Download Chapter 11 : Matter Notes

Chapter 11 : Matter
Notes
Mole (mol) is equal to
23
6.02x10
The mole was named in honor
of Amedeo Avogadro. He
determined the volume of
one mole of gas.
Molar Mass: mass in grams of 1
mole of any pure substance.
Equal to the atomic mass for an
element. Unit is g/mol.
Converting moles to Mass:
Number of moles
number of grams
1 mole
= mass(g)
Converting moles to particles.
Number of moles_ 6.02x1023_ particles =
particle
1 mole
Example: How many molecules are in 3.50 moles of
sucrose?
3.50 mol sucrose 6.20 x 1023 molecules of sucrose
1 mole sucrose
= 2.11 x1023molecules of sucrose
Converting Particles to Moles:
Number of particles
1 mole
= number of moles
6.02 x1023 particles
Example: How many moles are in 4.5 x 1023 atoms
of Zinc?
4.5 x 1023 atoms Zn
1 mole Zn
6.02 x1023 atoms Zn
= 7.48 moles Zn
Converting Moles to Mass:
Element:
How many grams are in 3 moles of
Manganese?
3.00 mol Mn
54.9g Mn = 165g Mn
1 mol Mn

Converting Moles to Mass:
Compound:
How many grams are in 3 moles of
K2CrO4?
1. Find molar mass of compound.
2 mol K 39.10 g K = 78.20 g K
1 mol K
1mol Cr 52.00g Cr = 52.00 g Cr
1 mol Cr
4 mol O 16.00 g O = 64.00 g O
1mol O
194.20 g K2CrO4

Converting Moles to Mass:
Compound:
1.
Multiply moles by molar mass:
3 moles K2CrO4 194.20 g K2CrO4
mol K2CrO4
= 582.6 g K2CrO4
Converting Mass to Moles:
Number of grams
1 mole
= moles
mass (g)
Element: How many moles are in 525g of Ca?
525g Ca
1 mol Ca = 13.1 mol Ca
40.06g Ca
Converting Mass to Moles:

1.
Compound: How many moles are in 325g
of Calcium Hydroxide [Ca(OH)2]?
Find the molar mass of compound:
1mol Ca 40.08g Ca = 40.08 g
1mol Ca
2mol O 16.00g O
= 32.00 g
1mol O
2mol H
1.008g H
= 2.026g
1mol H
74.096 g Ca(OH)2
Converting Mass to Moles:
Compound:
1.
Convert grams to moles:
325g Ca(OH)2
1mol Ca(OH)2
74.096 g Ca(OH)2
= 4.39 mol Ca(OH)2
Converting Mass to Particles:
Number of grams 1 mol
6.02 x 1023 particles
mass(g)
1 mol
= particles
Element: How many atoms are in 25 g of Gold?
25g Au 1 mol Au 6.02 x 1023 atoms Au
196.97g Au 1 mol Au
= 7.65 x 1023 atoms Au
Converting Mass to Particles:

1.
Compound: How many formula units are in
35.6 g of aluminum chloride? [AlCl3]
Find the molar mass of compound:
1 mol Al 26.98 g Al = 26.96 g
1mol Al
3 mol Cl 35.45g Cl = 106.35g
1 mol Cl
133.33g AlCl3
Converting Mass to Particles:
Compounds:
Convert grams to formula units:
35.6 g AlCl3 1 mol AlCl3 6.02 x 1023 formula units
133.33g AlCl3 1 mol AlCl3
= 1.61 x1023 formula units AlCl3
1.
Chapter 11 Matter
Notes
Section 11.4
Percent Composition:
Percent composition is the percent by mass
of any element in a compound.
Mass of element
X 100% = percent composition

Mass of compound
Percent Composition:
Example:
Calculate the percent composition of
water. H2O
2 mol H
1.01 g H
= 2.02 g H
1 mol H
18.02 g H2O

1 mol O
16.00 g O
1 mol O
= 16.00 g O
18.02 g H2O
Percent Composition: con’t.
2.02 g H
H
X 100%
=
11.2% H
X 100%
= 88.8% O
18.02g H2O
16.00 g O
O
18.02 g H2O
Empirical Formula

Empirical formula is a formula with the
smallest whole-number ratio of the
element.

Example: Determine the empirical formula
for methyl acetate which has 48.64%
Carbon, 8.16% Hydrogen, 43.20 % Oxygen.
Empirical Formula: con’t.
Find the number of moles of each
element
48.64g C 1 mol C
= 4.05 mol C
12.01 g C
8.16 g H
1 mol H
= 8.10 mol H
1.008 g H
43.20 g O 1 mol O
= 2.7 mol O
16.00 g O
1.
Empirical Formula: con’t.
1.
Divide each mole by the smallest
number:
4.05 mol C
= 1.5 mol C
2.7
8.10 mol H
= 3 mol H
2.7
2.7 mol O
= 1 mol O
2.7
Empirical Formula: con’t.
Multiply each mole by the smallest
number that will produce whole
numbers:
1.5 mol C x 2 = 3 C
3 mol H x 2 = 6 H
1 mol O x 2 = 2 O
1. Use the new whole numbers as
subscripts for each element in the
compound.
C3H6O2
1.
Molecular Formula:

Experimental molar mass = n
empirical formula mass

Determine the molecular formula for a
compound composed of 40.68% carbon,
5.08 % Hydrogen, and 54.24 % Oxygen
which has a molar mass of 118.1 g/mol.
Molecular Formula:
1.
Find the empirical formula:
40. 68 g C
1 mol C
= 3.387 mol C
12.01 g C
5.08 g H
1 mol H
= 5.04 mol H
1.008 g H
54.24 g O
1 mol O
= 3.39 mol O
16.00 g O
Molecular Formula:
( finding the empirical formula con’t.)
3.387 mol C = 1 mol C x 2 = 2
3.387
5.04 mol H = 1.5 mol H x 2 = 3
3.387
3.39 mol O = 1 mol O x 2 = 2
3.387
empirical formula: C2H3O2
Molecular Formula:
Find the molar mass of the empirical
formula:
2 mol C 12.01 g C = 24.2 g
1 mol C
3 mol H 1.008 g H = 3.024 g
1 mol H
2 mol O
16.00 g O = 32.00g
1 mol O
59.04 g C2H3O2
1.
Molecular Formula:
1.
Divide the molar mass given in the
problem by the molar mass of the
empirical formula to find n:
n = 118.1 g
59.04g
n = 2
Molecular Formula:
1.
Multiply all subscripts of the empirical
formula by n:
(C2H3O2)2
molecular formula: C4H6O4