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ObjectivesAtomic MassThe Mole Concept

Percentage CompositionEmpirical & Molecular Formulas

Molar Mass

Chemical ReactionsBalancing Chemical EquationsMass - Mass CalculationsLimiting ReactantPercent Yield

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Atomic mass is the mass of an atom in atomic mass units.

The atomic mass unit is exactly 1/12 of the mass of the carbon-12 isotope, which has been assigned the mass exactly 12.00000 amu.

The average atomic mass of all naturally occurring isotopes of a given element is called the atomic weight of the element. This number is unique for each element.

Key TermsKey Terms

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Use the Periodic Table to identify the atomic weight of each element and then round off to the nearest tenth.

Example #1

silicon 28.0855 = 28.1

sodium 22.98977 = 23.0

chromium 51.996 = 52.0

magnesium 24.305 = 24.3

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A mole of an element is that quantity of the element, which consists of its atomic weight expressed in grams.

1 mol Ag = 107.868

1 mol sulfur = 32.066

grams

grams

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A mole of an element contains the Avogadro Number of atoms. This number is equal to 6.022 x 1023 atoms.

1 mol Ag = atoms6.022 x 1023

1 mole = gram atomic weight

1 mole = 6.022 x 1023 atoms

1 mole = gram atomic weight

1 mole = 6.022 x 1023 atoms

1 mol Sn = g118.7 = atoms6.022 x 1023

1 mol K = g = atoms39.1 6.022 x 1023

g107.9 =

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Example #2

A sample of sulfur weighs 1.28 grams. How many moles of sulfur is this?

moles S=1.28 g S

Answer = 0.0399 mol S

1.28 g S xg

mol1

32.1

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Example #3

83.5 grams C =

4.19 x 1024 C atoms

83.5 g C x1 mol

12.0 gx

6.022 X 1023 atoms

1 mol

atoms

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1 mol = g-FW = molar mass

1 mol = 6.022 X 1023 molecules

1 mol = g-FW = molar mass

1 mol = 6.022 X 1023 molecules

The formula weight of a compound is the sum of the atomic masses of all atoms represented in the formula of the compound.

The formula weight of a compound expressed in grams is called the molar mass of the compound.

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Example #4

Calculate the molar mass of caffeine, C8H10N4O2.

MM C8H10N4O2 = 8(12.0) +10(1.0) +

4(14.0) +2(16.0) =

Answer: 194.0 g/mole

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Example #5How many moles of Na2CO3 are

represented by 1.00 x 103 grams of the compound?

mol Na2CO3=1.00 x 103 g Na2CO3

1.00 x 103 g xg

mol

MM Na2CO3 = 2(23.0) + 12.0 + 3(16.0)= 106.0 g/mol

1

106.0= 9.43 mol Na2CO3

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Percentage composition refers to the amount of each element in a compound expressed as a percent.

==

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Example #6Calculate the percentage composition of ammonia, NH3.

MM NH3 = 14.0 + 3(1.0) = 17.0

%N = 14.0

17.0X 100 = 82.4%

%H =

3.0

17.0X 100 = 17.6%

% H =

or

100% - 82.4% = 17.6%

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Example #7

What is the percentage of calcium in calcium acetate?

MM Ca(C2H3O2)2 = 40.1 + 4(12.0) + 6(1.0) + 4(16.0) =

%Ca = 40.1

158.1 g/mol

158.1 X 100 = 25.4%

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Example #8

How many grams of silver can be obtained from a 17.85 g sample of silver nitrate, AgNO3?

g Ag=17.85 g AgNO3

xg AgNO3

g Ag

MM AgNO3 = 107.9 + 14.0 + 3(16.0)= 169.9 g/mol

107.9

169.9= 11.34 g Ag17.85 g AgNO3

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A hydrate is a compound that contains water as part of its chemical composition.

CuSO4 5H2O Cupric sulfate, pentahydrate

Na2CO3 10H2O Sodium carbonate, decahydrate

The waterwater in the hydrate increases its molar mass.

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Example #9

Calculate the percentage of water in nickel(II) chloride, hexahydrate.

NiCl2 6H2OMM = 58.7 + 2(35.5) + 6(18.0)

= 237.7 g/mol

%H2O =

108.0

237.7X 100 =

45.44% H2O108.0

H2O

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The empirical formula of a compound indicates the simplest whole number mole ratio of atoms in a molecule.

The molecular formula of a compound shows the actual number of atoms in a molecule.

For benzene the molecular formula is C6H6.Its empirical formula would be CH, showing the one to one ratio between the atoms of carbon and hydrogen.

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Relationship Between Molecular Relationship Between Molecular Formulas and Empirical FormulasFormulas and Empirical Formulas

The molecular formula subscripts are always a whole number multiple of the empirical formula subscripts.

Butene = C4H8

Molecular formula

Empirical formula

Butene = CH2

So that C4H8 =

where n = MMEFW

(CH2)4

In general then MF = (EF)nStructural formula

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1. Calculate the number of moles of each element. (Conversion Problems)

2. Calculate the whole number mole ratio for each element by dividing by the smallest number of moles in part 1.

3. Use the whole number mole ratios as the subscripts in the empirical formula.

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Example #10Calculate the empirical formula

for the compound that consists of 77.73% iron and 22.27% oxygen.

