Chemistry 100(02) Fall 2013
Instructor: Dr. Upali Siriwardane
e-mail: upali@coes.latech.edu
Office: CTH 311
Phone 257-4941
Office Hours: M,W, 8:00-9:30 & 11:30-12:30 a.m
Tu,Th,F 8:00 - 10:00 a.m. Or by appointment
CHEM 100, Fall 2012 LA TECH
Test Dates:
September 30,
2013 (Test 1): Chapter 1 & 2
October 21,
2013 (Test 2): Chapter 3 & 4
November 13, 2013 (Test 3) Chapter 5 & 6
November 14, 2013 (Make-up test) comprehensive:
Chapters 1-6 9:30-10:45:15 AM, CTH 328
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Text Book & Resources
REQUIRED :
Textbook: Principles of Chemistry: A Molecular Approach,
2nd Edition-Nivaldo J. Tro - Pearson Prentice Hall and also
purchase the Mastering Chemistry
Group Homework, Slides and Exam review guides and
sample exam questions are available online:
http://moodle.latech.edu/ and follow the course information
links.
OPTIONAL :
Study Guide: Chemistry: A Molecular Approach, 2nd EditionNivaldo J. Tro 2nd Edition
Student Solutions Manual: Chemistry: A Molecular
Approach, 2nd Edition-Nivaldo J. Tro 2nd
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Chapter 3. Molecules, Compounds, and Chemical
Equations
3.1 Hydrogen, Oxygen, and Water…………………………….
3.2 Chemical Bonds……………………………………………
3.3 Representing Compounds: Chemical Formulas and Molecular Models..
3.4 An Atomic-Level View of Elements and Compounds……………..
3.5 Ionic Compounds: Formulas and Names……………………
3.6 Molecular Compounds: Formulas and Names………………………
3.7 Formula Mass and the Mole Concept for Compounds…………
3.8 Composition of Compounds……………………………..
3.9 Determining a Chemical Formula from Experimental Data………
3.10 Writing and Balancing Chemical Equations……………………
3.11 Organic Compounds……………………….
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Chapter 3. KEY CONCEPTS
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Classifying Substances as Atomic
Elements, Molecular Elements,
Molecular Compounds, or Ionic
Compounds (3.4)
Writing Formulas for Ionic
Compounds (3.5)
Naming Simple Ionic Compounds
(3.5)
Naming Ionic Compounds Containing
Polyatomic Ions (3.5)
Writing Molecular and Empirical
Formulas (3.3)
Naming Molecular Compounds (3.6)
Naming Molecular Compounds (3.6)
Naming Acids (3.6)
Calculating Formula Mass (3.7)
Using Formula Mass to Count
Molecules by Weighing (3.7)
Calculating Mass Percent
Composition (3.8)
CHEM 100, Fall 2013 LA TECH
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Using Mass Percent Composition as a
Conversion Factor (3.8)
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Using Chemical Formulas as
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Conversion Factors (3.8)
Obtaining an Empirical Formula from
Experimental Data (3.9)
Calculating a Molecular Formula from an
Empirical Formula and Molar Mass (3.9)
Obtaining an Empirical Formula from
Combustion Analysis (3.9)
Balancing Chemical Equations (3.10)
3-4
1) Glucose has a molecular formula of C6H12O6
(M.W. 180.16 g/mol).
a. How many grams of C are available in 1 mole of
glucose?
b. How many grams of H are available in 1 mole of
glucose?
c. How many grams of O are available in 1 mole of
glucose?
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Percentage Composition
description of a compound based on the percent
relative amounts of each element in the
compound
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% Element Composition in Compounds
from Formula
n x Gram Atomic weight
% mass = --------------------------------------- x 100
formula weight (GMW, GFW)
n = subscript of the element in the formula
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2) What are the mass % of carbon and hydrogen in
carbon dioxide CO2?
(CO2; M.W. = 44.01 g/mole)
% C:
% O:
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Example: What is the percent composition of
carbon in chloroform , CHCl3, a substance once
used as an anesthetic?
MM = 1(gaw)C + 1(gaw)H + 3(gaw)Cl
= (12.01 + 1.00797 + 3 x 35.453) amu = 119.377amu
1 x (12.011)
%C =
%H =
119.377
1(1.00797)
%C =
119.377
3(35.453)
119.377
CHEM 100, Fall 2013 LA TECH
x 100
= 10.061% C
x 100 = 0.844359% H
x 100 = 89.095% Cl
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3) Bicarbonate of soda (sodium hydrogen
carbonate) has ionic formula: NaHCO3. Find the
mass percentages (mass %) of Na, H, C, and O in
sodium hydrogen carbonate.
