Chemical Reactions, Chemical Reactions, Chemical Equations, and Chemical Equations, and StoichiometryStoichiometryBrown, LeMay Ch 3AP Chemistry

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3.2: Types of reactions3.2: Types of reactionshttp://web.fuhsd.org/kavita_gupta/

July.html

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3.3: Atomic, Molecular & Formula 3.3: Atomic, Molecular & Formula WeightsWeights

Atomic mass units:1 amu = 1.66054 x 10-24 g

or 1 g = 6.02214 x 1023 amu1 atom of 12C isotope defined as

weighing exactly 12 amuAverage atomic mass (or atomic

weight):12C: 98.892% x 12 amu =

13C: 1.108% x 13.00335 amu =

3

11.867+ 0.1441

AW = 12.011 amu…....Therefore 12.011 g C = 1

mol C

Molecular mass or weight (MM or MW): sum of atomic masses of each atom in the chemical formula. Ex: H2SO4

MW = (2)(1.0079) + 32.06 + (4)(16.00)

= 98.08 amu

Formula weight (FW): same as MW, except for ionic substances in which no “molecule” exists. Then, FW is the simplest integer ratio of moles of each element (ions) present. Ex: NaCl is a 3D array of ions

FW = 22.99 + 35.453 = 58.44 amul

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3.4 – 3.7: Mass Relationships3.4 – 3.7: Mass RelationshipsPercent Composition:

Ex: What is the percentage of oxygen in phosphoric acid?

5

100)(4

%43POHMW

OAWO

%31.65100)00.16(4974.30)0079.1(3

)00.16(4

Ex: What is the percent of O in CuSO4.5H2O?57.7%

The Mole and Molar Mass (MM)◦From Latin moles: “a mass”◦Avogadro’s Number = 6.022 x 1023 atoms in exactly 12.000 g of 12C = 1 mol

◦Moles important because they can be used for ratios. Mass can not be used for ratios. Ex. CO2 etc.

◦Converting: grams → moles → moleculesEx: How many molecules of H2O in 100.0 g H2O?

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number sAvogadro' massmolar

OH mol 1

OH molecules 10 6.022

OH g 18.0

OH mol 1

1

OH g 100.0

2

223

2

22

= 3.343 x 1024 molecules H2O

Ex: What is the empirical formula of a compound that is composed of 80.% Carbon and 20.% Hydrogen? If the molar mass is found to be 30 g/mol, what is the molecular formula?

Strategy: Assume 100. g of unknown

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C:H mole ratio is 6.7:20. ≈ 1:3. So, empirical formula is CH3,and the molecular formula is C2H6.Percent Composition Tutorial

C mol 7.6C g 12.011

C mol 1

1

C g 80.

H mol .20H g 1.0079

H mol 1

1

H g 20.

Practice Problem 1:Practice Problem 1:

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One of the most commonly used white pigments in the paint is a compound of Titanium and Oxygen that contains 59.9% Ti by mass. Determine the empirical formula of this compound.

Ans; TiO2

Practice problem2:Practice problem2:

Benzene contains only C and H and is Benzene contains only C and H and is 7.74% H by mass; the molar mass of 7.74% H by mass; the molar mass of benzene is 184 g/mol. What are the benzene is 184 g/mol. What are the empirical and molecular formula of this empirical and molecular formula of this compound?compound?

Ans: CH, CAns: CH, C1414HH1414

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EF from Combustion Data: EF from Combustion Data: To To find out the EF of a compound, the compound find out the EF of a compound, the compound is burned in the air. The equation for is burned in the air. The equation for combustion of these compounds is all follows.combustion of these compounds is all follows.For Hydrocarbons: CxHy + OFor Hydrocarbons: CxHy + O22 CO CO22 + H + H22OO

For Compounds Containing CHN: For Compounds Containing CHN: CxHyNz+ OCxHyNz+ O22 CO CO2 2 + H+ H22O+ NOO+ NO22

For Compounds containing C, H and OFor Compounds containing C, H and OCxHyOz + OCxHyOz + O22 CO CO22 + H + H22O O In such cases to figure out mass of O,In such cases to figure out mass of O,Mass of O = total mass of compound – (mass of Mass of O = total mass of compound – (mass of C + mass of H)C + mass of H)

Animation: Must See!! 10

Important Fact to remember Important Fact to remember for Combustion:for Combustion:

Complete combustion produces CO2 as opposed to incomplete combustion which produces CO.

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Practice Problem on EF from Practice Problem on EF from Combustion Data:Combustion Data:

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Many homes in rural America are heated by propane gas, a compound that contains only carbon and hydrogen. Complete combustion of a sample of propane produced 2.641 g of carbon dioxide and 1.442 g of water as the only products. Find the empirical formula of propane.

Answer: C3H8

Practice Problem Level 2: EF Practice Problem Level 2: EF from Combustion Datafrom Combustion Data

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Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water. The molar mass of cumene produces between 115 and 125 g/mol. Determine the empirical and molecular formulas.

Answer: C3H4, C9H12

Practice Problem 3:Practice Problem 3:

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A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of this compound?

Ans: CH2O

Limiting Reagent (or reactant):Ex: If 6.0 g hydrogen gas reacts with 40.0 g oxygen gas, what mass of water will be produced?

2 H2 (g) + O2 (g) → 2 H2O (g)

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22

22 H mol 3.0 976.2H g 2.0158

H mol 1

1

H g 6.0

22

22 O mol 25.1O g 32.00

O mol 1

1

O g 40.0

OH g 0.45OH mol 1

OH g 5818.01

O mol 1

OH mol 2

1

O mol 1.252

2

2

2

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O2 is limiting reagent, H2 is excess.

Limiting Reactant Movie

Percent yield & Percent error:Ex: If in the previous example, only 40.0 g water were formed, what is the percent yield and percent error?

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100yield lTheoretica

yield Actual Yield %

% 9.88100OH g 45.0

OH g 40.0

2

2

100 valueAccepted

result Measured - valueAccepted Error %

.%11100OH g 45.0

OH g 5.0100

OH g 45.0

OH g 40.0-OH g 45.0

2

2

2

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Determining Formula of a Hydrate: To determine the formula of a hydrate, a certain mass of hydrate is heated to drive off the water. Then the mass of water driven-off is calculated. Now the mole ratio of water to the compound is calculated. For more information visit the following website:

Movie on Determining formula of a Hydrate

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Questions to ponder over:1.What might be a good procedure to calculate percent of water in a hydrate?2.How do you identify hydrates? Do you remember how to name hydrates?

Practice Problem:Practice Problem:

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A student is given three unknowns labeled A, B, and C. The student is told they are Na2CO3.H2O; Na2CO3.7H2O; and Na2CO3.10H2O, not necessarily in that order.

a. When 2.000 gram of A was heated, 0.914 gram of anhydrous residue remained.Formula for sample A.Ans: Na2CO3.7H2O   Sodium carbonate heptahydrate

Practice Problem 2Practice Problem 2

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b. When 4.000 grams of B was heated, 1.48 gram of anhydrous residue remained.Formula for sample B.Chemical name for sample B.Ans: Na2CO3.10H2O   Sodium carbonate decahydrate

c. When 1.000 gram of C was heated, 0.855 gram of anhydrous residue remained.Formula for sample C.Chemical name for sample C.Ans: Na2CO3.H2O   Sodium carbonate monohydrate