EXAM 4 REVIEW CHEM 110 Page 1 of 13 Last Name Professor BEAMER First Name Which lab section are you in? M W R (Circle One) Note: As Part 1 is supposed to be memorized material, you can cross-check your answers with the information in your textbook and/or from your notes. PART 2: MATH 15a) Calculate the molar mass of Na2Cr2O7. Rewrite final answer 261.98 g/mol Na2Cr2O7 (Time Range: 3 to 5 min) g/mol g/mol 2 × Na 2 × 22.99 45.98 2 × Cr 2 × 52.00 104.00 7 × O 7 × 16.00 + 112.00 261.98 g/mol Na2Cr2O7 15b) Rewrite the molar mass of the substance above in conversion factor form: 261.98 g Na2Cr2O7 = 1 mol Na2Cr2O7 (Time Range: 1 min − all levels)
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EXAM 4 REVIEW CHEM 110
Page 1 of 13
Last Name Professor BEAMER
First Name Which lab section are you in?
M W R (Circle One)
Note: As Part 1 is supposed to be memorized material, you can cross-check your answers
with the information in your textbook and/or from your notes.
PART 2: MATH
15a) Calculate the molar mass of Na2Cr2O7. Rewrite final answer 261.98 g/mol Na2Cr2O7
(Time Range: 3 to 5 min)
g/mol g/mol
2 × Na 2 × 22.99 45.98
2 × Cr 2 × 52.00 104.00
7 × O 7 × 16.00 + 112.00
261.98 g/mol Na2Cr2O7
15b) Rewrite the molar mass of the substance above in conversion factor form:
261.98 g Na2Cr2O7 = 1 mol Na2Cr2O7
(Time Range: 1 min − all levels)
EXAM 4 REVIEW CHEM 110
Page 2 of 13
16) Calculate the number of molecules of NH3 in a 1.400 × 10−5 mole sample of NH3.
(Time Range: 3 to 6 min)
Solution/Explanation:
You are converting between particles (molecules) and moles (mol). Therefore, you need to use Avogadro’s Number.
You need to devise a conversion factor: 1 mol NH3 = 6.022 × 1023 NH3 molec
Always start with your initial value: 1.400 × 10−5 NH3 molec
1 mol NH3 = 6.022 × 1023 NH3 molec
initial value Avogadro’s Number Conversion
(
1.400 × 10-5 NH3 mol
1) (
6.022 × 1023 NH3 molec
1 mol NH3) = 8.431 × 1018 NH3 molec
Identify initial value!
EXAM 4 REVIEW CHEM 110
Page 3 of 13
17) A sample of CaCl2 contains 3.8 × 1025 formula units of CaCl2. Calculate the mass (in grams) of CaCl2 in the sample. The molar mass of CaCl2 is 110.98 g/mol.
(Time Range: 3 to 5 min)
Solution/Explanation:
You are converting between particles (molecules) and mass (grams). Therefore, you need to use Avogadro’s Number.
There is no direct pathway from particles to moles: mass moles particles
You must go through moles first. You need two conversion factors:
110.98 g CaCl2 = 1 mol CaCl2
6.022 × 1023 CaCl2 form = 1 mol CaCl2
initial value Avogadro’s Number Conversion
(
3.8 × 1025 CaCl2 form
1) (
1 mol CaCl2
6.022 × 1023 CaCl2 form ) = 63 mol CaCl2
molar mass
(
63 mol CaCl21
) (110.98 g CaCl21 mol CaCl2
) = 7003 g CaCl2 x7.0 × 103 g CaCl2x
Also acceptable:
initial value Avogadro’s Number Conversion molar mass
(
3.8 × 1025 CaCl2 form
1) (
1 mol CaCl2
6.022 × 1023 CaCl2 form ) (
110.98 g CaCl21 mol CaCl2
) = 7003 g CaCl2 x7.0 × 103 g CaCl2x
Identify initial value!
Identify final unit: mass (grams)
EXAM 4 REVIEW CHEM 110
Page 4 of 13
Reaction for Question 18
283.88 g/mol 63.01 g/mol 98.00 g/mol 108.01 g/mol
P4O10(s) + 12 HNO3(aq) 4 H3PO4(aq) + 6 N2O5(s)
Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5)
18) Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from
500.0 grams of nitric acid, HNO3(aq).
(Time Range: 6 to 12 min)
Solution/Explanation:
Steps to Theoretical Yield Calculations
Step 1: Convert the initial value (the KNOWN) to moles
Step 2: Use mole ratios to convert moles KNOWN moles UNK
Step 3: Convert moles UNK to mass UNK
STEP 1:
initial value molar mass
(500.0 g HNO3
1) (
1 mol HNO3
63.01 g HNO3 ) = 7.935 mol HNO3
* Use molar mass of the balanced equation to convert mass of HNO3 to moles of HNO3.
