Lecture 5. Chapter 3. Chemical Equations:

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Lecture 5. Chapter 3. Chemical Equations

• 2 H2 O2 ? 2 H2O

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Stoichiometry Calculations with Chemical
Formulas and Equations.

• Chemical Equations
• The equation for hydrogen burning in oxygen is
  represented as
• 2 H2 O2 ? 2 H2O
• The sign means reacts withThe arrow
  means produces

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2 H2 O2 ? 2 H2O

• The chemical substances on the left of the
  equation are the reactants.
• The chemical substances on the right of the
  equation are the products.
• The numbers in front of the formulas are the
  coefficients.

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Balanced equations

• Because of the Law of conservation of matter,
  the numbers of each type of atom must be the same
  on the left and the right side of the equation.
  The equation must be BALANCED.
• (4 Hs on each side, 2 Os on each side).

Note The coefficient multiplies through
everything in the substance that follows
2 H2 O2 ? 2 H2O
2 x 2 4 H 2 O 2 x 2 4H and 2 O
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View of reaction at the molecular level
4 Hs and 2 Os 4 Hs
and 2 Os Before reaction
after reaction
produces
O
H
H
O

H
H
O
O
two hydrogen one oxygen two
water molecules molecule
molecules
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Balancing equations

• Once we know the reactants and products in the
  reaction, we can write the unbalanced equation.
• e.g. CH4 O2 ? CO2 H2O
• To balance the equation, we must find the
  coefficients that will lead to a balanced
  equation.

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Balancing equations Change only the coefficients

• In balancing an equation, only coefficients
  can be changed. Changing subscripts is not
  allowed because this would lead to a change in
  the nature of the substances involved.

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• It is usually best to start by balancing the
  element(s) that occur in the fewest chemical
  formulas on either side of the equation.
• There always has to be the same number of
  carbon atoms on either side of the equation.
• Each C atom needs one O2 molecule to form CO2.
  
• The 4 Hs of CH4 need two O atoms one O2
  molecule to form 2 H2O
• CH4 2 O2 ? CO2 2 H2O
• (Note spaces between coefficients and formulas
  for substances).

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Molecular view of reaction
one C atom four O atoms one C
atom two O-atoms four H atoms
two O-atoms four
H-atoms


one methane two oxygen one
carbon two water molecule
molecules dioxide molecule
molecules
before after
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Combination and Decomposition reactions.

• Combination reaction Here two or more
  substances react to form one product
• 2 Mg O2(g) ? 2 MgO(s)
• In a decomposition reaction, one substance
  breaks down into two or more
• CaCO3(s) ? CaO(s) CO2(g)

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Combustion in air

• Combustion reactions are rapid reactions in
  air that produce a flame.
• Hydrocarbons consist only of C and H. When
  balancing an equation with combustion of these
  molecules, start with the C. The number of CO2
  molecules produced is the same as in the
  hydrocarbon
• C3H8 O2 ? 3 CO2 H2O
• (not yet balanced)

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• Next balance the H-atoms. The number of waters
  produced (4) is half the number of Hs (8) in the
  hydrocarbon
• C3H8 O2 ? 3 CO2 4 H2O

8 H 4 x 2 8 H
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Balance the O-atoms

• Finally, add up the number of O-atoms needed
  on the r.h.s. 6 4 10. Therefore need 5 O2
  molecules.
• C3H8 5 O2 ? 3 CO2 4 H2O

5 x 2 10 Os 3 x 2 6 Os 4 Os
10 O 10 O
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Molecular view of reaction
3 C atoms 10 O atoms 3 C atoms
4 O-atoms 8 H atoms
6 O-atoms 8 H-atoms


C3H8 5 O2 ? 3 CO2 4 H2O
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Combustion of O-containing derivatives

• One needs to remember here that there is an
  O on the left hand side, so the numbers of O2s
  needed is going to be less by that amount
• C2H5OH x O2 ? 2 CO2 3 H2O
• On r.h.s. are 4 3 7 O-atoms. But ethanol
  already has one O-atom, so need only 3 O2
  molecules
• C2H5OH 3 O2 ? 2 CO2 3 H2O
• 1 O 6 Os 4 Os
  3 Os
• 7 Os
  7 Os

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3.3. Formula weights.

• Chemical formulas and chemical equations
  contain quantitative information. They tell us
  how much of each reactant is needed for the
  reaction.

