4 OH–

Now you need to multiply the equations by appropriate factors so that the number of electrons lost in the oxidation half-reaction (LEO) is equal to the number of electrons gained in the reduction half-reaction (GER):

3 x [ 2 I- → I2 + 2 e- ]

2 x [ 3 e- + 2 H2O + MnO4- → MnO2 + 4 OH- ]

Sum the equations to obtain

6 I- + 6 e- + 4 H2O + 2 MnO4- → 3 I2 + 6 e- + 2 MnO2 + 8 OH-

Finally, simplify by subtracting out species that are identical on both sides:

6 I- + 4 H2O + 2 MnO4- → 3 I2 + 2 MnO2 + 8 OH-

Check our final equation above to confirm that it is balanced for atoms and balanced for charge.

Assignment # 8

Exercises:

Balance the following redox reactions. In each case

• (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction.

• (b) give the balanced net reaction.

• (c) identify the oxidizing agent and the reducing agent.

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1. Cl2(g) + S2O32-(aq) → Cl-(aq) + SO42-(aq) in acid solution.

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2. O3(g) + Br-(aq) → O2(g) + BrO-(aq) in basic solution.

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3. Balance the reaction, Br2(l) → Br-(aq) + BrO3-(aq) in basic solution. Hint: this is a disproportionation reaction!

Assignment # 9

Oxidation-Reduction

For each reaction below, identify the atom oxidized, the atom reduced, the oxidizing agent, the reducing agent, the oxidation half reaction, the reduction half reaction, and then balance the equation by the method of oxidation-reduction showing all electrons transfers.

 

1. Mg + HCl ( MgCl2 + H2

2. Fe + V2O3 ( Fe2O3 + VO

3. KMnO4 + KNO2 + H2SO4 ( MnSO4 + H2O + KNO3 + K2SO4

4. K2Cr2O7 + SnCl2 + HCl ( CrCl3 + SnCl4 + KCl + H2O

5. KMnO4 + NaCl + H2SO4 ( Cl2 + K2SO4 + MnSO4 + H2O + Na2SO4

6. K2Cr2O7 + H2O + S ( SO2 + KOH + Cr2O3

7. KClO3 + C12H22O11 ( KCl + H2O + CO2

8. H2C2O4 + K2MnO4 ( CO2 + K2O + Mn2O3 + H2O

9. Mn(NO3) 2 + NaBiO3 + HNO3 ( HMnO4 + Bi(NO3) 3 + NaNO3 + H2O

10. H2C2O4 + KMnO4 ( CO2 + K2O + Mn2O3 + H2O

Class Notes on Chemical Foundation

Below you will find a link to the first set of notes for class. These notes are on Chemical Foundation. This is basically a review of the first semester of general chemistry. In these notes, you will discuss matter, the scientific method, units of measurement, significant figures and calculation, dimensional analysis and density. You are required to print out the notes and work all problems (you must show work). You must have Adobe Acrobat Reader in order to view the files, if you do not please go to and download the reader.

Chemical Foundation Notes



Class Notes on Atoms, Molecules, and Ions

This is also a review of first semester general chemistry. In these notes, you will discuss the foundation of an atom in theory from Democritus thru Dalton, to Rutherford and Milikan’s oil drop experiment and briefly touch on Nuclear Chemistry core understanding of an alpha particle, beta particle and gamma particle. You will also discuss the Law of Multiple Proportions, Avagadro’s Hypothesis, and then move to molecules and ions. Looking closely at understanding the make-up of a compound, i.e. structural formula, cation, anion, polyatomics, and determining name of binary compounds.

Atoms, Molecules and Ions Notes



Class Notes on Stoichiometry

This is the beginning of class. The notes found here is the information discussed at the end of the second semester. These notes will be the bases of everything that you will do in AP Chemistry. It will be important that you get a firm understanding and grasp of the concepts discussed in this section. In these notes, you will see the “mole concept”, molar mass, percent abundance, percent composition, empirical formula and molecular formula, balancing chemical equations, predicting products in a chemical equation, calculating limiting reactants, and percent yield.

Stoichiometry



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