CHM 2045C                                    Name: _________________
Module Five II Homework Packet

Module Five Part II: Chemical Equations & Stoichiometry (Chapter 3)
A._____(05) Molecular Mass Calculation-Section 3.3

B ._____(05) Mole Calculations I-Sections 3.3 Answers

B1.____(05) Mole Calculation II-Section 3.3: Answers bcd
C._____(05) Percentage Composition Calculation-Section 3.10 Answers bcd

D._____(05) Empirical Formula Calc. from Lab Data-Section 3.11 Answers bcd

D1.____(05) Empirical Formula Calc. from % Comp-Section 3.11 Answers
________(30) Total M-5ii Homework Packet

M-5ii Required Homework:

______(30) M-5ii Pretest Hardcopy Homework Packet

______(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.

Homework #1 (M-5 A):

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 3.3 page 79

Additional Problems: #3.4 Page83

End of Chapter Page 106 Questions 3.44-3.54

Interactive Online Chem-i-Calc(Molar Mass & % Composition):

Module Five II Homework Packet

Part B: Mole Calculation 1        5 points

Your text in Section 3.1 does not address too well M-5B and M-5B1 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.

Solution: Homework #2 (M-5B):

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.

Solution: Homework #3 (M-5B):

B2: How many moles of Carbon are contained in 3.016 x 1022 atoms of Carbon (3 carat diamond)?

Part B1: Mole calculation II       5 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. Atoms–Mass Calculation

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

Mass–Molecule Calculation

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 1022 g/molecule

Homework #4 (M-5B1):

Calculate the mass in grams of 1.16 x 1022 molecules of nitrogen gas, N2

Homework #5 (M-5B1):

Calculate the number of entities in 0.641 g of oxygen gas, O2

Text reference: page 80
Conceptual Problem 3.5 page 83

For part B additional end of chapter problems may be found on page 106 questions 3.55-3.59

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 practice
using H2O.

1. Assume you have 1 mol of the compound.
2. One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore,

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

1. Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water. 1. Remember:         g H/100 g H2O  is % H   and     g O/100 g H2O  is % O

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 3.10-3.11 of the textbook,  pages _____, calculating the Percent Composition of a Compound. You will calculate the Percent Composition of one compound on M-5 Exam.

Homework Problem #6 (M-5C):

Calculate the percent composition of each element in the following compound:

Acetic Acid    HC2H3O2

Text Reference: Section 3.10 Pages 94-97
see worked example 3.16 pages 96-97  try problem 3.24 page 97

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 Eastern Michigan University, has placed a small fraction of his program online. Check out:
http://people.emich.edu/bramsay1/ccc-release/chem.html

For additional examples at the end of the chapter do the Percent Composition on page 108-109   Questions #3.98-3.99

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 % 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?

First:  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 #7 (M-5D) 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

Text Reference Section 3.10 see worked example 3.15 page 96
Additional problems: 3.22-3.23 p97; end chapter Q# 3.100-3.103 p109

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?.

Example #1
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 1.7551.640 = 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 may do a similar calculation in the lab: Determining the Empirical Formula     Homework #8 (M-5D1)  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?

Text Reference: Section 3.11 pages 97-99

See additional worked example 3.17 page 99;also work 3.25, 3.26, 3.27 page 99

There are empirical formula problems at the end of the chapter to practice your empirical formula calculation of a compound: page 109 problems #3.104-3.107.