Chemistry 112
Second Hour Exam
Name:____________
(4 points)
Please show all work for partial credit
All problems worth 12 points
1. Federal Law prohibits more than 44 mg of nitrate (NO3-) per liter of water in drinking
water.
A. If the water in my well only has 4 mg of nitrate per liter of drinking water, how
many grams of nitrate are in a single drop of my well water? (Assume a drop of water
is .05 mL)
B. How many molecules of nitrate are in that drop of water?
Molar mass of NO3- = 14.01 + 3(16) = 62.01 g/mol
C. How many atoms of oxygen from nitrate are in that drop of water?
2. A compound is 40% Carbon, 53.3% Oxygen and 6.7% Hydrogen, what is the
empirical formula of this compound?
Multiplying 100g by the various % we have: 40g C, 53.3g O and 6.7 g H
Dividing by molar mass
40g C / 12.01g/mol = 3.33 mol C
53.3 g O /16.00 g/mol = 3.33 mol O
6.7 g H /1.008g/mol = 6.65 mol H
Dividing by the lowest # of moles we have
C = 3.333/3.33 = 1
O = 3.33/3.33 =1
H = 6.66/3.33 =2 C:O:H = 1:1:2
2
3. Write a balanced Molecular, complete ionic and net ionic chemical equation for the
following two reactions:
Solid aluminum hydroxide plus aqueous hydrochloric acid yields aqueous aluminum
chloride and water.
Molecular:
Complete Ionic:
Net Ionic:
Al(OH)3(s) + 3HCl(aq) 6 AlCl3(aq) + 3H2O(l)
Al(OH)3(s) + 3H+(aq) + 3Cl- (aq) 6 Al3+(aq) + 3 Cl- (aq) + 3H2O(l)
Al(OH)3(s) + 3H+(aq) 6 Al3+(aq) + 3H2O(l)
Solid iron plus oxygen gas yields solid iron(III) oxide
Molecular:
Complete Ionic:
Net Ionic:
Fe(s) + O2(g) 6 Fe2O3(s)
Fe(s) + O2(g) 6 Fe2O3(s)
Fe(s) + O2(g) 6 Fe2O3(s)
4. Aspirin can be synthesized from salicylic acid by the following reaction:
C7H6O3
+
C4H6O3
6
C9H8O4 +
C2H4O2
Salicylic Acid
Acetic Anhydride Aspirin
Acetic Acid
If I start with 10 grams each of salicylic acid and acetic anhydride, and my reaction has
a 95% yield, how much aspirin can I make?
Molar Masses:
Salicylic
7(12.01)+6(1.008)+3(16) = 138.118 g/mole
Acetic Anhydride 4(12.01)+6(1.008)+3(16) = 102.088 g/mol
Aspirin
9(12.01)+8(1.008)+4(16) = 180.154 g/mol
If Salicylic is limiting:
Salicylic acid gives the smallest amount of product, so it must be the limiting reactant,
and the theoretical yield is 13.04 g
3
5. Define the following terms:
Empirical Formula
A formula that gives the ratios of atoms to each other in a compound, but not the actual
number of atoms in the molecule.
Solvent
The liquid that the solute is dissolved in.
Weak Electrolyte
An ionic material that ionizes weakly when it dissolves in water so it can only carry a
weak current.
Molarity
Molarity(M) = moles soute/ liters solution.
Complete Ionic equation
A balanced chemical equation in which all aqueous ionic molecules are broken up into
their individual component ions.
Oxidation ½ reaction
A balanced ½ reaction of an oxidation/reduction reaction in which the electrons appear
on the right hand side of the equation. (Or, in which the oxidation number of a reactant
increases.)
6. When the following solutions are mixed together, what precipitate (if any) will form?
FeSO4 (aq) + KCl(aq)
No Reaction (both products are aqueous)
CaCl2 (aq) + Na2SO4 (aq)
CaSO4 (s) (Sulfates slightly soluble)
Hg2(NO3)2 (aq)+ CuSO4 (aq)
Hg2SO4 (s) (Sulfates slightly soluble - Hg salts insoluble)
K2CO3 (aq)+ MgI2 (aq)
MgCO3(s) (Carbonates are slightly soluble)
4
7. Give at least 5 of the rules I listed in class for assigning oxidation numbers:
1. The oxidation number for an atom in the elemental form is 0.
2. The oxidation number for F is always -1.
3. The oxidation number for O in a covalent compound is -2, except in peroxides when
it is -1.
4. The oxidation number of H is +1 in covalent compounds (But it can be -1 in ionic
metal hydrides)
5. The oxidation number for an atom in a monoatomic ion is equal to the charge of the
ion.
6. The sum of the oxidation numbers for all atoms in a molecule or ion must equal the
net charge on that molecules or ion.
I gave +2 bonus points if you gave all 6 rules!
8. Balance the following redox reactions:
Cl2(g) + Al(s) 6 Al3+(aq) + Cl-(aq) (ACIDIC conditions)
Cl26ClCl262Cl2e- + Cl262Clx3
6e + 3Cl2 6 6Cl-
Al6Al3+
Al6Al3+
Al6Al3+ + 3eX2
2Al62Al3+ + 6e-
6e- + 3Cl2 + 2Al 62Al3+ + 6e- + 6Cl3Cl2 (g)+ 2Al(s) 62Al3+(aq) + 6Cl-(aq)
Lots of people forgot to put the physical forms into the final equation. I didn’t take off on
this test, but next time I will!
NO2-(aq) + Al(s) 6NH3(g) + AlO2-(aq) (BASIC conditions)
NO2-(aq) 6 NH3(g)
Al(s) 6AlO2-(aq)
NO2-(aq) 6 NH3(g) + 2H2O
2H2O + Al(s) 6AlO2-(aq)
+
7H + NO2 (aq) 6 NH3(g) + 2H2O
2H2O + Al(s) 6AlO2-(aq) + 4H+
+
6e + 7H + NO2 (aq) 6 NH3(g) + 2H2O 2H2O + Al(s) 6AlO2-(aq) + 4H+ +3eX1
X2
+
6e + 7H + NO2 (aq) 6 NH3(g) + 2H2O 4H2O + 2Al(s) 62AlO2-(aq) + 8H+ +6e6e- + 7H+ + NO2-(aq) + 4H2O + 2Al(s)6 NH3(g) + 2H2O + 2AlO2-(aq) + 8H+ +6eNO2-(aq) + 2H2O + 2Al(s)6 NH3(g) + 2AlO2-(aq) + 1H+
1 OH- + NO2-(aq) + 2H2O + 2Al(s)6 NH3(g) + 2AlO2-(aq) + 1H+ + 1 OH1 OH- + NO2-(aq) + 2H2O + 2Al(s)6 NH3(g) + 2AlO2-(aq) + 1H2O
1 OH- (aq) + NO2-(aq) + 1H2O(l) + 2Al(s)6 NH3(g) + 2AlO2-(aq)