Transcript Chemistry 103

Chemistry 103
Lecture 13
Outline
I. The MOLE continued….
II. Determining Chemical Formulas


Percent Composition (review)
Empirical/Molecular Formulas
Atomic masses in Grams
107.9g
of Ag.
How
many Ag
atoms?
32.07g of S.
How many S
atoms?
12.01g of C. How
many atoms of C?
Avogadro’s Number

Solution = 6.022 x 1023

Avogadro’s number is equal
to 1 mole
Makes working with large numbers
easier

Molar Mass from Periodic Table
Molar mass
• Is the atomic
mass
expressed in
grams
Molar Mass
The molar mass
• Is the mass of one mole of
an element or compound
• Is the atomic mass
expressed in grams
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Some One-Mole Quantities
32.07 g
55.85 g
58.44 g
294.20 g
342.30 g
Conversion Factors

1 mole = 6.022 x 1023
(when the question asked for “individual”
numbers of something)

1 mole = molar mass
(relationship between mass and number)
The Mole
Molar mass
Mass in grams
1 mole
6.022 x 1023
Individual particles
Conversion Factors

1 mole = 6.022 x 1023

1 mole = molar mass

Mole ratio = (“n” mole part)/ (1 mole total)
Subscripts State Atoms and Moles
1mole aspirin
9 mol C
8 mol H
4 mol O
Learning Check

Calculate the number of moles of aspirin in
52.1 g of aspirin (C9H8O4).
52.1g C9H8O4 (1 mol C9H8O4)
180.16g C9H8O4.
Learning Check

Calculate the number of moles of hydrogen in
52.1 g of aspirin (C9H8O4).
52.1g C9H8O4 x (1 mol C9H8O4) (8moles H)
180.16g C9H8O4 1 mol C9H8O4
Learning Check

Calculate the number of grams of hydrogen
in 52.1 g of aspirin (C9H8O4).
Learning Check

Calculate the individual number of hydrogen
atoms in 52.1 g of aspirin (C9H8O4).
Learning Check
Allyl sulfide C6H10S is a
compound that has the odor of
garlic. How many moles of
C6H10S are in 225 g?
Learning Check
Allyl sulfide C6H10S is a
compound that has the odor of
garlic. How many moles of C
atoms are in 225g of C6H10S?
Learning Check
Allyl sulfide C6H10S is a
compound that has the odor of
garlic. How many grams of C
are in 225g of C6H10S?
Learning Check
How many H2O molecules are in 24.0 g H2O?
a) 4.52 x 1023
b) 1.44 x 1025
c) 8.02 x 1023
Learning Check
How many H atoms are in 24.0 g H2O?
a) 4.01 x 1023
b) 1.60 x 1024
c) 8.02 x 1023
Learning Check
Determine the mass of 6 molecules of O2 in gram
units.
Percent Composition
Percent composition
• Is the percent by mass of each element in a
formula.
Mass element in compound x 100%
Mass of compound
Percent Composition
Example:
Calculate the percent composition of CO2.
CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol
12.01 g C
44.01 g CO2
x 100
=
27.29 % C
32.00 g O
x 100
44.01 g CO2
=
72.71 % O
100.00 %
Learning Check
What is the percent composition of lactic acid,
C3H6O3, a compound that appears in the blood
after vigorous activity?
Solution
STEP 1
3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
36.03 g C + 6.048 g H + 48.00 g O
STEP 2
%C = 36.03 g C x 100 =
90.08 g
40.00% C
%H = 6.048 g H x 100 =
90.08 g
6.714% H
%O = 48.00 g O x 100 =
90.08 g
53.29% O
Types of Formulas
The molecular formula
 Is the true or actual number of the atoms in a
molecule
The empirical formula

Is the simplest whole number ratio of the atoms
(this is the formula for ionic compounds)
H2O2
molecular formula
HO
empirical formula
Some Molecular and Empirical Formulas

