Stoichiometry

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# Stoichiometry - PowerPoint PPT Presentation

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 10 23 particles 6.022 x 10 23 is Avogadro’s number , named after

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Presentation Transcript
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 after

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

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
How it works

Two new conversion factors:

Molar mass:

How many atoms are in 3.84 mol of B?
• 3.84
• 2.31 x 1025
• 2.31 x 1024
• 6.38 x 10-24
How many moles do 6.84 x 1021 atoms of B make?
• 6.84 x 1021 mol
• 0.0114 mol
• 4.12 x 1045 mol
• 6.84 x 10-21 mol
How many moles do 45.0 g of Al make?
• 1.67 mol
• 1214 mol
• 7.47 x 10-23 mol
• 2.71 x 1025 mol
What is the mass of 2.60 mol of O?
• 8.90 g
• 0.163 g
• 2.60 g
• 41.6 g
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
Molecules, Atoms and Moles

Consider UF6

• 0.50 mol UF6 contains …
Language for nonmolecular compounds …

1 mol CO2 contains 6.022 x 1023molecules

1 mol NaCl contains 6.022 x 1023formula units

Percent Composition

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

Consider FeO …

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

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.

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.

Example:

A compound has Ca, S, and O.

Ca: 29.45%

S: 23.55%

O: 47.00%

What is the empirical formula?

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.

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?

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

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?