chapter #3 lecture part 1[1]

17
Atoms, Molecules, and Ions Stoichiometry Stoichiometry = “Chemical Arithmetic” Stoichiometry of Elements Stoichiometry of Compounds Stoichiometry of Chemical Reactions

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Page 1: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Stoichiometry

• Stoichiometry = “Chemical Arithmetic”

• Stoichiometry of Elements

• Stoichiometry of Compounds

• Stoichiometry of Chemical Reactions

Page 2: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Macro Worldgrams

Page 3: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

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

Page 4: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

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

Page 5: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Page 6: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Lecture Problem

• Calculate the mass, in grams, of 4.50 moles of calcium atoms.

Page 7: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Lecture Problem

• Calculate the number of moles of calcium atoms in 612 grams of calcium.

• Calculate the number of calcium atoms in 612 grams of calcium.

Page 8: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

Page 9: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Molar Mass Problem

• Calculate the molar mass of sucrose, C12H22O11.

Page 10: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound

x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C =2 x (12.01 g)

46.07 gx 100% = 52.14%

%H =6 x (1.008 g)

46.07 gx 100% = 13.13%

%O =1 x (16.00 g)

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

Page 11: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Empirical Formula

• Identifies the elements present and shows the simplest whole-number ratio of the atoms in a substance.

Page 12: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Data Utilized To Determine An Empirical Formula

• Weight of each element present in a given sample of the compound.

• Weight percentage of each element in a compound.

• Weight of products generated from a reaction involving the compound of interest.

Page 13: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Empirical Formula Problem

• Determine the empirical formula of a compound containing 11.66 grams of iron and 5.01 grams of oxygen.

Page 14: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Empirical Formula Problem

• Determine the empirical formula of a compound containing 62.1% carbon, 5.21% hydrogen, 12.1% nitrogen, and 20.7% oxygen.

Page 15: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Molecular Formula

• Identifies the elements present and shows the exact number of atoms of each element present per molecule.

Page 16: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Data Utilized To Determine A Molecular Formula

• Empirical Formula

&

Molar Mass of Compound

Page 17: Chapter #3   Lecture Part 1[1]

Atoms,Molecules,

and Ions

Molecular Formula Problem

• Cadaverine, a foul smelling substance produced by the action of bacteria on meat, contains 58.55% carbon, 13.81% hydrogen and 27.40% nitrogen. The molar mass of the compound is 102.2 grams/mole. Determine the molecular formula of cadaverine.