3 - 1 Stoichiometry Law of Conservation of Matter Balancing Chemical Equations Mass Relationships in Chemical Reactions Limiting Reactants Theoretical,

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StoichiometryStoichiometry

Law of Conservation of Matter

Balancing Chemical Equations

Mass Relationships in Chemical Reactions

Limiting Reactants

Theoretical, Actual and Percent Yields

Empirical Formulas

Molecular and Structural Formulas

Law of Conservation of Matter

Balancing Chemical Equations

Mass Relationships in Chemical Reactions

Limiting Reactants

Theoretical, Actual and Percent Yields

Empirical Formulas

Molecular and Structural Formulas

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StoichiometryStoichiometry

StoichiometryStoichiometryThe study of quantitative relationships between substances undergoing chemical changes.

Law of Conservation of MatterLaw of Conservation of MatterIn chemical reactions, the quantity of matter does not change.

The total mass of the products must equal that of the reactants.

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Chemical equationsChemical equations

Chemist’s shorthand to describe a reaction.

It shows:•All reactants and products•The state of all substances•Any conditions used in the reaction

CaCO3 (s) CaO (s) + CO2 (g)

ReactantReactant Products Products

A balanced equation shows the relationshipbetween the quantities of all reactants and products.

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Balancing chemical equationsBalancing chemical equations

Each side of a chemical equation must have the same number of each type of atom.

CaCO3 (s) CaO (s) + CO2 (g)

Reactants Products1 Ca 1 Ca

1 C 1 C

3 O 3 O

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Balancing chemical Balancing chemical equationsequations

Step 1Step 1Count the number of atoms of each element on each side of the equation.

Step 2Step 2 Determine which atom numbers are not balanced.

Step 3Step 3 Balance one atom at a time by using coefficients in front of

one or more substances.

Step 4Step 4Repeat steps 1-3 until everything is balanced.

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Balance the followingBalance the following

HCl + Ca _____> CaCl2 + H2

Step 1 & 2 1 H 2 H - - not balancednot balanced 1 Cl 2 Cl - - not not

balancedbalanced 1 Ca 1 Ca

Step 3 22HCl + Ca _____> CaCl2 + H2

Step 4 2 H 2 H 2 Cl 2 Cl 1 Ca 1 Ca

The balanced equation22HCl + Ca _____> CaCl2 + H2

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C2H6 + O2 _____> CO2 + H2O

Step 1 & 2 2 C 1 C - - not balancednot balanced 6 H 2 H - - not not

balancedbalanced 2 O 3 O - not - not

balancedbalanced

Step 3 Balance biggest molecule first, C2H6

C2H6 + O2 _____> 22CO2 + 33H2O

Step 4 2 C 2 C 6 H 6 H 2 O 7 O - - not not

balancedbalanced

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C2H6 + O2 _____> 22CO2 + 33H2O

Step 3a Now let’s balance O2

C2H6 + 3.53.5O2 _____> 2 CO2 + 3

H2O

Step 4a 2 C 2 C 6 H 6 H 7 O 7 O

You can’t have 3.5 O2 so double the equation!

Balanced EquationBalanced Equation22 C2H6 + 77 O2

_____> 4 4 CO2 + 66 H2O

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Example.Example.Decomposition of ureaDecomposition of urea

(NH2)2CO + H2O ______> NH3 + CO2

2 N 1 N < < not balancednot balanced6 H 3 H < <

not balancednot balanced1 C 1 C2 O 2 O

We need to double NH3 on the right.(NH2)2CO + H2O ______> 22NH3 + CO2

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Another exampleAnother example

CH3OH + PCl5 _____> CH3Cl + POCl3 +

H2O 1 C 1 C

4 H 5 H1 O 2 O1 P 1 P5 Cl 4 Cl

We need another Cl on the right.Increase CH3Cl then recheck.

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Another example.Another example.

CH3OH + PCl5 _____> 22CH3Cl + POCl3 +

H2O 1 C 2 C

4 H 8 H1 O 2 O1 P 1 P5 Cl 5 Cl

Another C is needed on the left so doubleCH3OH.

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Another example.Another example.

22CH3OH + PCl5 _____> 22CH3Cl + POCl3 + H2O

2 C 2 C8 H 8 H2 O 2 O1 P 1 P5 Cl 5 Cl

Now its balanced!

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2 H2 + O2 -----> 2 H2O 2 H2 + O2 -----> 2 H2O

You need a balancedequation and you WILL

work with moles.

You need a balancedequation and you WILL

work with moles.

Mass relationshipsMass relationshipsin chemical reactionsin chemical reactions

StoichiometryStoichiometry - The calculation of quantities of reactants and products in a chemical reaction.

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Stoichiometry,Stoichiometry,General steps.General steps.

