Chemistry 100

Fall, 2000                              Name: __________________
Third Exam
Exam begins here. There are 30 questions.

Questions 1-4 are worth 2 pts each.

Three types of equations are used to describe reactions in aqueous solutions.

1. The molecular equation shows the overall reaction but not necessarily the actual forms of the reactants and products in solution.

2. The complete ionic equation represents all reactants and products that are strong electrolytes (completely dissociate) as ions. All reactants and products are included.

3. The net ionic equation includes only those components that undergo a change. Spectator ions are not included.

4. What subatomic particle is transferred during an oxidation- reduction reaction?


The next two questions are worth 6 points each.

Write the complete molecular, complete ionic, and net ionic equations for each of the following reaction equations. Show spectator ions and cross them through (italicized here) to indicate that they are spectator ions.


HCl(aq) + RbOH(aq) ---> RbCl(aq) + H2O(l)

H+(aq) + Cl-(aq) + Rb+(aq) + OH-(aq) ---> Rb+(aq) + Cl-(aq) + H2O(l)

H+(aq) + OH-(aq) ---> H2O(l)

H2SO4(aq) + 2CsOH(aq) ---> Cs2SO4(aq) + 2H2O(l)

2H+ + SO42-(aq) + 2Cs+(aq) + 2OH-(aq) ---> 2Cs+(aq) + SO42-(aq) + 2H2O(l)

2H+ + 2OH-(aq) ---> 2H2O(l)

Identify the reaction category (precipitation, acid-base, or oxidation-reduction) to which the following reactions belong.

These 3 questions are worth 2 points each.

7. H2O2(aq) ---> H20(l) + O2(g)


8. AgNO3(aq) + CuCl2(aq) ---> Cu(NO3)2 + AgCl(s)


9. Ca(OH)2(aq) + HNO3(aq) ---> Ca(NO3)2(aq) + H2O(l)


For each of the following metals, how many electrons will the metal lose when it ionizes (forms its ion)?

The next 6 questions are worth 1 point each.

10. sodium 1

11. barium 2

12. aluminum 3

For each of the following nonmetals, how many electrons will the nonmetal gain when it ionizes?

13. nitrogen 3

14. sulfur 2

15. chlorine 1

The next 6 questions rely upon your knowledge of Avogadro's Number, moles, molecules, and atoms. REQUIRED: Show your calculations.

These problems are worth 3 points each.

16. How many atoms are there in 2 moles of He?

1 mole contains 6.022 x 1023 atoms so 2 moles must contain 2 x 6.022 x 1023 which is 12.04 x 1023, and in scientific notation this is 1.204 x 1023.

17. Calculate the mass of one atom of lithium (Li). Show the mass both in a.m.u. and in grams (g).

The atomic mass on the Periodic Table is a.m.u. for a single atom so the mass of Li in a.m.u. is 6.941.

The mass of 6.941 is grams when discussing a mole of Li. A mole contains 6.022 x 1023 atoms. The mass of a single atom would have units of g/atom. So one needs to divide the mass in grams by the number of atoms.

6.941 g/mole


6.022 x 1023 atoms/mole

The answer is: 1.15 x 10-23 g/atom.

18. A sample of "stuff" is found to weigh 54 grams. The "stuff" contains 2 hydrogen atoms for each oxygen atom in the compound. The simplest formula shows 2 hydrogens to 1 oxygen. How many moles of this "stuff" is in the 54 grams?

The simplest formula shows that the compound must be H2O. THe molar mass of H2O is 18 g.

Divide the total number of grams by the molar mass and one gets a value of 3 moles.

19. What is the mass in grams of 0.5 mole of CaCO3?

The mass of one mole of CaCO3 is 100 g so 0.5 mole must equal 50 g.

20. Calculate the number of moles of NaOH in 100 g of NaOH.

First calculate the molar mass of NaOH which is: 23 + 16 + 1 = 40 g.

Next set up a ratio;

                      1 mole      X moles
                      ------  =  ---------
                       40 g        100 g
The answer is: 2.5 moles

21. Calculate the number of molecules present in 2.62 x 10-6 moles of water.

1 mole of water (or any substance for that matter) always contains 6.022 x 1023 molecules

Set up a ratio:

                        1 mole         2.62 x 10-6 moles
                      -----------  =  ------------------
                       6.022 x 1023       X molecules
The answer is: 1.58 x 1018 molecules

The following problems involve determining the chemical formula for an unknown compound based upon its chemical analysis.

These two problems are worth 6 points each.

