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Ch. 3 Stoichiometry: Calculations with Chemical Formulas. Law of Conservation of Mass. Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged. Stoichiometry. The quantitative nature of chemical formulas and chemical reactions. Reactants.

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law of conservation of mass
Law of Conservation of Mass
  • Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged.
stoichiometry
Stoichiometry
  • The quantitative nature of chemical formulas and chemical reactions
reactants
Reactants
  • The chemical formulas on the left of the arrow that represent the starting substances

2H2 + O2 2H2O

Reactants

products
Products
  • The substances that are produced in the reaction and appear to the right of the arrow

2H2 + O2 2H2O

Products

slide6

Because atoms are neither created nor destroyed in any reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

balancing chemical equations
Balancing Chemical Equations

CH4 + O2 CO2 + H2O

C=1 C=1

H=4 H=2

O=2 O=3

balancing chemical equations1
Balancing Chemical Equations

CH4 + O2 CO2 + 2H2O

C=1 C=1

H=4 H=2 X 2 =4

O=2 O=3

balancing chemical equations2
Balancing Chemical Equations

CH4 + O2 CO2 + 2H2O

C=1 C=1

H=4 H=2 x 2 = 4

O=2 O= 2 + 2x1 = 4

balancing chemical equations3
Balancing Chemical Equations

CH4 + 2O2 CO2 + 2H2O

C=1 C=1

H=4 H=2 x 2 = 4

O=2 x 2 = 4 O= 2 + 2x1 = 4

combustion reactions
Combustion Reactions
  • Rapid reactions that produce a flame.
  • Most combustion reactions involve O2 as a reactant
  • Form CO2 and H2O as products
combustion reactions1
Combustion Reactions

C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)

C= 3 C=1 X 3 = 3

H=8 H=2 X 4 = 8

O= 2 X 5 = 10 O=(2 X3)+(1X4)=10

combination reactions synthesis
Combination Reactions (synthesis)
  • 2 or more substances react to form 1 product.
combination reactions synthesis1
Combination Reactions (synthesis)

2Mg(s) + O2(g) 2MgO(s)

Mg=1 x 2=2 Mg= 1 x 2=2

O= 2 O=1 x 2 = 2

decomposition reaction
Decomposition Reaction
  • 1 substance undergoes a reaction to produce 2 or more substances
decomposition reaction1
Decomposition Reaction

CaCO3 (s)  CaO (s) + CO2(g)

Ca=1 Ca=1

C=1 C=1

O=3 O=1+2=3

3 methods of measuring
3 Methods of Measuring
  • Counting
  • Mass
  • Volume
example 1
Example 1
  • If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?
example 11
Example 1
  • Count: 1 dozen apples = 12 apples
  • Mass: 1 dozen apples = 2.0 kg apples
  • Volume: 1 dozen apples = 0.20 bushels

apples

Conversion Factors:

  • 1 dozen2.0 k.g1 dozen

12 apples 1 dozen 0.20 bushels

example 12
Example 1
  • 0.50 bushel x 1 dozen x 2.0 kg =

0.20 bushel 1 dozen

= 5.0 kg

avogadro s number
Avogadro’s Number
  • Named after the Italian scientist Amedo Avogadro di Quaregna
  • 6.02 x 10 23
mole mol
Mole (mol)
  • 1 mol = 6.02 x 10 23 representative particles
  • Representative particles: atoms, molecules ions, or formula units (ionic compound)
mole mol1
Mole (mol)
  • Moles= representative x 1 mol

particles 6.02 x 10 23

example 2 atoms mol
Example 2 (atoms  mol)
  • How many moles is 2.80 x 10 24 atoms of silicon?
example 2
Example 2
  • 2.80 x 10 24 atoms Si x 1 mol Si

6.02 x 10 23 atoms Si

= 4.65 mol Si

example 3 mol molecule
Example 3 (mol  molecule)
  • How many molecules of water is 0.360 moles?
example 3
Example 3
  • 0.360 mol H2O x 6.02 x 10 23 molecules H2O 1 mol H2O

=2.17 molecules H2O

the mass of a mole of an element
The Mass of a Mole of an Element
  • The atomic mass of an element expressed in grams = 1 mol of that element = molar mass

