Announcements

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Announcements

• This Friday 5/16 there will be no lecture--its
  my moms birthday. No thats not why
• Chemistry Dept Scholars Day activities
• 12-2pm Student research posters in Chem Bldg
• 2-5pm Student research seminars in SL 140

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Average 72.2 St. Dev 12.5
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Average 151.9 StDev 22.7
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An Ice Cream Sundae Analogy for Limiting
Reactions
Fig. 3.10
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Molecular Formula
Molecules
Atoms
Avogadros Number
6.022 x 1023
Moles
Moles
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Chemical Equations
Qualitative Information
Reactants
Products
States of Matter (s) solid (l)
liquid (g) gaseous (aq) aqueous
2 H2 (g) O2 (g) 2 H2O (g)
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Chemical Equation Calculation - I
Atoms (Molecules)
Avogadros Number
6.02 x 1023
Molecules
Reactants
Products
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Chemical Equation Calculation - II
Mass
Atoms (Molecules)
Molecular Weight
Avogadros Number
g/mol
6.02 x 1023
Molecules
Reactants
Products
Moles
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Mass - Mole Relationships of a Compound
For an Element
For a Compound
Mass (g) of Element
Mass (g) of compound
Moles of Element
Amount (mol) of compound
Amount (mol) of compound
Molecules (or formula units of compound)
Atoms of Element
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Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem Tungsten (W) is the element used as the
filament in light bulbs, and
has the highest melting point of any element
3680oC. How many moles of
tungsten, and atoms of the
element are contained in a 35.0 mg sample of the
metal? Plan Convert mass into moles by dividing
the mass by the atomic weight of the
metal, then calculate the number of atoms by
multiplying by Avogadros
number! Solution Converting from mass of W to
moles Moles of W 35.0
mg W x 0.00019032
mol
1.90 x
10 - 4 mol NO. of W atoms 1.90 x 10 - 4 mol
W x


1.15 x 1020 atoms of Tungsten
1 mol W 183.9 g W
6.022 x 1023 atoms 1 mole of W
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Calculating the Moles and Number of Formula
Units in a Given Mass of Cpd.
Problem Trisodium Phosphate is a component of
some detergents. How many moles
and formula units are in a 38.6 g sample? Plan
We need to determine the formula, and the
molecular mass from the atomic masses
of each element multiplied by the
coefficients. Solution The formula is Na3PO4.
Calculating the molar mass M
3x Sodium 1 x Phosphorous 4 x Oxygen
3 x 22.99 g/mol 1 x 30.97 g/mol
4 x 16.00 g/mol 68.97
g/mol 30.97 g/mol 64.00 g/mol 163.94 g/mol
Converting mass to moles
Moles Na3PO4 38.6 g Na3PO4 x (1 mol Na3PO4)

163.94 g Na3PO4 0.23545 mol
Na3PO4
Formula units 0.23545 mol Na3PO4 x 6.022 x 1023
formula units
1 mol Na3PO4
1.46 x 1023 formula units
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Flow Chart of Mass Percentage Calculation
Moles of X in one mole of Compound
M (g / mol) of X
Mass (g) of X in one mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100
Mass of X
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Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem Sucrose (C12H22O11) is common table
sugar. ( a) What is the mass percent of each
element in sucrose? ( b) How many grams of
carbon are in 24.35 g of sucrose?

(a) Determining the mass percent of each
element mass of C 12 x 12.01
g C/mol 144.12 g C/mol
mass of H 22 x 1.008 g H/mol
22.176 g H/mol mass of O 11 x
16.00 g O/mol 176.00 g O/mol

342.296 g/mol Finding
the mass fraction of C in Sucrose C
Total mass
of C 144.12 g C
mass of 1 mole of sucrose
342.30 g Cpd
Mass Fraction of C

0.421046 To find mass of C 0.421046 x
100 42.105
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Calculating Mass Percents and Masses of Elements
in a Sample of Compound - II
(a) continued
Mass of H
x 100 x 100
6.479 H Mass of O
x 100
x 100
51.417 O (b) Determining the
mass of carbon Mass (g) of C mass of
sucrose X( mass fraction of C in sucrose) Mass
(g) of C 24.35 g sucrose X
10.25 g C
mol H x M of H 22 x
1.008 g H mass of 1 mol sucrose
342.30 g
mol O x M of O 11 x
16.00 g O mass of 1 mol sucrose
342.30 g
0.421046 g C 1 g sucrose
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Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a
compound that agrees
with the elemental analysis! The
smallest set of whole numbers of
atoms. Molecular Formula - The formula of the
compound as it exists,
it may be a multiple of the Empirical
formula.
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Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
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Some Examples of Compounds with the Same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 ,
C3H6 , C4H8 OH or HO
H2O2 S