The molecular formula is the same or a multiple
of the empirical.
Learning Check
1. What is the empirical formula for C4H8?
a) C2H4
b) CH2
c) CH
2. What is the empirical formula for C8H14?
a) C4H7
b) C6H12
c) C8H14
Learning Check
1. What is the empirical formula for C4H8?
a) C2H4
b) CH2
c) CH
2. What is the empirical formula for C8H14?
a) C4H7
b) C6H12
c) C8H14
Learning Check
A compound has an empirical formula
SN. If there are four atoms of N in one
molecule, what is the molecular formula?
1) SN
2) SN4
3) S4N4
Learning Check
A compound has an empirical formula
SN. If there are four atoms of N in one
molecule, what is the molecular formula?
1) SN
2) SN4
3) S4N4
Empirical Formula

Using Experimental data to find the empirical
formula of a compound.
Law of Definite Proportions

Experimental studies (decomposition reactions)
led to the conclusion that the percentage of each
element present in a given compound does not
vary - Law of Definite Proportions (Dalton)
Experiment - Law of Definite Proportions


Measure Mass of a compound (an oxide of tin)
Experimentally “decompose” into oxygen and
tin. Measure mass of each.
Law of Definite Proportions
4 different experiments with a compound that contains both Sn and Oxygen only.
Mass Sn(g)
Mass tin oxide(g)
Mass O(g)
Mass Sn/Mass O
5.00g
(78.7%)
6.35g
1.35g (21.3%)
3.70
10.0g
(78.7%)
12.7g
2.7g (21%)
3.7
23.7g
(78.7%)
30.1g
6.4g (21%)
3.7
73.4g
(78.8%)
93.2g
19.8g (21.2%)
3.71
Compound of Sn and O

Row 1: mass Sn = 5.00 g mass O = 1.35g

Row 3: mass Sn = 23.7g mass O = 6.4g
Empirical Formula Problems




1). Convert percentages given to grams
2). Convert grams to moles
3). Divide by the smallest number in order to
ascertain the whole number ratio of atoms of
different elements in the compound
4). Clear any obvious fractions in step 3.
Learning Check
A compound contains 7.31 g Ni and
20.0 g Br. Calculate its empirical
(simplest) formula.
Learning Check
A particular compound is 60.0% C, 4.5%
H and 35.5% O. Calculate its empirical
(simplest) formula.
Percentages given




Assume 100 g sample (percentages easily
translate into gram quantities)
Convert grams into moles
Divide by the smallest number of moles
present.
If you do not have whole numbers at this
point, clear the fraction by multiplying every
number by the whole number that
accomplishes this end.
Percentages given




Assume 100 g sample (percentages easily
translate into gram quantities)
Convert grams into moles
Divide by the smallest number of moles
present.
If you do not have whole numbers at this
point, clear the fraction by multiplying every
number by the whole number that
accomplishes this end.
Percentages given




Assume 100 g sample (percentages easily
translate into gram quantities)
Convert grams into moles
Divide by the smallest number of moles
present.
If you do not have whole numbers at this
point, clear the fraction by multiplying every
number by the whole number that
accomplishes this end.
Percentages given




Assume 100 g sample (percentages easily
translate into gram quantities)
Convert grams into moles
Divide by the smallest number of moles
present.
If you do not have whole numbers at this
point, clear the fraction by multiplying every
number by the whole number that
accomplishes this end.
Converting Decimals to Whole Numbers
When the number of moles for an element is a decimal,
all the moles are multiplied by a small integer to obtain
whole number.
Empirical Formula for Aspirin
Learning Check
A pure phosphorus/oxygen compound is
43.7% “P” and the remainder “O”.

What is the empirical formula of this compound?
Relating Molecular and Empirical
Formulas
A molecular formula
 Is a multiple (or equal) of its empirical
formula
 Has a molar mass that is the empirical
mass multiplied by a whole number
molar mass
= a whole number
empirical mass
 Is obtained by multiplying the empirical
formula by a whole number
Some Compounds with Empirical
Formula CH2O
Molecular Formula from Empirical

If the molar mass of the compound with
empirical formula P2O5 is 284g/mol, what is
the molecular formula?
Learning Check
A compound has a molar mass of 176.1g
and an empirical formula of C3H4O3.
What is the molecular formula?
A) C3H4O3
B) C6H8O6
C) C9H12O9