CHM 1025C Name: _________________
Module Five II Homework Packet
Module Five Part II: Chemical Equations & Stoichiometry (Chapter 8 and 9)
B1.____(10) Mole Calculation II-Section 8.4 old sample pretest: Answers bcd
______(30) Total = ______%
M._____(50) Multiple Choice Exam Chapters 8, 9,
M-5ii Required Homework:
______(30) M-5ii Pretest Hardcopy Homework Packet (Submit on Exam#3Day)
______(25) M-5ii Multiple Choice (MC) Practice (Blackboard Online)
Module Five-Part A: Molecular Mass Calculation 5 points
The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.
• The atomic mass of carbon is 12.01 amu per atom.
• Therefore, the molar mass of carbon is 12.01 g/mol .
• Since nitrogen occurs naturally as a diatomic, N2, the molar mass of nitrogen gas is two times 14.01 g or 28.02 g/mol.
Calculating Molar Mass
• The molar mass of a substance is the sum of the molar masses of each element.
• What is the molar mass of copper(II) nitrite, Cu(NO2)2?
• The sum of the atomic masses is as follows:
63.55 + 2(14.01 + 16.00 + 16.00) =
63.55 + 2(46.01) = 155.57 amu per molecule
• The molar mass for Cu(NO2)2 is 155.57 g/mol.
Using a periodic
chart calculate the molar mass of the following:
1. Calculate the molecular mass of Acetic Acid, HC2H3O2.
2. Calculate the formula unit mass of Ammonium Chromate, (NH4)2CrO4 .
3. Calculate the molecular mass of glucose, C6H12O6.
Reference: Section 8.3
Additional Problems: #13-#16 Pages 244-245
Interactive Online Chem-i-Calc(Molar Mass & % Composition):
Module Five II Homework Packet
Part B: Mole calculation 20 points
Your text has a lot of focus to convert masses and volumes to mole to actual particles. Section 8.2 (Mole Calculation I) relates the number of particles (Avagadro’s Number) to the number of moles. In Section 8.4 (Mole Calculation II) the moles are related to mass starting with the number of particles. Then vice versa. Section 8.5 defines Molar Volume. We will wait till Module 6 to do Sections 8.5 (Molar Volume) and 8.6(Mole Calculation III)
This is the only time in the course you will need to use Avogadro's number to relate actual particles to masses and volumes. In the lab we move from masses to mole and never focus on the number of particles.
What is a mole?
• The mole (mol) is a unit of measure for an amount of a chemical substance (chemist dozen).
• A mole is Avogadro’s number of particles, which is 6.02 x 1023 particles.
1 mol = Avogadro’s number = 6.02 x 1023 units (atoms, Molecules, ions, e1-)
We can use the mole relationship to convert between the number of particles and the mass of a substance
Example B1: How many sodium atoms are in 0.240 mol Na?
Question : _____?______Na Atoms = 0.240 mol Na
1. We want atoms of Na.
2. We have 0.240 mol Na.
3. 1 mole Na = 6.02 x 1023 atoms Na.
B1: How many Carbon atoms are contained in 0.01667 moles of Carbon ( 1 carat diamond)?
Example B2: How many moles of aluminum are in 3.42 x 1021 atoms Al?
Question : _____?______Mole Al = 3.42 x 1021 atoms Al
1. We want moles Al.
2. We have 3.42 x 1021 atoms Al.
3. 1 mol Al = 6.02 x 1023 atoms Al.
B2: How many moles of Carbon are contained in 3.016 x 1022 atoms of Carbon (3 carat diamond)?
Part B: Mole calculation II 10 points
We will be using the unit analysis method again.
Recall the following steps:
Mole Calculations II
Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance.
6.02 x 1023 particles = 1 mol = molar mass
If we want to convert particles to mass, we must first convert particles to moles, and then we can convert moles to mass.
Example B3: What is the mass of 1.33 moles of titanium, Ti?
1. We want grams.
2. We have 1.33 moles of titanium.
3. Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti.
Example B4: What is the mass of 2.55 x 1023 atoms of lead?
1. We want grams.
2. We have atoms of lead.
3. Use Avogadro’s number and the molar mass of Pb.
= 87.9 g Pb
Example B5: How many F2 molecules are present in 2.25 g of fluorine gas?
1. We want molecules F2.
2. We have grams F2.
3. Use Avogadro’s number and the molar mass of F2.
= 3.56 x 1022 molecules F2
Mass of an Atom or Molecule
• What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is 64.07 g/mol.
• We want mass/molecule SO2, we have the molar mass of sulfur dioxide.