Molar mass S

Molar mass Hg

Molar mass C

Molar mass Fe

slide29

6.02 x 10 23 atoms S

6.02 x 10 23 atoms Hg

6.02 x 10 23 atoms C

6.02 x 10 23 atoms Fe

example 4 mol gram
Example 4 (mol  gram)
  • If you have 4.5 mols of sodium, how much does it weigh?
example 4
Example 4
  • .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na

1 mol Na

example 5 grams atoms
Example 5 (grams  atoms)
  • If you have 34.3 g of Iron, how many atoms are present?
example 5
Example 5
  • 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms

55.8 g Fe 1 mol Fe

=3.70 x 10 23 atoms Fe

the mass of a mole of a compound
The Mass of a Mole of a Compound
  • To find the mass of a mole of a compound you must know the formula of the compound
  • H2O  H= 1 g x 2

O= 16 g

18 g = 1 mole = 6.02 x 10 23

molecules

example 6 gram mol
Example 6 (gram  mol)
  • What is the mass of 1 mole of sodium hydrogen carbonate?
example 6
Example 6
  • Sodium Hydrogen Carbonate = NaHCO3
  • Na=23 g
  • H=1 g
  • C=12 g
  • O=16 g x3
  • 84 g NaHCO3 = 1 mol NaHCO3
mole volume relationship
Mole-Volume Relationship
  • Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical
avogadro s hypothesis
Avogadro’s Hypothesis
  • States that equal volumes of gases at the same temperature and pressure contain the same number of particles
  • Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible
standard temperature and pressure stp
Standard Temperature and Pressure (STP)
  • Volume of a gas changes depending on temperature and pressure
  • STP= 0oC (273 K)

101.3 kPa (1 atm)

standard temperature and pressure stp1
Standard Temperature and Pressure (STP)
  • At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume
conversion factors
Conversion Factors
  • AT STP
  • 1 mol gas22.4 L gas

22.4 L gas 1 mol gas

example 7
Example 7
  • At STP, what volume does 1.25 mol He occupy?
example 71
Example 7
  • 1.25 mol He x 22.4 L He = 28.0 L He

1 mol He

example 8
Example 8
  • If a tank contains 100. L of O2 gas, how many moles are present?
example 81
Example 8
  • 100. L O2 X 1 mol O2 = 4.46 mol O2

22.4 L O2

calculating molar mass from density
Calculating Molar Mass from Density
  • The density of a gas at STP is measured in g/L
  • This value can be sued to determine the molar mass of gas present
example 9
Example 9
  • A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.
example 91
Example 9
  • 1 mol gas x 22.4 L gas X 3.58 g gas =

1 mol gas 1 L gas

Molar Mass= 80.2 g

percent composition
Percent Composition
  • The relative amounts of the elements in a compound
  • These percentages must equal 100
percent composition1
Percent Composition
  • %element = mass of element x 100

mass of compound

example 10
Example 10
  • Find the mass percentage of each element present in Al2 (CO3)3
example 101
Example 10
  • Al2(CO3)3
  • Al= 27 g x 2 = 54 g / 234 g x 100=23%
  • C= 12 g x 3 = 36 g/ 234 g x 100= 15%
  • O = 16 g x 9 = 144 g / 234 g x 100=62%

234 gAl2(CO3)3

empirical formula
The simplest whole number ratio of atoms in a compound

The formula obtained from percentage composition

Ex CH , CH4, H2O, C3H8

NOT C2H4, or C6H12O6 these could be simplified

Empirical Formula
example 111
Example 11
  • Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical fromula.
example 112
Example 11

Assume that you have 100 grams of the compound therefore:

