**CHM 1020 ****Path 4 Chapter 9 Study Packet Part II**

**Chapter 9 Part II: ****Chemical
Equations **

C. Molecular Mass
Calculation-** ****Answers**

D._Mole Calculations
I-**Sections 8.2 old sample pretest: ****Answers**
bcd

D1.Mole Calculation II-Section 8.4 **old sample pretest: ****Answers**
bcd

E._Percentage
Composition Calculation-**Section 8.7
old sample pretest ****Answers**
bcd

E1.Empirical Formula
Calc. from Lab Data-**Section 8.8
old preterst ****Answers**
bcd

E2.Empirical Formula
Calc. from % Comp-**Section 8.8
old pretest ****Answers**
bcd

**______
Total**

**Chapter 9-Part C:**** Molecular Mass Calculation
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}.

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

http://people.emich.edu/bramsay1/ccc-release/chem.html
(Requires
JAVA)

**Chapter
9 Part D Study Packet**

**Part D: Mole calculation **

**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 Avogadros number of particles,
which is 6.02 x 10 ^{23} particles.**

**1 mol = Avogadros 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 D1: Mole calculation II **

**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.**

** **

**AtomsMass 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
Avogadros number and the molar mass of Pb. **

** **

** ****= 87.9 g Pb **

**MassMolecule 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
Avogadros 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 Avogadros 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**,

**Part D2:
Percent Composition **

**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. Lets
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**

** 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 D3:
Empirical Formula Calculation
**

**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**