77.73 g Fe xg

mol1

55.85= 1.392 mol

22.27 g O xg

mol1

16.00= 1.392 mol

1.392 mol

1.392 mol

=1

= 1

EF = FeO

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Example #11

Calculate the empirical formula of the compound which consists of 10.91 g phosphorus and 14.09 g oxygen.

10.91 g P xg

mol130.97

= 0.3523 mol

14.09 g O xg

mol16.00

= 0.8806 mol

0.3525 mol

0.3525 mol

=1

= 2.50

x 2 = 2

x 2 =

5

EF = P2O5

1

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Example #12

If the molar mass of the compound in example # 11 is 283.88 g/mol, what is it’s molecular formula?

From example #11: EF = P2O5

EFW P2O5 = 2(30.97) + 5(16.00) = 141.94

Since MF = (EF)n and n=

MMEFW

= 283.88141.94

= 2

The MF = (P2O5) 2 = P4O10= P4O10

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A beginning material in a chemical reaction is called a reactant.

Any material formed as a result of the chemical change is called a product of the reaction.

reactants productsyield

formproduce

change to

Chemical ReactionsChemical Reactions

A chemical reaction is a change in matter in which new substances with new properties are formed.

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A word equation represents the reactants and the products in a chemical reaction with their names.

A chemical equation represents the reactants and products in a chemical reaction with their symbols and formulas.

aluminum(s) +

oxygen(g)

aluminum oxide(s)

4Al(s) +

3O2(g) 2Al2O3(s)

(s) = solid(l) = liquid

(g) = gas(aq) = aqueous

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Determine the reactants, the products, and the physical states involved.

Use the correct symbols or formulas of the reactants and products to write the unbalanced chemical equation.

Balance the equation by applying the law of conservation of matter to determine the

correct coefficients.

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The law of conservation of matter states that matter cannot be created nor destroyed in any chemical reaction.

Implication:Implication:

The number of atoms of reactants must be equal to the number of atoms of products.

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Balancing an equation is the process of using multiplication to make the number of atoms of products equal to the number of atoms of reactants.

A coefficient is a number placed in front of some chemical species to represent that number of chemical units.

iron(s) + oxygen(g) iron(III) oxide(s)Fe(s) +O2(g) Fe2O3(s)

Fe(s) + O2(g) Fe2O3(s)34 2

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The following elements exist in the gaseous state as diatomic (two atom) molecules and not as single atom substances.

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Examples:Balance each equation.

#13.#13. Hydrogen gas burns in oxygen gas to produce water vapor.

H2(g) + O2(g) H2O(g)22

#14.#14. Pure iridium metal reacts with oxygen gas to produce solid iridium(III) oxide.

Ir(s) + O2(g) Ir2O3(s)24 3

#15.#15. Butane gas, (C4H10), burns in air to produce carbon dioxide and water vapor.

C4H10(g) + O2(g) CO2(g) + H2O(g)5132

2( )4

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

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The stoichiometrystoichiometry of a chemical reaction deals with calculationscalculations involving the amountsamounts of reactantsreactants and productsproducts in the reaction.Such calculations are often referred to as mass-massmass-mass problems if only masses are involved.

These problems can also be referred to as mass-volumemass-volume, volume-massvolume-mass, or volume-volume-volumevolume problems depending on the data provided by the problem.

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The coefficientscoefficients of a balanced chemical equation indicate the mole ratiomole ratio between any reactant and/or product.

44Fe(s) + 33O2(g) 22Fe2O3(s)

ImplicatioImplicationn

4 moles 4 moles ironiron

react with 3 moles3 moles of oxygenoxygento produce exactly 2 moles 2 moles of ferric ferric

oxideoxide.

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44Fe(s) + 33O2(g)

Refer to the equation described below to answer the following oral questions.

22Fe2O3(s)

4 mol 3 mol 2 mol

8 mol 6 mol 4 mol

1 mol2 mol 1.5 mol

9 mol12 mol 6 mol

4 mol 6 mol 2 mol Too much oxygen3 mol in excess

6 mol 7 mol 3 mol Too much oxygen

Only 4.5 mol oxygen used

2.5 mol in excess

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2.2. Convert Convert the given datagiven data to molesmoles using the molar massmolar mass of the given substance.

3.3. Use the coefficientscoefficients of the balanced equation to convert moles givenconvert moles given to

molesmoles asked forasked for.

4.4. Use the molar massmolar mass of the substancesubstance asked asked forfor to convert to gramsgrams.

grams givengrams given

molmol mol askedmol asked gramsgramsmolar mass coefficients molar mass

1.1. Write a balanced chemical balanced chemical equationequation.

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Example #16

Cobalt metal reacts with hydrochloric acid according to the following unbalanced equation. If you begin with 2.56 g of cobalt metal and excess hydrochloric acid, what mass of cobalt(II) chloride can be obtained?