11 Balance the chemical equation.

33 Convert masses to moles.

22 Calculate formula masses.

44Use chemical equation toget the needed answer.

Convert back to mass if needed.55

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Mole calculationsMole calculations

The balanced equation shows the reacting ratio between reactants and products.

2C2C22HH66 + 7O + 7O22 4CO 4CO22 + 6H + 6H22OO

For each chemical, you can determine the•moles of each reactant consumed •moles of each product made

If you know the formula mass, mass quantities can be used.

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Mole-gram conversionMole-gram conversion

How many moles are in 14 grams of N2 ?

Formula mass = 2 N x 14.01 g/mol= 28.02 g /mol

moles N2

= 14 g x 1 mol /28.02 g = 0.50 moles

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Mass calculationsMass calculations

We don’t directly weigh out molar quantities.

We can use measured masses like kilograms, grams or milligrams.

The formula masses and the chemical equations allow us to use either mass or molar quantities.

We don’t directly weigh out molar quantities.

We can use measured masses like kilograms, grams or milligrams.

The formula masses and the chemical equations allow us to use either mass or molar quantities.

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How many grams of hydrogen will be produced if 10.0 grams of calcium is added to an excess of hydrochloric acid?

2HCl + Ca ______> CaCl2 + H2

Note:•We produce one H2 for each calcium.•There is an excess of HCl so we have

all we need.

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2HCl + Ca ____> CaCl2 + H2

First - Determine the number of moles of calcium available for the reaction.

Moles Ca = grams Ca / FM Ca

= 10.0 g

= 0.25 mol Ca

1 mol40.08 g

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2HCl + Ca _____> CaCl2 + H2

10 g Ca = 0.25 mol Ca

According to the chemical equation, we get one mole of H2 for each mole of Ca.

So we will make 0.25 moles of H2.

grams H2 produced = moles x FW H2

= 0.25 mol x 2.016 g/mol

= 0.504 grams

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OK, so how many grams of CaCl2 were made?

2HCl + Ca _____> CaCl2 + H2

10 g Ca = 0.25 mol Ca

We would also make 0.25 moles of CaCl2.

g CaCl2 = 0.25 mol x FM CaCl2

= 0.25 mol x 110.98 g / mol CaCl2

= 27.75 g CaCl2

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Limiting reactantLimiting reactant

In the last example, we had HCl in excess.

Reaction stopped when we ran out of Ca.

Ca is considered the limiting reactant.limiting reactant.

Limiting reactantLimiting reactant - the material that is in the shortest supply based on a balanced chemical equation.

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Limiting reactant exampleLimiting reactant example

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ExampleExample

For the following reaction, which is limiting if you have 5.0 g of hydrogen and 10 g oxygen?

Balanced Chemical ReactionBalanced Chemical Reaction

2H2 + O2 ________> 2H2O

You need 2 moles of H2 for each mole of O2.

Moles of H2 5 g = 2.5 mol

Moles of O2 10g = 0.31 mol1 mol32.0 g

1 mol2.0 g

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ExampleExample

Balanced Chemical ReactionBalanced Chemical Reaction

2H2 + O2 2H2O

You need 2 moles of H2 for each mol of O2

You have 2.5 moles of H2 and 0.31 mol of O2

Need a ratio of 2:1but we have a ratio of 2.5 : 0.31 or 8.3 : 1.

Hydrogen is in excess and oxygen is thelimiting reactant.

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Theoretical, actual Theoretical, actual and percent yieldsand percent yields

Theoretical yieldTheoretical yieldThe amount of product that should be formed according to the chemical reaction.

Actual yieldActual yieldThe amount of product actually formed.

Percent yieldPercent yieldRatio of actual to theoretical yield, as a %.

Quantitative reactionQuantitative reactionWhen the percent yield equals 100%.

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YieldYield

Less product is often produced than expected.

Possible reasonsPossible reasons

• A reactant may be impure.

• Some product is lost mechanically since the product must be handled to be measured.

• The reactants may undergo unexpected reactions - side reactions.

• No reaction truly has a 100% yield due to the limitations of equilibrium.

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Percent yieldPercent yield

The amount of product actually formed divided by the amount of product calculated to be formed, times 100.

% yield = x 100

In order to determine % yield, you must be able to recover and measure all of the product in a pure form.

Actual yieldTheoretical yield

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% Yield example% Yield example

Example.Example. The final step in the production of aspirin is the reaction of salicylic acid with acetic anhydride.

48.6 g of aspirin is produced when 50.0 g of salicylic acid and an excess of acetic anhydride are reacted. What is the % yield?

