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

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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 x particles x is Avogadro’s number, named after Amedeo Avogadro. 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 = g/mol Unit sign: mole = mol sometimes, molar mass = M

How it works Two new conversion factors: Avogadro’s number: Molar mass:

How many atoms are in 2.60 mol of Ne?

How many atoms are in 3.84 mol of B? x x x

How many moles do 4.00 x atoms of Ne make?

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

How many moles do 45.0 g of Ne make?

How many moles do 45.0 g of Al make? mol mol x mol x mol

What is the mass of 2.60 mol of C?

What is the mass of 2.60 mol of O? g g g g

How many moles are represented by 1.00 g of C?

How many atoms are in 1.00 g of O?

What is the mass of one Au atom?

How many C atoms are in a 1.00 caret diamond?

Compounds and Moles 1 mole of a compound contains x molecules or “units” Molar mass = sum of atomic molar masses –H 2 O –CO 2 –Fe(NO 3 ) 3

Molecules, Atoms and Moles Consider UF 6 –0.50 mol UF 6 contains …

Language for nonmolecular compounds … 1 mol CO 2 contains x molecules 1 mol NaCl contains x formula units

Percent Composition What fraction of a compound is made up of one of its elements? Consider FeO 2 … (** added the 2)

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?

Example: 20.48% Zn 79.51% I

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 C 2 H 4 Percent composition tells us mol H/mol C = 2 The empirical formula is CH 2. The molecular formula is C 2 H 4.

Example: A hydrocarbon has 82.65% C and 17.34% H ----Molar mass is 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: CaSO 4 2H 2 O is hydrated calcium sulfate CaSO 4 is anhydrous calcium sulfate

Determining the number of waters of hydration g CuSO 4 x H 2 O is heated to drive off the water. The resulting anhydrous CuSO 4 has a mass of g. What is the value of x?