S8 P
P4
Cl
Cl2 CH2O
(carbohydrates)
C6H12O6
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Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
0.2469 mol Na
Moles of Cr 6.420 g Cr x
0.12347 mol Cr Moles of O 7.902 g
O x 0.4939 mol O
1 mol Na 22.99 g Na
1 mol Cr 52.00 g Cr
1 mol O 16.00 g O
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Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by
smallest subscript)
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers
Na2CrO4 Sodium Chromate
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Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem The sugar burned for energy in cells of
the body is Glucose (M 180.16 g/mol), elemental
analysis shows that it contains 40.00 mass C,
6.719 mass H, and 53.27 mass O. (a)
Determine the empirical formula of glucose.
(b) Determine the molecular formula. Plan We are
only given mass , and no weight of the compound
so we will assume 100g of the compound,
and becomes grams, and we can do as
done previously with masses of the
elements. Solution Mass Carbon
40.00 x 100g/100 40.00 g C Mass
Hydrogen 6.719 x 100g/100 6.719g H
Mass Oxygen 53.27 x 100g/100 53.27 g
O
99.989 g Cpd
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Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles
Moles of C Mass of C x
3.3306 moles C Moles of H Mass of H x
6.6657 moles H Moles
of O Mass of O x 3.3294
moles O Constructing the preliminary formula
C 3.33 H 6.67 O 3.33 Converting to integer
subscripts, divide all subscripts by the
smallest C 3.33/3.33 H 6.667 / 3.33 O3.33 /
3.33 CH2O
1 mole C 12.01 g C
1 mol H 1.008 g H
1 mol O 16.00 g O
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Determining the Molecular Formula
from Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula The
formula weight of the empirical formula is
1 x C 2 x H 1 x O 1 x 12.01 2 x 1.008 1
x 16.00 30.03
M of Glucose empirical formula mass
Whole-number multiple

6.00 6
180.16 30.03
Therefore the Molecular Formula is
C 1 x 6 H 2 x 6 O 1 x 6 C6H12O6
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Adrenaline Is a Very Important Compound in the
Body - I

• Analysis gives
• C 56.8
• H 6.50
• O 28.4
• N 8.28
• Calculate the Empirical Formula

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Adrenaline - II

• Assume 100g!
• C 56.8 g C/(12.01 g C/ mol C) 4.73 mol C
• H 6.50 g H/( 1.008 g H / mol H) 6.45 mol H
• O 28.4 g O/(16.00 g O/ mol O) 1.78 mol O
• N 8.28 g N/(14.01 g N/ mol N) 0.591 mol N
• Divide by 0.591
• C 8.00 mol C 8.0 mol C or
• H 10.9 mol H 11.0 mol H
• O 3.01 mol O 3.0 mol O C8H11O3N
• N 1.00 mol N 1.0 mol N

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Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
m 2
m 2
CnHm (n ) O2 n CO(g) H2O(g)
Fig. 3.4
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Ascorbic Acid ( Vitamin C ) - I Contains C , H ,
and O

• Upon combustion in excess oxygen, a 6.49 mg
  sample yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate its Empirical formula!
• C 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
• 2.65 x 10-3 g C
• H 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
• 2.92 x 10-4 g H
• Mass Oxygen 6.49 mg - 2.65 mg - 0.30 mg
• 3.54 mg O

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Vitamin C Combustion - II

• C 2.65 x 10-3 g C / ( 12.01 g C / mol C )
• 2.21 x 10-4 mol C
• H 0.295 x 10-3 g H / ( 1.008 g H / mol H )
• 2.92 x 10-4 mol H
• O 3.54 x 10-3 g O / ( 16.00 g O / mol O )
• 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C 1.00 Multiply each by 3 3.00 3.0
• H 1.32
  3.96 4.0
• O 1.00
  3.00 3.0

C3H4O3
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Determining a Chemical Formula from
Combustion Analysis - I
Problem Erthrose (M 120 g/mol) is an
important chemical compound as
a starting material in chemical synthesis, and
contains Carbon Hydrogen, and
Oxygen. Combustion analysis of
a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O. Plan We find the masses
of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of
Carbon and Hydrogen are subtracted from
the sample mass to get the mass of
Oxygen. We then calculate moles, and construct
the empirical formula, and from the
given molar mass we can calculate the
molecular formula.
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Determining a Chemical Formula from Combustion
Analysis - II
Calculating the mass fractions of the elements
Mass fraction of C in CO2