H 1.0 amu
O 16.0 amu
Weight of H2O 16.0 1.0 1.0 18.0 amu
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Formula and molecular weights

• The formula weight (F. Wt.) is the sum of the
  atomic weights of each atom in its chemical
  formula. If the chemical formula is that of a
  molecule, then the formula weight is referred to
  as a molecular weight (M. Wt.).

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Formula weights

• e.g. M. Wt. of glucose
• C6H12O6
• C 12.0 amu H 1.0 amu, O 16.0 amu
• So F. Wt.
• (6 x 12.0) (12 x 1.0) (6 x 16.0)
• 180 amu.

glucose
Exercise 3.5 F. Wt. of sucrose C12H22O11
(342.0) Ca(NO3)2 164.1 amu.
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Percentage composition from formulas

• When a new compound is analyzed to obtain its
  elemental composition, this is compared with the
  percentage composition obtained from its formula
• element
• (no. atoms of element) x (At. Wt. element) x
  100
• F. Wt. of compound
• C12H22O11 (sucrose) F. Wt. 342.0 amu.
• C (12 x 12.0) x 100/342 42.1
• H (22 x 1.0) x 100/342 6.4
• O (11 x 16.0) x 100/342 51.5
• check 42.1 6.4 51.5 100.0.

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The Mole.

• 3.4. Avogadros number and the Mole.
• Amu are far too small to be useful in everyday
  chemistry. We use instead the Mole (abbreviated
  mol) which is basically the F. Wt. or M. Wt.
  expressed in grams. e.g. Water has a formula
  weight of 18.0 amu, so
• 1 mole of water weighs 18.0 grams.

Lorenzo Romano Amedeo Carlo Avogadro, conte di
Quaregna e di Cerreto (1776 - 1856)
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One mole of water

• Water has
  a formula weight of 18.0
• amu, so 1 mole of
  water weighs 18.0
• grams. How do we
  arrive at this?
• Conversion factors
  1 amu 1.6 x 10-24 g,
• 1 mole contains
  6.022 x 1023 molecules,
• 1 molecule of H2O
  weighs 18.0 amu
• 18.0 amu x 6.022 x 1023 molecules x 1.6 x
  1024 g
• 1 molecule 1 mol
  1 amu
• 18.0 g/mol
  

18.0 grams of water
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Avogadros number

• Note 1 mol of 12C contains 6.022 x 1023 12C
  atoms. 1 mol of C2H4 contains 6.022 x 1023 C2H4
  molecules. 1 mol of glucose contains
• 6.022 x
  1023 molecules

1 mol of water 18.0 grams of water
1 mol of water contains 6.022 x 1023 water
molecules. How many H-atoms does it contain?
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Think about dozens instead of moles
One dozen water molecules..
contains one dozen oxygen atoms, but two dozen
hydrogen atoms. One mol of water contains one mol
of oxygen atoms but two moles of Hydrogen atoms
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Numbers of atoms in molecules

• One mol of ethylene (C2H4) contains 6.022 x 1023
• ethylene molecules, but
• One mol of
  ethylene (C2H4)
• contains 4
  x 6.022 x 1023
• 
  2.4088 x 1024 H-atoms
• One mol of
  ethylene (C2H4)
• contains 2
  x 6.022 x 1023
• 
  1.2044 x 1024 C-atoms

ethylene
C2H4
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Counting BBs

• Suppose you sell BBs for air-guns in boxes
  each containing 1000 BBs. How would you put 1000
  BBs in each box?

Box must contain 1000 BBs (one BB weighs
0.34 g)
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Weighing as a means of counting

• Avogadros number (6.022 x 1023) is a very
  large number, but it is still just a number, like
  a dozen is a number. Just as with BBs, there are
  too many molecules in even a tiny amount of a
  substance for us to count them. We have to weigh
  them. And we know that in one mol of substance
  there are 6.022 x 1023 molecules. So chemists
  work in moles, knowing 1 mole contains 6.022 x
  1023 molecules.