22. A 10.00 g sample of copper metal is heated strongly in air. After reacting, the sample weighs 12.52 g. The additional weigh comes from oxygen which reacted with the copper metal. Determine the molecular formula using the % composition method.

The compound is made of Cu and O. According to the Law of Conservation of Matter the mass of Cu going in equals the mass of Cu coming out so, 10.00 g of Cu come out and the difference of 2.52 g is oxygen.

Next calculate the number of moles of Cu and O.

               10.00 g
              -----------  =  0.157 mole
              63.5 g/mole  

               2.52 g
              -----------  =  0.157 mole
              16 g/mole
So the ratio is 0.157 to 0.157 which is equivalent to 1:1 and the formula therefore is CuO.

23. A compound with an empirical (simplest reduced) formula CH was found by experiment to have a molar mass of 78. What is the molecular formula of the compound?

The empirical formula is a reduced form of a formula where common factors for each atom are divided out. The molecular formula is the actual formula for the molecule being studied.

The empirical formula has a mass of 13. The molar mass of the molecular compound is 78 so divide 78 by 13 to obtain the number of CH units in the molecular compound. 78/13 = 6 so there must have been a common coefficient of 6 for the C and the H. The molecular formula is C6H6.

The next problems involve chemical quantities.

24. (4 pts) What is the limiting reactant for a chemical reaction? Explain.

The limiting reactant is the reactant in the reaction which is less than the balanced ratio of moles needed to react with the other reactant(s).

The reaction will continue until the limiting reactant is used up then the reaction will stop.

25. (4 pts) What is meant by theoretical yield and by actual yield when discussing a chemical reaction?

The theoretical yield is the yield calculated using the balanced chemical equation as a guide. This is the amount of product that should be produced.

The actual yield is the amount of product recovered after an experiment is conducted. As a rule the actual yield is less than but can equal the theoretical yield.

26. (6 pts) The theoretical yield of a chemical reaction is 1.352 g of barium sulfate. After running an experiment, the researcher measured the amount of barium sulfate produced and had 1.279 g. What was the % yield of this reaction?

                            1.279 g
               % yield =   --------- x 100  =  94.6 %
                            1.352 g
27. (6 pts) When glucose, C6H12O6, is burned completely in air, CO2 and water are produced. Write a balanced chemical equation for this reaction. Then calculate the theoretical yield of CO2 is 1.00 g of glucose is burned completely.

First the balanced chemical equation must be determined so that the number of moles of CO2 produced per mole of glucose can be determined. Actually that is not necessary in this reaction since all carbon coming into the reaction ends up in a single product. Glucose contains 6 moles of carbon atoms in each mole of glucose so the conplete combustion will produce 6 moles of CO2 per mole of glucose.

Next find the number of moles equivalent to 1 g of glucose. The molar mass is 180 g/mole. So:

                     180 g          1.00 g
                   ---------  =   -----------
                    1 mole         x moles
X = 5.56 x 10-3 moles of glucose.

6 x 5.56 x 10-3 moles of CO2 are produced from 5.56 x 10-3 moles of glucose. This equals 3.33 x 10-2 moles of CO2.

28. (2 pts each part) Calculate the mass in grams of:

A) 0.000305 mole mercury (Hg) 6.12 x 10-2 g

B) 125 moles ammonia (NH3) 2.125 x 103 g

C) 0.01205 mole of sodium peroxide (Na2O2) 0.94 g

29. (6 pts) Using :

FeO(s) + C(s) ---> Fe(l) + CO2(g)

Calculate the number of moles of CO2(g) produced when 0.125 mole of FeO is reduced to Fe.

First balance the equation.

2FeO(s) + C(s) ---> 2Fe(l) + CO2(g)

We see that the ratio of FeO to CO2 is 2:1. So 0.125 mole of FeO will produce 0.125/2 mole of CO2.

0.125/2 = 0.0625 mole of CO2 will be produced.

30. (6 pts) Use words to rewrite the following chemical sentence stating what the equation represents in terms of individual molecules and in terms of moles of molecules.

B2O3 + CaF2(s) ---> BF3(s) + CaO(s)

Again, the first step is balancing.

B2O3 + 3CaF2(s) ---> 2BF3(s) + 3CaO(s)

In molecules, the sentence reads:

1 molecules of boron oxide react with 3 molecules of calcium fluoride to produce 2 molecules of boron trifluoride and 3 molecules of calcium oxide.

In moles, the sentence reads:

1 moles of boron oxide react with 3 moles of calcium fluoride to produce 2 moles of boron trifluoride and 3 moles of calcium oxide.