Molar mass:
%Na:
% C:
% O:
%H:
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Example: What is the percent composition
of chloroform, CHCl3, a substance once
used as an anesthetic?
%C = 10.061% C
%H = 0.844% H
%Cl = 89.095% Cl
100.00
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Mass percent of element in C 6H12O6
Molar Mass = 180.16 g/mol
6 x 12
%C =
x 100 = 40.00% C
180.16
12 x 1.01
%H=
180.16
x 100 = 6.73% H
6 x 16.00
%O =
x 100 = 53.29% O
180.16
Total =100.02%
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4) NH3 (M.W. 17.03 g/mole) and NH4NO3 (F.W. 80.05
g/mole) used as nitrogen fertilizer. Which one will
provide more nitrogen for the same weight?
NH3
CHEM 100, Fall 2013 LA TECH
NH4NO3
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What is Empirical Formula?
Simple whole number ratio of each atom
expressed in the subscript of the
formula.
Molecular Formula = C6H12O6 of glucose
Empirical Formula = CH2O
Empirical formula is calculated from % composition
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How do you get Empirical
Formula from %
composition and vice
versa?
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Example: The burning of fossil fuels in
air produces a brown-colored gas, a
major air pollutant, that contains 2.34 g
of N and 5.34 g of O. What is the
empirical formula of the compound?
%N =
x 100 = 30.5% N
%O =
x 100 = 69.5% O
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Empirical Formula from % composition
%N = 30.5% N%O = 69.5% O
To get Relative # Atoms
(%/gaw), Divide by Smaller and by Multiply Integer
N 30.5 30.5/14.0067 = 2.18
O 69.5 69.5/15.9994 = 4.34
2.18/2.18 = 1.00
4.34/2.18 = 1.99
1.00
1
1.99
2
Empirical Formula NO2
Empirical Formula Weight = 46.0
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Molecular from Empirical Formula
Molecular Formula = n x empirical Formula
Molecular weight
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Molecular Formula from Empirical
Empirical Formula Weight 30
Molecular Formula = (CH 2 O) n = (CH 2 O) 6
Molecular Formula = C6H12O6 of glucose
Molecular Formula Weight = 180
Molecular Formula Weight 180
n
180
= ------------------------- = ---- = 6
CHEM 100, Fall 2013 LA TECH
Empirical Formula Weight 30
30
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Example: A colorless liquid used in
rocket engines, whose empirical
formula is NO2, has a molar mass (MW)
of 92.0. What is the molecular formula?
FW = 1(gaw)N + 2(gaw)O = 46.0
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5) A carbohydrate contains 38.71% weight of C,
9.71% weight of H and 51.58% weight of O. What
is the empirical formula?
Moles of C:
Moles of H:
Moles of O:
Mole ration of C:= H:= O: =
Simple mole ratio of C:= H:= O: =
Empirical formula =
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6) Combustion Analysis gives the following mass
%: 26.7% C; 2.2% H; 71.1% O
If a separate analysis determined the molecular
mass of the compound to be 90 g/mole.
Determine the Empirical Formula and the
Molecular Formula of the compound.
 Moles of C:
 Moles of H:
 Moles of O:
 Mole ration of C:= H:= O: =
 Simple mole ratio of C:= H:= O: =
 Empirical formula =
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6) Combustion Analysis gives the following mass
%: 26.7% C; 2.2% H; 71.1% O
If a separate analysis determined the molecular
mass of the compound to be 90 g/mole.
Determine the Empirical Formula and the
Molecular Formula of the compound.
 Empirical formula =
 Empirical formula mass
 Molar mass/empirical formula mass =
 Molecular formula =
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Combustion Analysis
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7) When 5.00 g of a compound containing only
carbon and hydrogen is burned completely, 15.7
g of CO2 and 6.45 g of H2O is produced. What is
the empirical formula?
 Moles of C:
 Moles of H:
 Mole ratio of C:= H:=
 Simple mole ratio of C:= H:=
 Empirical formula of the compound =
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Example Benzoic acid is known to
contain only C, H, and O. A 6.49-mg samp e
of benzoic acid was burned comp etely in a
C-H ana yzer. The increase in the mass o
each absorption tube showed that 16.4-mg
of CO2 and 2.85-mg of H2O formed. What is
the empirical formula of benzoic acid?
