Empirical Formula

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Empirical & Molecular
Formulas
Unit 4: Stoichiometry
Chapter 10 – The Mole
2/6/2014
 Learning Target:
 Understand how to determine the percent
composition of a compound.
 Learning Outcome:
 Know how to calculate the percent composition of a
compound.
Practice
• Suppose you spend 6 hours texting friends
each day.
What percentage of each day do you spend
texting?
6 ÷ 24 = 0.25 hours
0.25 x 100% = 25%
Percentage Composition
• Definition
o
The mass of each element in a compound
compared to the entire mass of the compound and
multiplied by 100 percent
2 ways to find % composition
1) Total mass of compound
Percentage of sodium in sodium chloride
22.99 g Na __ x 100% = 39.3% Na
58.44 g NaCl
2) Experimentally a compound’s mass is measured
and then decomposed into its individual
element.
Practice
• Find the percentage composition of a
compound that contains 1.94 g of carbon, 0.48
g of hydrogen, and 2.58 grams of sulfur in a
5.00 g sample of that compound.
Practice
A sample of unknown compound with a mass
of 0.847 g has the following composition: 50.51
percent fluorine and 49.49 percent iron. When
this compound is decomposed into its
elements, what mass of each element would be
recovered?
•Can we find the number of moles of
carbon, hydrogen and sulfur?
2/7/2014
• Learning Target:
– Understand how to determine empirical and
molecular formulas of a substance.
• Learning Outcome:
– Know how to calculate the empirical and molecular
formula of a substance using percent composition.
Empirical Formula
• Definition
• A formula that gives the simplest whole-number ratio
of the atoms of elements
• For example, H2O2 is the molecular formula for
hydrogen peroxide.
• HO is the empirical formula for hydrogen peroxide.
Determining the Empirical
Formula
• Now that you know how to calculate
percentage composition, you can use that ratio
of masses to find the ratio of atoms which is a
chemical formula
• Ratio of Masses  Ratio of Atoms
(% composition)  chemical formula
Steps for Finding Empirical
Formula
• Percentage  Mass (g)
• Mass (g)  # of Moles
• Once you have moles, find the smallest whole
number ratio between the moles.
Example
• Determine the empirical formula of a compound
containing 5.75 g Na, 3.5 g N, and 12.0 g O.
• Since you have mass, convert mass → moles
o
Na: 5.75 g Na x 1 mol Na = 0.25 mol
22.99 g
o
N: 3.5 g N x 1 mol N = 0.25 mol
14.00 g
o
O: 12.0 g O x 1 mol O = 0.75 mol
16.00 g
Example Continued
• Now divide each mole value by the smallest
number of moles:
• Na: 0.25/0.25 = 1
• N: 0.25/0.25 = 1
• O: 0.75/0.25 = 3
• Empirical Formula = Na N O3
Practice
Determine the empirical formula of a compound
containing 0.928 g of gallium and 0.412 g of
phosphorus.
Practice
Determine the empirical formula of a compound
containing 2.644 g of gold and 0.476 g of
chlorine.
• Determine the empirical formula of a
compound containing 1.723 g of
carbon, 0.289 g of hydrogen, and 0.459
g of oxygen.
Molecular Formula
• Definition
o
The formula that gives the actual number of atoms
of each element in a molecular compound
• The molecular formula is always a wholenumber multiple of the empirical formula.
Determining Molecular Formula
• Determine molecular formula by comparing
the molar mass of an unknown compound with
the molar mass of the empirical formula.
Example
• Find the molecular formula of a compound that
contains 42.56 g of palladium and 0.80 g of
hydrogen. The molar mass of the compound is
216.8 g/mol.
Step1 Convert to moles
42.56 g Pd x 1 mole Pd = 0.40 mol Pd
106.42 g
o 0.80 g H x 1 mole H = 0.79 mol H
1.01 g
o
Example Continued
Step 2: Find mole ratio:
Pd: 0.40/0.40 = 1
H: 0.79/0.40 = 2
• Step 3: Write empirical formula:
Pd H2
Example Continued
• We know that the molar mass of the
compound (given in question) = 216.8 g/mol
• Step 4: Find the empirical molar mass
If empirical formula is PdH2 , then the
empirical molar mass = 106.42g/mol Pd +
2(1.01g/mol H) = 108.44 g/mol
Continued Once Again
• Step 5 Find, Molar Mass ÷ Formula Mass:
216.8 g/mol = 1.995  2
108.44 g/mol
**MOST IMPORTANT STEP
Step 6:
FINALLY!!! Now adjust the molar formula
PdH2 becomes Pd2H4
Practice Problem
Octane (C8), a compound of hydrogen and
carbon, has a molar mass of 114.26 g/mol.
If the compound contains 18.17g/mol
hydrogen, what is its molecular formula?
Last Practice
• β-carotene, a compound found in carrots, can be
broken down to form vitamin A. The empirical
formula for β-carotene is C5H7. The molar mass
of β-carotene is 536 g/mol. What is the
molecular formula for β-carotene?
3/12/12 PROBLEM
•
A component of protein called serine has an
approximate molar mass of 105.11 g/mole. If
the percent composition is as follows, what is
the molecular formula of serine?
C = 34.95 % H= 6.844 % N= 13.59 % O = 46.56 %
3/12/2012 Problem
Sample (3.585g) contains 38.72% C, 9.62% g
of H, 51.66% O and its molar mass is
62g/mol. What is molecular formula of this
substance?
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