empirical and molecular formulas
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Empirical and Molecular Formulas. Section 11.4 Chemistry. Objectives. Explain what is meant by the percent composition of a compound. Determine the empirical and molecular formulas for a compound from mass percent and actual mass data. Key Terms. Percent composition Empirical formula - PowerPoint PPT PresentationTRANSCRIPT
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Empirical and Molecular FormulasSection 11.4Chemistry
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Objectives
Explain what is meant by the percent composition of a compound.
Determine the empirical and molecular formulas for a compound from mass percent and actual mass data.
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Key Terms
Percent compositionEmpirical formulaMolecular formula
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Percent Composition
Percent by mass of each element in a compound.
Mass of Element x 100Mass of Compound
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Calculate percent composition of H in H2O
Molar mass of water: 18.02 g/mol
Determine mass of H in 1 mol of H2O
Example
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Mass of H
1.01 x 2 = 2.02 g H in 1 mol water
Atomic mass of H
from periodic table
Number of H in H2O
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Percent Composition of H
2.02 g of H x 100 = 11.2% H18.02 g of H2O
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Empirical Formula
Formula for a compound with the smallest whole-number ratio of elements.
Percent composition can be used to find the chemical formula.
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Empirical Formula
When given percent composition, assume: The total mass of the compound is
100 g The percent composition of the
element is equal to the mass in grams of the element.
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Example
A compound has a percent composition of 40.05% S and 59.95% O. So in 100 g of the compound,
40.05 g are S and 59.95 g are O.
Find the amount of mol for each element.
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Empirical Formula
40.05g S x 1 mol S = 1.249 mol S32.07g S
59.95g O x 1 mol O = 3.747 mol O
16.00g O
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Empirical Formula
The element with the smallest number of mol gets the subscript 1.
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Empirical Formula
S has a subscript 1Then divide the mol O by the mol S 3.747 mol O/ 1.249 mol S = 3 mol
O Then write your empirical
formula using your subscripts: SO3
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Practice Problems
Pg. 333
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Molecular Formula
Formula that specifies the actual number of atoms of each element in one molecule or formula unit of a substance.
n = molecular formula mass empirical formula mass
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Example
Compound is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.
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Example
40.68 g C x 1 mol C = 3.387 mol C
12.01 g C5.08 g H x 1 mol H = 5.04 mol H
1.008 g H 54.24 g O x 1 mol O = 3.390 mol
O16.00 g O
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Example
3.387 mol C/ 3.387 = 1 mol C5.040 mol H/ 3.387 = 1.49 = 1.5 mol
H3.390 mol O/ 3.387 = 1.001 = 1 mol ORatio of C : H : O = 1 : 1.5 : 1
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Example
Empirical Formula: C2H3O2
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ExampleTo find molecular formula, calculate
n.n = molecular formula mass empirical formula mass Molecular mass is in the problem!Calculate molar mass of empirical
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Example
n = 118.1 = 2 59.04
Multiply the subscripts of the empirical by n to find the molecular formula.
Molecular Formula: C4H6O4
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Practice Problems
Pg 335
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Homework
Section 11.4 Problems 27-29 on page 877