0.2729 g C
/ 1 g CO2 Mass fraction of H in H2O


0.1119 g H / 1 g H2O Calculating masses of C
and H Mass of Element mass of compound x
mass fraction of element
mol C x M of C mass of 1 mol CO2
1 mol C x 12.01 g C/ 1 mol C 44.01 g
CO2
mol H x M of H mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H 18.02
g H2O
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Determining a Chemical Formula from
Combustion Analysis - III
0.2729 g C 1 g CO2
Mass (g) of C 1.027 g CO2 x
0.2803 g C Mass (g) of H 0.4194 g H2O x
0.04693 g H Calculating
the mass of O Mass (g) of O Sample mass -(
mass of C mass of H )
0.700 g - 0.2803 g C - 0.04693 g H 0.37277 g
O Calculating moles of each element C
0.2803 g C / 12.01 g C/ mol C 0.02334 mol C
H 0.04693 g H / 1.008 g H / mol H 0.04656 mol
H O 0.37277 g O / 16.00 g O / mol O
0.02330 mol O C0.02334H0.04656O0.02330 CH2O
formula weight 30 g / formula 120 g /mol / 30 g
/ formula 4 formula units / cpd C4H8O4
0.1119 g H 1 g H2O
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Some Compounds with Empirical Formula CH2O
(Composition by Mass 40.0 C, 6.71 H, 53.3O)
Molecular M Formula
(g/mol) Name Use or Function
CH2O 30.03 Formaldehyde
Disinfectant Biological

preservative C2H4O2 60.05
Acetic acid Acetate polymers vinegar

( 5 solution) C3H6O3
90.08 Lactic acid Causes milk
to sour forms
in muscle
during exercise C4H8O4 120.10
Erythrose Forms during sugar

metabolism C5H10O5
150.13 Ribose Component of
many nucleic
acids and
vitamin B2 C6H12O6 180.16
Glucose Major nutrient for energy

in cells
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Two Compounds with Molecular Formula C2H6O
Property Ethanol
Dimethyl Ether
M (g/mol) 46.07
46.07 Color
Colorless
Colorless Melting point - 117oC
- 138.5oC Boiling point
78.5oC -
25oC Density (at 20oC) 0.789 g/mL
0.00195 g/mL Use
Intoxicant in In
refrigeration
alcoholic beverages
H H
H H H C C O
H H C O C H
H
H H
H
Table 3.4
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When 66.6 g of O2 gas is mixed with 27.8 g of NH3
gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is
produced by the following reaction 2CH4 2NH3
3O2 2HCN 6H2O Given 17.03 g
NH3/mol, 32.00 g O2/mol, 27.03 g HCN/mol, 16.04 g
CH4/mol What is the percent yield of HCN in this
reaction?
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Theoretical Yield Which Reactant is Limiting?

• 1) calculate moles (or mass) of product formed by
  complete reaction of each reactant.
• 2) the reactant that yields the least product is
  the limiting reactant.
• 3) the theoretical yield for a reaction is the
  maximum amount of product that could be generated
  by complete consumption of the limiting reactant.

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2CH4 2NH3 3O2 2HCN 6H2O 66.6 g
of O2 2.08 mol O2 27.8 g of NH3 1.63 mol
NH3 25.1 g of CH4 1.56 mol CH4 Which reactant
is limiting? 2.08 mol O2 can yield 1.39 mol (or
37.5 g) HCN 1.63 mol NH3 can yield 1.63 mol (or
44.1 g) HCN 1.56 mol CH4 can yield 1.56 mol (or
42.2 g) HCN O2 is the limiting reactant.
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O2 is the limiting reagent thus, the theoretical
yield is based on 100 consumption of O2. 2.08
mol O2 can yield 1.39 mol (or 37.5 g) HCN

• yield actual yield

100
theoretical yield
yield 36.4 g HCN
97.1
100
37.5 g HCN
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When 66.6 g of O2 gas is mixed with 27.8 g of NH3
gas and 25.1 g of CH4 gas, 36.4 g of HCN gas is
produced by the following reaction 2CH4 2NH3
3O2 2HCN 6H2O Given 17.03 g
NH3/mol, 32.00 g O2/mol, 27.03 g HCN/mol, 16.04 g
CH4/mol What is the percent yield of HCN in this
reaction? How many grams of NH3 remain?