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Fractions of moles and numbers of atoms

• Exercise 3.8. Calculate the number of H atoms
  in 0.350 moles of C6H12O6.
• Conversion 1 mole 6.022 x 1023 molecules
• 1 molecule 12 H-atoms
• 0.35 moles x 6.022 x 1023 molecules x 12 atoms
  
• 1 mole
  1 molecule
• 25.3 x 1023 2.53 x 1024 H-atoms

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Table 3.2 Mole Relationships

• Substance F.Wt. Molar Mass No. of particles
• 
  (in one mole)
• Atomic N 14.0 14.0 g/mol 6.022 x 1023
• N2 gas 28.0 28.0 g/mol 6.022 x 1023
• N atoms in
• 1 mole N2 1.2044 x 1024 Ns
• Ag(s) 107.9 107.9 g/mol 6.022 x 1023
• Ag ions 107.9 107.9 g/mol 6.022 x 1023
• BaCl2 208.2 208.2 g/mol 6.022 x
  1023 Ba2
• 1.2044 x 1024 Cl-
• AlCl3 133.3 133.3 g/mol 6.022 x 1023
  Al3
• 
  1.806 x
  1024 Cl-

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Some problems

• How many moles of H atoms are there in 1.0
  moles of NH4Cl ?
• How many H atoms are there in 1.0 moles of NH4Cl
  ?
• How many H atoms are there in 2 moles of NH4Cl
  ?
• How many H atoms are there in 0.3724 moles of
  NH4Cl ?

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Interconverting masses and numbers of particles

• e.g. How many Cu
  atoms in
• a Cu penny, weight
  3.0 g
• (assume the penny
  is pure
• copper).
• copper penny
• Conversion Cu 63.5 g/mol 1 mol 6.022 x
  1023 atoms (note units)
• 3.0 g x 6.022 x 1023 atoms x 1 mol
  
• 1 mol 63.5
  g
• 2.8 x 1022 atoms

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3.5 Empirical Formulas from Analyses.

• When Chemists discover new compounds, they may
  analyze them to get their percentage elemental
  composition. Thus, if we analyze a compound and
  find that it contains 73.9 Hg and 26.1 Cl by
  weight, we can work out the molar ratio.
• Note that when we say something is 73.9 Hg,
  that means 100.0 g of the substance would contain
  73.9 g Hg, and similarly 26.1 g Cl.

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• Assume that we have 100g sample. Thus, the
  percentages mean that we have 73.9 g Hg and 26.1
  g Cl.
• Conversion factors (from periodic table)
• Hg, 1 mol 200.6 g ( atomic wt.)
• Cl, 1 mol 35.5 g ( atomic wt.)
• So the molar ratio is given by
• Hg 73.9 g x 1 mol/200.6 g 0.368 mol
• Cl 26.2 g x 1 mol/35.5 g 0.735 mol

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• To get the molar ratio, divide through by the
  lowest molar ratio present
• Hg 0.368 mol/0.368 mol 1.00
• Cl 0.735 mol/0.368 mol 1.997
• 
  ( 2.00)
• We can say that within experimental error, the
  empirical formula of the compound is HgCl2.
• Example 3.13 Ascorbic acid is 40.92 C, 4.58
  H, and 54.40 O by mass. What is the empirical
  formula?

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Example 3.13 Ascorbic acid

• Conversion factors C, 1 mol 12.0 g H, 1
  mol 1.0 g O, 1 mol 16.0 g.
• C 40.92 g x 1 mol/12.0 g 3.407 mol
• H 4.58 g x 1 mol/1.0 g 4.54 mol
• O 54.40 g x 1 mol/16.0 g 3.406 mol
• Divide through by lowest number of moles
  (3.406)
• C 3.407 mol/3.406 mol 1.00
• H 4.54 mol /3.406 mol 1.33
• O 3.406 mol /3.406 mol 1.00
• Multiply through by 3 to get all near whole
  numbers.
• Empirical formula is C3H4O3.

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Molecular formula from empirical Formula.

• If we know the molecular weight of a
  compound, we can then convert the empirical
  formula into a molecular formula .
• We calculate the empirical formula weight,
  which is the formula weight calculated from the
  empirical formula. We then obtain a whole number
  multiple by dividing the molecular weight by the
  empirical formula weight, and it is this multiple
  that we use to multiply through the subscripts in
  the empirical formula.
• Mesitylene has an empirical formula of
  C3H4. The experimentally determined M.Wt. is
  121.0 amu. What is the molecular formula?

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An example mesitylene

• Mesitylene has an empirical formula of C3H4.
  The experimentally determined M.Wt. is 121.0 amu.
  What is the molecular formula of mesitylene?
• C3H4 empirical formula weight
• 3 x 12 4 x 1 40 amu
• Multiple Experimental M.Wt
• empirical formula
  weight
• 121amu /40 amu
• 3.02 ( 3.0 )
• Multiple 3.0, so molecular formula C9H12