Co(s) + HCl(aq) CoCl2(aq) + H2(g)22

g CoCl22.56 g Co =

2.56 g Co xg

mol

MM CoCl2 = 58.9 + 2(35.5) = 129.9 g/mol

58.9

1x

mol Co

mol CoCl21

1x

mol

g129.9

1

5.65 g CoCl2

=

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Example #17aC3H8(g) + O2(g) CO2(g)

+H2O(g)55

g O2454 g C3H8 =

454 g C3H8 x g

mol

MM C3H8 = 12.0) + 8(1.0) = 44.0 g/mol

44.0

1x

mol C3H8

mol O25

1x

mol

g32.0

1

1,650 g O2

=

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Propane, C3H8, can be used as a fuel in your home, car, or barbeque grill because it is easily liquefied and transported. If 454 g of propane is burned, [a] what mass of oxygen is required and how much [b] carbon dioxide and [c] water are formed?

3(

MM O2 = 2(16.0) = 32.0 g/mol

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1,360 g CO2

Student Presentation

Example #17b

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Example #17c

743 g H2O

Student Presentation

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The reactant determiningreactant determining the amountamount of productproduct formed formed in a chemical reaction is called the limitinglimiting reactantreactant.

The other reactant(s)other reactant(s) is/are said to be in excessexcess.

The amountamount of productproduct that should be formedshould be formed in a chemical reaction is called theoretical yieldtheoretical yield.

Theoretical yieldTheoretical yield also corresponds to the maximummaximum amount of productamount of product possible.

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Example #18A tool set consists of 4 wrenches, 3 screwdrivers, and

2 pliers. The manufacturer has in stock 1,000 pliers, 2,000 screwdrivers, and 1,500 wrenches.

A.) Can an order for 500 tool sets be filled out?

B.) How many tool sets can be made?

C.) How many of which tools will remain after the maximum number of tool sets are made?

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1,000 pliers / 2 pliers per set = 500 sets

2000 screwdrivers / 3 screwdrivers per set = 666 sets

1500 wrenches / 4 wrenches per set = 375 sets375 sets

Once 375 sets375 sets have been put together, allall the wrenches wrenches will be used upused up so that 0 wrenches0 wrenches will remainremain. More sets cannot be made even though some pliers and screwdrivers remain.

Making the 375 tool sets375 tool sets with 3 screwdrivers per set will requirewill require only 1,125 screwdrivers1,125 screwdrivers and this means that (2,000 - 1,125) = 875 screwdrivers 875 screwdrivers will remainremain.

Making the 375 tool sets375 tool sets with 2 pliers per set, usesuses up 750 pliers750 pliers. This means that (1,000 - 750) = 250 250 plierspliers will remainremain.

Limiting reactant

Reactants in excess

Theoretical yield

Reactants in excess that remain

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Example #19Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperature. The other products are solid copper and water vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

NH3(g) + CuO(s) N2(g) + Cu(s) + H2O(g)22 3333 33

L. R.

18.1 g NH3 xg

mol

MM NH3 = 14.0 + 3( 1.0) = 17.0 g/mol

17.0

1x

mol NH3

mol N21

2=

MM N2 = 2(14.0) = 28.0 g/mol

0.5324 mol N2

g N2 produced

90.4 g CuO xg

mol

79.5

1x

mol CuO

mol N21

3x

mol

g28.0

1=

MM CuO = 63.5 + 16.0 = 79.5 g/mol

0.3790 mol N2

Choose smaller value to determine the L. R.

10.6 g N2

L.R. = CuO =

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The maximum amount of product possible, theoreticaltheoretical yieldyield, is seldom obtainedseldom obtained in a chemical reaction due to side reactions and other complications.

The actual yieldactual yield of product is the amount of product that is obtained in a chemical reaction and it is always less than the theoretical yield.

The actual yield of product is often expressed as the percent yieldpercent yield.

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Example #20Methanol, CH3OH, is used as a fuel in race cars and is a potential replacement for gasoline. It can be prepared by combining gaseous carbon monoxide with hydrogen gas. A student reacts 68.5 g CO(g) with 8.60 g H2(g). Calculate the T.Y. of methanol and if 35.7 g CH3OH are actually produced, what is the %yield of methanol?

H2(g) + CO(g) CH3OH(l)22

g CH3OH8.60 g H2=

8.60 g H2 xg

mol

MM H2 = 2(1.0) = 2.0 g/mol

2.0

1x

mol H2

mol CH3OH1

2x

mol

g32.0

1=

MM CH3OH = 12.0 + 4(1.0) + 16.0 =

68.8 g CH3OH

68.5 g CO = g CH3OH

68.5 g CO xg

mol

28.0

1x

mol CO

mol CH3OH1

1x

mol

g32.0

1=

MM CO = 12.0 + 16.0 = 28.0 g/mol

78.3 g CH3OH

Choose smaller number as the

T. Y.

T.Y. = 68.8 g CH3OH

32.0 g/mol

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Example #21

%Yield =35.7 g

68.8 gX 100

%Yield = 51.9%