• Use Avogadro’s number and the molar mass of SO2 as follows:
= 1.06 x 10–22 g/molecule
Calculate the mass in grams of 1.16 x 1022 molecules of nitrogen gas, N2
Calculate the number of entities in 0.641 g of oxygen gas, O2
For part B additional problems may be found on page 245 #17-22.
Part C: Percent Composition 5 points
The percent composition of a compound lists the mass percent of each element.
For example, the percent composition of water, H2O, is 11% hydrogen and 89% oxygen.
All water contains 11% hydrogen and 89% oxygen by mass.
Calculating Percent Composition of water
There are a few steps to calculating the percent
composition of a compound. Let’s
2(1.01 g H) + 1(16.00 g O) = molar mass H2O
2.02 g H + 16.00 g O = 18.02 g H2O
Example #2: Percent Composition Problem
TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3.
First calculate the molar mass of TNT
7(12.01 g C)
+ 5(1.01 g H)
+ 3 (14.01 g N + 32.00 g O) = g C7H5(NO2)3
84.07 g C
+ 5.05 g H
+ 42.03 g N
+ 96.00 g O
227.15 g C7H5(NO2)3
Second Calculate the % Composition
Module 5 Part C is covered in Section 8.7 of the textbook, pages 235-236, calculating the Percent Composition of a Compound. Example 8.12 on page 236 demonstrates the process with a four element compound. You will calculate the Percent Composition of one compound on M-5 Exam.
Calculate the percent composition of each element in the following compound:
Acetic Acid HC2H3O2
There is an interactive web site that will allow you to check your
work for Parts A & C. The author of ChemiCalc (which I posses two site
licenses if you want a copy), Dr. Bert Ramsay of
For additional examples at the end of the chapter do the Percent Composition on page 245 Questions #39-46
Part D: Empirical Formula Calculation 5 points
What is the Empirical Formula?
The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.
The molecular formula of benzene is C6H6.
- The empirical formula of benzene is CH.
The molecular formula of octane is C8H18.
- The empirical formula of octane is C4H9
Calculating Empirical Formulas from Lab Data
We can calculate the empirical formula of a compound from its composition data upon analysis in the lab.
We can determine the mole ratio of each element from the mass to determine the empirical formula of radium oxide, Ra?O?.
A 1.640 g sample of radium metal was heated to produce 1.755 g of radium oxide. What is the empirical formula?
– We have 1.640 g Ra and 1.755 – 1.640 = 0.115 g O
– The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol.
- We get Ra0.00726O0.00719.
- Simplify the mole ratio by dividing by the smallest number,0.00719 into both 0.00726 and .00719.
- We get Ra1.01O1.00
- Although we use at three significant figures throughout the calculation, empirical formulas are rounded to one significant figure resulting in whole number ratios
- Answer: RaO is the empirical formula
We will do a similar calculation in the lab: Determining the Empirical Formula
Calculating Empirical Formulas from % Composition
We can also use percent composition data to calculate empirical formulas.
• Assume that you have 100 grams of sample.
Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula?
If we assume 100 grams of sample,
we have 92.2 g carbon
and 7.83 g hydrogen
Second: Calculate the moles of each element.
Third: The ratio of elements in acetylene is C7.68H7.75.
Fourth: Divide by the smallest number to get the following formula:
Homework #1 Empirical Formula via Lab Data:
In an experiment 2.410 gram of copper oxide produced 1.925 g of copper metal after heating with oxygen gas. What is the empirical formula of copper oxide?
Homework #2 Empirical Formula via % Composition:
Given the following percent compositions, determine the empirical formula of the following compound: 54.5% carbon, 9.15% hydrogen, and 36.3% oxygen
Module 5 Part D is a calculation of the Empirical and Molecular Formula from both the Lab Data and the Percent Composition which is described on pages 237-239.
Example 8.13, pages 237-238, demonstrates the process for calculation of empirical formula from lab data.
Example 8.14 page 239 demonstrates this process for the calculation from Percent Composition.
For Section 8.9 pages 239-240 You must know from this section what the difference is between an Empirical Formula and a true Molecular Formula of a compound and this will be tested in the Multiple choice section.
Section 11.10 is Hydrates. This section introduces the concept hydrates and Anhydrates which we may do as one of our M-5 crucible labs. Most crystalline salts come as hydrates, meaning water molecules are part of the crystalline structure. You should be able to interpret a hydrate formula where the dot between the two compounds reads "in combination with". CuSO4 ∙ 5 H2O reads one formula unit of CuSO4 in combination with 5 molecules of water. Example 11.8 page 332 is a sample problem of interpreting the formula of a hydrate from lab data.
There are empirical formula problems at the end of the chapter to practice your empirical formula calculation of a compound: pages 246 problems #47-56.