**CHM 1025C ** Name: _________________

**Module Five II Homework Packet**

**Module Five
Part II: ****Chemical Equations &
Stoichiometry (Chapter 8 and 9)**

A._____(05) Molecular Mass
Calculation-**Section8.3 ****Answers** a

B._____(10) Mole Calculations
I-**Sections 8.2 old sample pretest: ****Answers**
bcd

B1.____(10) Mole Calculation II-Section 8.4 **old sample pretest: ****Answers**
bcd

C._____(05) Percentage
Composition Calculation-**Section 8.7
old sample pretest ****Answers**
bcd

D._____(05) Empirical Formula
Calc. from Lab Data-**Section 8.8
old preterst ****Answers**
bcd

D1.____(05) Empirical Formula
Calc. from % Comp-**Section 8.8
old 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, N _{2}, 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(NO _{2})_{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(NO _{2})_{2} is
155.57 g/mol.**

**Homework:**

**Using a periodic
chart calculate the molar mass of the following:
**1. Calculate the molecular
mass of Acetic Acid

2. Calculate the
formula unit mass of Ammonium Chromate**,
(NH _{4})_{2}CrO_{4} .**

3. Calculate the
molecular mass of glucose,
**C _{6}H_{12}O_{6}**

Reference: Section 8.3

Additional Problems: #13-#16 Pages 244-245

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

http://people.emich.edu/bramsay1/ccc-release/chem.html

**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 10 ^{23} particles.**

**1 mol = Avogadro’s number = 6.02 x 10^{23} units
(atoms, Molecules, ions, e^{1-})**

**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 10 ^{23} atoms
Na.**

**Solution:**

**Homework #1:**

**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 10 ^{21} atoms Al?**

** ****Question :****
_____?______Mole Al = 3.42 x 10 ^{21} atoms Al**

**1.
****We want moles Al.**

**2.
****We have 3.42 x 10 ^{21} atoms Al.**

**3.
****1 mol Al = 6.02 x 10 ^{23} atoms
Al.**

**Solution:**

**Homework #2:**

**B2: How many moles of Carbon are contained in 3.016 x 10 ^{22}
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 10 ^{23}
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 10 ^{23}
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 F _{2}
molecules are present in 2.25 g of fluorine gas?**

**1. ****We
want molecules F _{2}.**

**2. ****We
have grams F _{2}.**

**3. ****Use
Avogadro’s number and the molar mass of F _{2}._{ }**

** **

= **3.56 x 1022
molecules F _{2}**

**Mass of an Atom or Molecule**

•
**What is the mass of a single
molecule of sulfur dioxide? The molar
mass of SO _{2} is 64.07 g/mol.**

•
**We want mass/molecule SO _{2}, we have the molar mass of sulfur dioxide.**

•
**Use Avogadro’s number and the
molar mass of SO _{2 }as follows: **

** **

** ****= 1.06 x 10^{–}^{22} g/molecule_{ }**

**Homework #3:**

Calculate the mass in grams of **1.16 x 10 ^{22}**
molecules of

**Homework #4:**

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

**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, H _{2}O, 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 H _{2}O.**

**Assume you have 1 mol of the compound.****One mole of H**_{2}O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore,

** 2(1.01
g H) + 1(16.00 g O) = molar mass H _{2}O**

**2.02
****g H + 16.00 g O = 18.02 g H _{2}O**

**Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water.**

** **

**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, C _{7}H_{5}(NO_{2})_{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 C _{7}H_{5}(NO_{2})_{3}
**

** 84.07 g C **

**+ 5.05 g H **

**+
42.03 g N **

__+
96.00 g O__

** 227.15 g C _{7}H_{5}(NO_{2})_{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.**

**Homework:**

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

**Acetic
Acid HC _{2}H_{3}O_{2}**

**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
**

**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 C _{6}H_{6}.**

- **The
empirical formula of benzene is CH.**

**The molecular formula of octane is C _{8}H_{18}.**

- **The
empirical formula of octane is C _{4}H_{9}**

**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
**

**– 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 Ra _{0.00726}O_{0.00719}.
**

- **Simplify
the mole ratio by dividing by the smallest number,0.00719
into both 0.00726 and .00719.**

- **We
get Ra _{1.01}O_{1.00} **

- **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?**

**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 C _{7.68}H_{7.75}.**

**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". CuSO _{4} ∙ 5 H_{2}O
reads one formula unit of CuSO_{4} 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.**