STEP 2:
mole ratio
(7.935 mol HNO3
1) (
6 mol N2O5
12 mol HNO3 ) = 3.968 mol N2O5
* Use the coefficients of the balanced equation to convert moles of HNO3 to moles of N2O5.
STEP 3:
molar mass
(3.968 mol N2O5
1) (
108.05 g N2O5
1 mol N2O5 ) = x428.7 g N2O5 OR 4.287 × 102 g N2O5x
* Convert moles of N2O5 to mass of N2O5 using molar mass.
* You can always put answers to calculations in standard SCINOT. (Make sure the SCINOT is
correct.)
See next page to see “all-in-one-step” calculation:
Identify initial value!
EXAM 4 REVIEW CHEM 110
Page 5 of 13
Reaction for Question 18
283.88 g/mol 63.01 g/mol 98.00 g/mol 108.01 g/mol
P4O10(s) + 12 HNO3(aq) 4 H3PO4(aq) + 6 N2O5(s)
Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5)
18) Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from 500.0 grams of nitric acid, HNO3(aq).
(Time Range: 6 to 12 min)
Solution/Explanation:
Steps to Theoretical Yield Calculations
Step 1: Convert the initial value (the KNOWN) to moles
Step 2: Use mole ratios to convert moles KNOWN moles UNK
Step 3: Convert moles UNK to mass UNK
INITIAL VALUE STEP 1 STEP 2 STEP 3
initial value molar mass mole ratio molar mass
(500.0 g HNO3
1) (
1 mol HNO3
63.01 g HNO3 ) (
6 mol N2O5
12 mol HNO3 ) (
108.05 g N2O5
1 mol N2O5 ) = x428.7 g N2O5 OR 4.287 × 102 g N2O5x
* Again, I have no preference as to whether you break down this solution into three steps or choose the “all-in-one-step” method.
Identify initial value!
EXAM 4 REVIEW CHEM 110
Page 6 of 13
19) Assume that a scientist perform this experiment and obtains a yield of 411.7 grams of N2O5. Calculate the %-yield value for this
experiment. Use the value obtained in Question 18 for the theoretical yield. Note: You must rewrite the %-yield equation first with variables only as your first step.
(Time Range: 2 to 4 min)
Solution
%-yield = experimental yield
theoretical yield × 100 =
experimental yield
theoretical yield × 100
= 411.7 g N2O5
428.7 g N2O5 × 100
= 96.03 % yield N2O5
EXAM 4 REVIEW CHEM 110
Page 7 of 13
PART 3: NON-CALCULATOR
Resources:
You will be given a periodic table (symbols only)
You will be given the memorized table of polyatomic ions chart
You will be given the Activity Series Table (Top Dog)
You will be given the Solubility Rules (all of them)
No calculators for this section
Timing: 10 min (mastery) 25 min (competence)
20) Indicate whether the following elements are solid (s), liquid (l), or gas (g).
20a) Cu (#29) (s) 20b) Hg (#80) (l)
(Time Range: 1 min maximum – all levels)
21) List the charges of the following elements when they become ions:
21a) nitrogen (#7) −3 or 3− 21c) potassium (#19) +1 or 1+
21b) sulfur (#16) −2 or 2− 21d) iodine (#53) −1 or 1−
(Time Range: 1 min 30 s maximum – all levels)
IMPORTANT: For positive ions, you must write the “+” symbol!
Question 21a has been corrected. It should be −3 (not +3). Sorry.
EXAM 4 REVIEW CHEM 110
Page 8 of 13
22) Circle the metals that require a Roman Numeral in their names when they are present in ionic compounds.
Na (#11) Fe (#26) Sn (#50)
Al (#13) Zn (#30) Ba (#56)
(Time Range: 2 min maximum – all levels)
Na Group 1A (alkali metals) always have a charge of 1+. Therefore, Roman numerals are not
required when naming salts that contain Group 1A metals.
Fe Transition metal. With three exceptions, transitions metals always require a Roman
numeral in their names to indicate their charge.
Sn Post-Transition metal. All post-transition metals always require a Roman numeral in
their names to indicate their charge.
Al Group 3A. This is a little tricky. Even though there are transition metals in Group 3A,
aluminum itself is not a transition metal. Therefore, it does not require a Roman numeral
to indicate its charge.
Zn Zinc is one of the three exceptions of the transition metals. Zinc always has a charge
2+ in ionic compounds. Therefore, a Roman numeral is not required. Here are the three
exceptions again:
Ag (silver) is always 1+ When present in ionic compounds,
these three metals do not require a
Roman numeral when naming them. Cd (cadmium) is always 2+ in ionic compounds
Zn (zinc) is always 2+ in ionic compounds
Ba Group 2A (alkali earth metals) always have a charge of 2+. Therefore, Roman numerals
are not required when naming salts that contain Group 2A metals.
EXAM 4 REVIEW CHEM 110
Page 9 of 13
23) Circle the compounds that are soluble in water.