Chapter 7 Section 3

Download Report

Transcript Chapter 7 Section 3

Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Definitions:
Percent Composition: the percentage by mass of each element
in a compound
Empirical Formula a chemical formula that shows the composition
of a compound in terms of the relative numbers and kinds of atoms in
the simplest ratio
Molecular Formula a chemical formula that shows the number
and kinds of atoms in a molecule, but not the arrangement of the atoms
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Percent Composition
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
How to calculate the percent composition of a compound.
1. Calculate the mass of each element in the compound. For example if
the compound contains 3 oxygen atoms the mass of the oxygen in the
compound is 16.0 x 3 = 48
2. Calculate the gram molar mass of the whole compound. For example if
the compound was CaSO3, the total mass would be 40.1 from calcium + 32.0
from sulfur and 48.0 from oxygen. The total would be 40.1 + 32.0 + 48.0 for
A total mass of 120.1.
3. Follow the formula on your Reference Table:
Mass of the part is step 1, mass of the whole is step 2.
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Practice you percent composition skills
1. Calculate the percent composition of sodium chloride, NaCl
Na – 39.7%
Cl- 60.3 %
2. Calculate the percent composition of silver nitrate,AgNO3
Ag - 63.5%
N - 8.2%
O - 28.3%
3. Calculate the percent composition of magnesium hydroxide, Mg(OH)2
Mg – 41.7%
O – 54.9%
H - 3.4%
4. What is the mass percentage of water in the hydrate CuSO4• 5 H2O
Water is 36.1% of the compound
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Homework Page 860 #47, #48 and #49
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Difference between empirical formula and molecular formula
Empirical Formula: Simplest ratio of elements
Molecular Formula: “True formula” Does not have to be simplest ratio
C6H12O6
Molecular
Reduced by factor of 6
CH2O
Empirical
H2 O
Both empirical and molecular
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Calculating Empirical Formula
1. Set up the problem by each element
2. divide each element quantity by its gram molar mass from the
periodic table
3. Look at each answer and divide the smallest answer into each
elements answer.
4. This will produce the ratio of each element in the empirical formula
5. If ratio is not whole numbers increase all answers by a factor
to create whole numbers
6. Write empirical formula
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
A compound is found to contain 54.5% carbon, 9.1% hydrogen, and 36.4%
oxygen. Determine the simplest formula.
Carbon
Hydrogen
Oxygen
54.5
12
9.1
1.0
36.4
16.0
4.54
2.27
9.1
2.27
2.27
2.27
2
4
1
C2H4O
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Find the empirical formula of a compound found to contain 26.56% potassium,
35.41% chromium, and the remainder oxygen.
Potassium
Chromium
Oxygen
26.56
39.1
35.41
52.0
38.03
16.0
0.68
0.68
0.68
0.68
2.34
0.68
1.0
1.0
3.5
Increase by factor of 2 to make all whole numbers
K2Cr2O7
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Homework page 861 #69 - #73
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Calculating Molecular Formula
1. Calculate the empirical formula if not given.
2. Calculate the empirical formulas gram formula mass
3. Divide the molecular formulas gram formula mass by the molecular
formulas gram formula mass. This will give you a factor number that
you must increase the ratio of the empirical formula by.
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Determine the molecular formula of a compound with an empirical formula
of NH2 and a formula mass of 32.06 amu.
NH2 gram formula mass is 16.0 amu
Molecular formulas gram formula mass is 32.06 amu
32.06
16.0
= 2
You must increase the ratio of NH2 by a factor of 2
N2H4
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition
Chapter 7 – The Mole and Chemical Composition
Sec 3 - Formulas and Percentage Composition