HOC6H4COOH(s) + (CH3CO)2O(l)

salicylic acid acetic anhydride

CH3OC6H4COOH(s) + CH3COOH(l)

aspirin acetic acid

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Number of moles of salicylic acid used:

One mole of aspirin should be produced for each mole of salicylic acid consumed. Number of grams of aspirin that should have been produced -- theoretical yield:

(0.362 mol aspirin)( 180 g/mol) = 65.2 g aspirin

= 0.362 mole of salicylic acid 1 mol138 g

50.0 g

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% Yield for this reaction

Theoretical yield = 65.2 gActual yield = 48.6 g

% Yield = x 100

= 74.5%

Yields less than 100% are very common in industrial processes.

48.6 g65.2 g

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Empirical formulaEmpirical formula

This type of formula shows the ratios of the number of atoms of each kind in a compound.

For organic compounds, the empirical formula can be determined by combustion analysis.

Elemental analyzerElemental analyzerAn instrument in which an organic compound is quantitatively converted to carbon dioxide and water -- both of which are then measured.

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Elemental analyzerElemental analyzer

furnace

CO2

trapH2Otrap

O2

sample

A sample is ‘burned,’ completely converting it to CO2and H2O. Each is collected and measured as a weightgain. By adding other traps elements like oxygen, nitrogen, sulfur and halogens can also be determined.

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Elemental analysisElemental analysis

ExampleExampleA compound known to contain only carbon, hydrogen and nitrogen is assayed by elemental analysis. The following information is obtained.

Original sample mass = 0.1156 gMass of CO2 collected = 0.1638 gMass of H2O collected = 0.1676 g

Determine the % of each element in the compound.

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Mass of carbonMass of carbon

Mass of hydrogenMass of hydrogen

Mass of nitrogenMass of nitrogen

0.1638 g CO2

12.01 g C44.01 g CO2

= 0.04470 g C

0.1675 g H2O2.016 g H

18.01 g H2O= 0.01875 g H

0.1156 g - 0.04470 g C - 0.01875 g H = 0.05215 g N

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Since we know the total mass of the original sample, we can calculate the % of each element.

% C = x 100% = 38.67 %

% H = x 100% = 16.22 %

% N = x 100% = 45.11 %

0.04470 g0.1156 g

0.01875 g0.1156 g

0.05215 g0.1156 g

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Empirical formulaEmpirical formulaThe simplest formula that shows the ratios of the number of atoms of each element in a compound.

ExampleExample - the empirical formula for hydrogen peroxide (H2O2) is HO.

We can use our percent composition information from the earlier example to determine an empirical formula.

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From our earlier example, we found that our compound had a composition of:

If we assume that we have a 100.0 gram sample, then we can divide each percentage by the elements atomic mass and determine the number of moles of each.

% C = 38.67 %% H = 16.22 %% N = 45.11 %

% C = 38.67 %% H = 16.22 %% N = 45.11 %

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16.22 g H1 mol H

1.008 g H= 12.09 mol H

38.67 g C1 mol C

12.01 g C= 3.220 mol C

45.11 g N1 mol N

14.01 g N= 3.220 mol N

( )

( )

( )

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The empirical formula is then found by looking for the smallest whole number ratio.

C 3.220 / 3.220 = 1.000

H 16.09 / 3.220 = 4.997

N 3.220 / 3.220 = 1.000

The empirical formula is CHThe empirical formula is CH55NN

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Molecular formulaMolecular formula

Molecular formulaMolecular formula - shows the actual number of each type of atom in a molecule.

• They are multiples of the empirical formula.

• If you know the molecular mass, then the molecular formula can be found.

For our earlier example, what would be the molecular formula if you knew that the molecular mass was 62.12?

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Molecular formulaMolecular formula

Empirical formula CH5N Empirical formula mass 31.06 u

Molecular mass 62.12

Ratio: 62.12 / 31.06 = 2

The molecular formula is C2H10N2

Note: This does not tell you have the atoms are arranged in the compound!

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Structural formulaStructural formula

These are used to show how atoms are attached in a molecule.

ExampleExampleBoth of the following structural formula would have a molecular formula of C2H6O

H

CH

H

C

H

H

O H H C

H

H

O C

H

H

H

ethyl alcohol dimethyl ether

These chemicals have very different properties.

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We use a variety of ways to represent structural formula.

Condensed structural formulas are commonly used for organic molecules.

They list a carbon and then what is attached to it. The next carbon in the chain is then listed.

ethyl alcohol - CH3CH2OHdimethyl ether - CH3OCH3

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Models may also be used to help view a molecule.

ethyl alcohol dimethyl ether

Ball-and-stick

Space-filling

3 - 1 Stoichiometry Law of Conservation of Matter Balancing Chemical Equations Mass Relationships in Chemical Reactions Limiting Reactants Theoretical,

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