(16.4-mg of CO2 )(12.01-mg C)
#mg C = = 4.48-mg C
(44.01-mg CO2)
4.48-mg C
%C = 100 = 68.9% C
6.49-mg sample
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Example Benzoic acid is known to
contain only C, H, and O. A 6.49-mg samp e
of benzoic acid was burned comp etely in a
C-H ana yzer. The increase in the mass o
each absorption tube showed that 16.4-mg
of CO2 and 2.85-mg of H2O formed. What is
the empirical formula of benzoic acid?
(2.85-mg of H2O )(2.02-mg H)
#mg H = = 0.319-mg H
(18.02-mg H2O)
%C =
0.319-mg H
100 = 4.92% H
6.49-mg sample
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Example Benzoic acid is known to
contain only C, H, and O. A 6.49-mg sampl e
of benzoic acid was burned comp etely in a
C-H ana yzer. The increase in the mass of
each absorption tube showed that 16.4-mg
of CO2 and 2.85-mg of H2O formed. What is
the empirical formula of benzoic acid?
68.9% C 4.92% H
% O = (100 - (68.9% C + 4.92% H) = 26.2% O
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Chemical Equations
P4O10 (s) + 6H2O (l) = 4 H3PO4(l)
Reactants enter into a reaction.
Products are formed by the reaction.
Parantheses represent physical state
Stoichiometric Coefficients are numbers in front of chemical
formula formula
gives the amounts (moles) of each substance used and each
substance produced.
Equation Must be balanced!
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Chemical Reactions
Could be described in words
Chemical equation:
Reactants?
Products?
Reaction conditions?
=, ......> , <==> or ?
Stoichiometric coefficients?
Number in front of substances representing
moles, atoms, molecules
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Balanced Chemical Equation
Representation of a chemical reaction which
uses coefficients (prefix numbers known as
stoichiometric coefficients) to represent the
relative amounts of reactants and products
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Writing and Balancing
Chemical Equations
Write a word equation.
Convert word equation into formula equation .
Balance the formula equation by the use of prefixes
(coefficients) to balance the number of each type
of atom on the reactant and product sides of the
equation.
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Types of Chemical Reactions
0+4 >04
Combination
The spheres represent
atoms or groups of atoms.
00 0+0
Decomposition
4+00 J4+0
Displacement
40+00JO +
Exchange
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Reaction of H2 and I2
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Synthesis or Combination Reactions
Formation of a compound from simpler
compounds or elements.
Decomposition reactions
Compound breaks up to from simp er compounds
or el ements.
Displacement reactions
Single displ acement: In a compound an el ement
is replaced by another element.
Exchange reactions
Double displacement: n a compound group or ion
is replaced by another group or ion in another
compound.
.
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Formation Reactions
Formation of a compound from elements at
standard state.
Combustion Reactions
Compound reacts with oxygen to produce oxid es:
water and
carbon dioxide for organic
compounds
Acid/Base (Neutralization)Reactions
An acid and a base react to form water and salt
( most ionic compounds are sa lts)
Precipitation Reactions
when two aqueous salt sol utions are mixed an
insolu ble salt is formed when two aqueous salt
solutions are mixed .
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Balancing Chemical Equations
. Check for Diatomic Molecules - H 2 - N2- O2 - F2 - Cl2
- Br2 - I2
f these elements appear
By Themselves in an equation, they Must be written
with a subscript of 2
2. Balance Metals
3. Balance Nonmetals
1. Balance Oxygen
2. Balance Hydrogen
6.Recount All Atoms
7. If EVERY coefficient will red uce, rewrite in the
simplest
whole-number
ratio.
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Balance following reactions:
a ) MgO(s) + Si(s) Mg(s) + SiO 2 (s)
First balance Metals (Mg and Si): MgO(s) + Si(s) Mg(s) + SiO2(s)
Then balance Nonmetals(O): MgO(s) + Si(s) Mg(s) + SiO 2(s)
Recount All Atoms:
Reactant side: Mg, O, Si.
Product side: Mg, O, Si.
b)
P4O10(s)
+ H 2 O(l) H 3 PO 4 (l)
First balance P and H: P4O10(s) + H2O(l) H3PO4(l)
Then balance O: P 4 O 10 (s) + H 2 O(l) H 3 PO 4 (l)
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8) Balance following reactions:
Combustion reactions of organic compounds
f) C 3 H 8 (g) + O 2 (g) = CO 2 (g) + H 2 O(l)
g)
C4H10
(g) + O 2 (g) = CO 2 (g) + H 2 O(l)
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