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Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry
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Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Dec 22, 2015

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Page 1: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Stoichiometry

Study of mass and chemical amounts.

1. Compound stoichiometry

2. Reaction stoichiometry

Page 2: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Connecting the Macro and Atomic Scales: The Mole

Chemical macroscopic counting unit:

1 mole contains 6.022 x 1023 particles

6.022 x 1023 is Avogadro’s number, named afterAmedeo Avogadro.

Avogadro’s number is not special, like pi, but is invented by us for our convenience.

Page 3: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Why it works.

• The mass of 1 mol of an element is equal to its atomic mass in grams

• Molar mass is in g/mol• Molar mass of Al = 26.98 g/mol

• Unit sign: mole = mol sometimes, molar mass = M

Page 4: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How it worksTwo new conversion factors:

Avogadro’s number:

Molar mass:

23

23

6.022 x 10 particles 1 mol

1 mol 6.022 x 10 particles

26.98g Al 1 mol Al

1 mol Al 26.98 g Al

Page 5: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many atoms are in 2.60 mol of Ne?

Page 6: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many atoms are in 3.84 mol of B?

1. 3.84

2. 2.31 x 1025

3. 2.31 x 1024

4. 6.38 x 10-24

Page 7: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many moles do 4.00 x 1022 atoms of Ne make?

Page 8: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many moles do 6.84 x 1021 atoms of B make?

1. 6.84 x 1021 mol

2. 0.0114 mol

3. 4.12 x 1045 mol

4. 6.84 x 10-21 mol

Page 9: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many moles do 45.0 g of Ne make?

Page 10: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many moles do 45.0 g of Al make?

1. 1.67 mol

2. 1214 mol

3. 7.47 x 10-23 mol

4. 2.71 x 1025 mol

Page 11: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

What is the mass of 2.60 mol of C?

Page 12: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

What is the mass of 2.60 mol of O?

1. 8.90 g

2. 0.163 g

3. 2.60 g

4. 41.6 g

Page 13: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many moles are represented by 1.00 g of C?

Page 14: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many atoms are in 1.00 g of O?

Page 15: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

What is the mass of one Au atom?

Page 16: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

How many C atoms are in a 1.00 caret diamond?

Page 17: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Compounds and Moles

• 1 mole of a compound contains 6.022 x 1023 molecules or “units”

• Molar mass = sum of atomic molar masses– H2O

– CO2

– Fe(NO3)3

Page 18: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.
Page 19: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Molecules, Atoms and Moles

Consider UF6

– 0.50 mol UF6 contains …

Page 20: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Language for nonmolecular compounds …

1 mol CO2 contains 6.022 x 1023 molecules

1 mol NaCl contains 6.022 x 1023 formula units

Page 21: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.
Page 22: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Percent Composition

What fraction of a compound is made up of one of its elements?

Consider FeO2 … (** added the 2)

mass of one elementPercent composition = × 100%

total mass

Page 23: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.
Page 24: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Determining Formulas from Percent Composition

The key: the ratio of atoms in the formula is the same as the ratio of moles of those elements

Page 25: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Determining Formulas from Percent Composition

The key: the ratio of atoms in the formula is the same as the ratio of moles of those elements

If you determine the ratio of moles, you know the formula.

Page 26: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Determining Formulas from Percent Composition

The key: the ratio of atoms in the formula is the same as the ratio of moles of those elements

If you determine the ratio of moles, you know the formula.

But! You get the empirical formula.

Page 27: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Example:A compound has Ca, S, and O.

Ca: 29.45%

S: 23.55%

O: 47.00%

What is the empirical formula?

Page 28: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Example:

20.48% Zn79.51% I

Page 29: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Example:A compound has Ca, S, and O.

Ca: 29.45%

S: 23.55%

O: 47.00%

What is the empirical formula?

Page 30: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Empirical vs. Molecular Formulas

Example: Ethene is C2H4

Percent composition tells us mol H/mol C = 2

The empirical formula is CH2.

The molecular formula is C2H4.

Page 31: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Example:

A hydrocarbon has 82.65% C and 17.34% H

----Molar mass is 58.12 g/mol

----What are the empirical and molecular formulas?

Page 32: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.
Page 33: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Hydrated Compounds

Solids in which molecules of water are trapped and become part of the compound.

Ex: Gypsum: CaSO4 • 2H2O

is hydrated calcium sulfate

CaSO4 is anhydrous calcium sulfate

Page 34: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.

Determining the number of waters of hydration

1.023 g CuSO4 • x H2O is heated to drive off the water. The resulting anhydrous CuSO4 has a mass of 0.654 g.

What is the value of x?

Page 35: Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry.