Mass Relationships in Chemical Reactions

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Mass Relationships in Chemical Reactions

• Chapter 3

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Micro World atoms molecules
Macro World grams
Atomic mass is the mass of an atom in atomic mass
units (amu)
By definition 1 atom 12C weighs 12 amu
On this scale 1H 1.008 amu 16O 16.00 amu
3.1
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Natural lithium is 7.42 6Li (6.015 amu) 92.58
7Li (7.016 amu)
Average atomic mass of lithium
6.941 amu
3.1
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The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly 12.00 grams of 12C
1 mol NA 6.0221367 x 1023
Avogadros number (NA)
3.2
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eggs
shoes
Molar mass is the mass of 1 mole of
in grams
marbles
atoms
1 mole 12C atoms 6.022 x 1023 atoms 12.00 g 1
12C atom 12.00 amu
1 mole 12C atoms 12.00 g 12C 1 mole lithium
atoms 6.941 g of Li
For any element atomic mass (amu) molar mass
(grams)
3.2
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One Mole of
S
C
Hg
Cu
Fe
3.2
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x
1 amu 1.66 x 10-24 g or 1 g 6.022 x 1023 amu
molar mass in g/mol
M
NA Avogadros number
3.2
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How many atoms are in 0.551 g of potassium (K) ?
1 mol K 39.10 g K
1 mol K 6.022 x 1023 atoms K
0.551 g K
8.49 x 1021 atoms K
3.2
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Molecular mass (or molecular weight) is the sum
of the atomic masses (in amu) in a molecule.
For any molecule molecular mass (amu) molar
mass (grams)
1 molecule SO2 64.07 amu 1 mole SO2 64.07 g
SO2
3.3
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How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O (3 x 12) (8 x 1) 16 60 g C3H8O
1 mol C3H8O molecules 8 mol H atoms
1 mol H 6.022 x 1023 atoms H
72.5 g C3H8O
5.82 x 1024 atoms H
3.3
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KE 1/2 x m x v2 v (2 x KE/m)1/2 F q x v x B
3.4
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Percent composition of an element in a compound
n is the number of moles of the element in 1 mole
of the compound
52.14 13.13 34.73 100.0
3.5
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Types of Formulas

• Empirical Formula
• The formula of a compound that expresses the
  smallest whole number ratio of the atoms present.
• Ionic formula are always empirical formula
• Molecular Formula
• The formula that states the actual number of
  each kind of atom found in one molecule of the
  compound.

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To obtain an Empirical Formula

• 1. Determine the mass in grams of each element
  present, if necessary.
• 2. Calculate the number of moles of each
  element.
• 3. Divide each by the smallest number of moles to
  obtain the simplest whole number ratio.
• If whole numbers are not obtained in step 3),
  multiply through by the smallest number that will
  give all whole numbers
• Be careful! Do not round off numbers
  prematurely

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• A sample of a brown gas, a major air pollutant,
  is found to contain 2.34 g N and 5.34g O.
  Determine a formula for this substance.
• require mole ratios so convert grams to moles
• moles of N 2.34g of N 0.167 moles of N
• 14.01 g/mole
• moles of O 5.34 g 0.334 moles of O
• 16.00 g/mole
• Formula

(HONORS only)
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Calculation of the Molecular Formula

• A compound has an empirical formula of NO2. The
  colourless liquid, used in rocket engines has a
  molar mass of 92.0 g/mole. What is the molecular
  formula of this substance?
• empirical formula mass 14.012 (16.00) 46.01
  g/mol
• n molar mass 92.0 g/mol emp.
  f. mass 46.01 g/mol
• n 2
• N2O4

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Empirical Formula from Composition

• A substance has the following composition by
  mass 60.80 Na 28.60 B 10.60 H
• What is the empirical formula of the substance?
• Consider a sample size of 100 grams
• This will contain 60.80 grams of Na, 28.60
  grams of B, and 10.60 grams H
• Determine the number of moles of each
• Determine the simplest whole number ratio

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Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
6.0 g C 0.5 mol C
1.5 g H 1.5 mol H
g of O g of sample (g of C g of H)
4.0 g O 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
3.6
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Mass Changes in Chemical Reactions

• Write balanced chemical equation
• Convert quantities of known substances into moles
• Use coefficients in balanced equation to
  calculate the number of moles of the sought
  quantity
• Convert moles of sought quantity into desired
  units

3.8
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Other units

• Molarity
• Moles solute / L solution
• Gases
• 22.4 L 1 mole of ANY GAS at STP

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Methanol burns in air according to the equation
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
molar mass CH3OH
molar mass H2O
coefficients chemical equation
209 g CH3OH
235 g H2O
3.8
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Limiting Reagents
3.9
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Method 1

• Pick A Product
• Try ALL the reactants
• The lowest answer will be the correct answer
• The reactant that gives the lowest answer will be
  the limiting reactant

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Limiting Reactant Method 1
LimitingReactant

• 10.0g of aluminum reacts with 35.0 grams of
  chlorine gas to produce aluminum chloride. Which
  reactant is limiting, which is in excess, and how
  much product is produced?
• 2 Al 3 Cl2 ? 2 AlCl3
• Start with Al
• Now Cl2

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g
AlCl3 27.0 g Al 2 mol Al
1 mol AlCl3
49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g
AlCl3 71.0 g Cl2 3 mol Cl2
1 mol AlCl3
43.9g AlCl3
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Method 2

• Convert one of the reactants to the other
  REACTANT
• See if there is enough reactant A to use up the
  other reactants
• If there is less than the GIVEN amount, it is the
  limiting reactant
• Then, you can find the desired species

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In one process, 124 g of Al are reacted with 601
g of Fe2O3
Calculate the mass of Al2O3 formed.
367 g Fe2O3
124 g Al
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9
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Use limiting reagent (Al) to calculate amount of
product that can be formed.
234 g Al2O3
124 g Al
3.9
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Finding Excess Practice

• 10.0g of aluminum reacts with 35.0 grams of
  chlorine gas 2 Al 3 Cl2 ? 2 AlCl3
• We found that chlorine is the limiting reactant,
  and 43.8 g of aluminum chloride are produced.

35.0 g Cl2 1 mol Cl2 2 mol Al 27.0 g Al
71 g Cl2 3 mol Cl2 1
mol Al
7.8 g Al USED!
10.0 g Al 7.8 g Al 2.2 g Al EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant!
Once you determine the LR, you should only start
with it!
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Theoretical Yield is the amount of product that
would result if all the limiting reagent reacted.
Actual Yield is the amount of product actually
obtained from a reaction.
3.10
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Chemistry In Action Chemical Fertilizers
Plants need N, P, K, Ca, S, Mg