Hg = 73.9 %  73.9 g

Cl= 26.1%  26.1 g

example 113
Example 11

Step 2: Change grams of your compound to moles

Hg = 73.9 g x 1 mol =0.368 mol Hg

200.6g

Cl= 26.1 g x 1 mol = 0.735 mol Cl

35.5 g

example 114
Example 11

Step 3: Find the lowest number of moles present

Hg = 73.9 g x 1 mol =0.368 mol Hg

200.6g

Cl= 26.1 g x 1 mol = 0.735 mol Cl

35.5 g

0.368 < 0.735

example 115
Example 11

Step 4: Divide by the lowest number of moles to obtain whole numbers

Hg = 73.9 g x 1 mol = 0.368 mol = 1

200.6g 0.368 mol

Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2

35.5 g 0.368 mol

example 116
Example 11

Step 5: Put the whole numbers into the empirical formula

Hg = 73.9 g x 1 mol = 0.368 mol = 1

200.6g 0.368 mol

Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2

35.5 g 0.368 mol

HgCl2

molecular formulas
Molecular Formulas
  • The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula
  • We can obtain the molecular formula from the empirical formula IF we know the molecular weight of the compound.
example 121
Example 12
  • The empirical formula of ascorbic acid is C3H4O3. The molecular weight of ascorbic acid is 176 amu. Determine the molecular formula.
example 122
Example 12

Step 1: First determine the molecular weight of the empirical formula

C3H4O3

C= 12 amu x 3

H= 1 amu x 4

O= 16 amu x 3

88 amu

example 123
Example 12

Step 2: Divide the molecular weight of the molecular formula by the molecular weight of the empirical formula

C3H4O3 176 amu = 2

C= 12 amu x 3 88 amu

H= 1 amu x 4

O= 16 amu x 3

88 amu

example 124
Example 12

Step 3: Multiply the empirical formula by the number calculated in step 2

176 amu = 2

88 amu

(C3H4O3) x 2 = C6H8O6

quantitative information from a balanced equation
Quantitative Information from a Balanced Equation

2 H2 (g) + O2 (g)  2 H2O (l)

2 molecules 1 molecule 2 molecules

Or since we can’t count out 2 molecules

2 mol 1 mol 2 moles

The coefficients in a chemical reaction can be interpreted as either the relative number of molecules (formula units) involved in the reaction OR the relative number of moles

example 13 mol mol
Example 13 (mol mol)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many moles of O2 do you need to react with 5 moles of C4H10?

example 13
Example 13

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many moles of O2 do you need to react with 5 moles of C4H10?

5 mol C4H10 x 13 mol O2 = 32.5 mol O2

2 mol C4H10

example 14 g g
Example 14 (gg)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many grams of O2 do you need to react with 50.0 g of C4H10?

example 14 g g1
Example 14 (gg)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

How many grams of O2 do you need to react with 50.0 g of C4H10?

50.0 g C4H10 x 1 mol C4H10 x 13 mol O2 x 32 g O2=179 g O2

58 g C4H10 2 mol C4H10 1 mol O2

limiting reactants reagents
Limiting Reactants (Reagents)
  • The reactant that is completely consumed
  • It determines or limits the amount of product that forms
  • The other reactant(s) are called excess reagents
example 14 limiting reactants
Example 14 (limiting reactants)

2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

If you have 25.0 g O2 and 25.0 g of C4H10, what is the limiting reactant?

example 14 limiting reactants1
Example 14 (limiting reactants)
  • 2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2

58 g C4H10 2 mol C4H10

25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2

32 g O2 13 mol O2

0.481 mol < 1.72 mol C4H10 is the limiting reactant

theoretical yield
Theoretical Yield
  • The quantity of the product that is calculated to form when all of the limiting reactant reacts.
example 15 theoretical yield
Example 15 (Theoretical Yield)
  • 2C4H10(l) + 13 O2(g)  8CO2(g) + 10H2O(l)

25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2

58 g C4H10 2 mol C4H10

25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2

32 g O2 13 mol O2

0.481 mol < 1.72 mol C4H10 is the limiting reactant Calculate the theoretical yeild

example 15 theoretical yield1
Example 15 (Theoretical Yield)

25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2

32 g O2 13 mol O2

0.481 mol CO2 X 44 g CO2= 21.2 g CO2

1 mol CO2

If all of the limiting reactant (25.0 g C4H10) reacts than 21.2 g of CO2 will form.

percent yield
Percent Yield

Percent Yield = Actual Yield X 100

Theoretical yield

example 15 yield
Example 15 (% Yield)
  • A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield?
example 15 yield1
Example 15 (% Yield)

A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield?

87.9 g / 105 g x 100 = 83.7%

(actual) (theoretical)

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