Lecture September 14
Goal for today: Chapter 3 Finish Chap. 13
(up to p. 107)
HW and Quiz due September 19th - through
Section 3.6 p. 97 - naming compounds
Reading Assignment - we
start Chapter 7 on Friday
CHEM131 - Fall 11 - September 14
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Example (I)
How many CaCO3 molecules are in 0.5632 g?
Molar mass CaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09
# molecules = 0.5632g (1 mol/100.09 g) (6.022 x 1023 molecules/
mol) =
➘
Answer: 3.389 x 1021 molecules CaCO3
What is the percentage (by mass) of Ca in CaCO3?
Mass of Ca / mass of CaCO3 = 40.08/100.09
Answer: 40.046% by mass
How many Ca atoms are in 0.5632 g of CaCO3?
# Ca atoms = [(0.5632 x .40046)/40.078] x 6.022x1023
➚
Answer: 3.389 x 1021 atoms of Ca
CHEM131 - Fall 11 - September 14
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Percent Composition and
Chemical Formula
Using Oxalic acid H2C2O4 as an example
molar mass:
2 x 1.008 =
2 x 12.011 =
4 x 16.00 =
Total
=
2.016
24.022
64.00
90.04 g/mol
Can calculate
%O from the
H and C
% H = (2.016/90.04)x100 = 2.239%
% C = (24.022/90.04)x100 = 26.68%
% O= (64.00/90.04)x100 =71.08%
CHEM131 - Fall 11 - September 14
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Question?
What mass of carbon is in 25.0 g of
oxalic acid?
mass C = 25.0 g acid (1 mol acid/90.04 g acid)
(2 mol C/1 mol acid) (12.011 g C/1 mol C) =
6.67 g C
OR (from previous slide C 26.68%)
25.0 (.2668 C) = 6.67 g C in oxalic acid
CHEM131 - Fall 11 - September 14
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Empirical Formulas
Using the % by mass
A compound containing calcium and
oxygen is 71.47 % Ca, what is the
empirical formula of the compound?
1) Assume 100.00 g of compound.
Ca:O
1:1
CaO
2) Calculate the moles of Ca
moles Ca = 71.47 g (1 mol/40.08 g) = 1.783 mol
3) Find the mass and then moles of oxygen
moles O = (100.00-71.47)g (1 mol/16.00 g) = 1.783
4) Find the ratio of the elements ☞ formula
CHEM131 - Fall 11 - September 14
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Empirical Formula Example
A compound with a molar mass of 82 g/mol
has been determined to contain 14.6% C, 39.0%
O and 46.3% F, what are the empirical and
molecular formulas?
Assume 100 g sample
14.6 g/12.01 = 1.22 moles C
39.0 g/16.00 = 2.44 moles O
46.3/19.00 = 2.44 moles F
C:O:F = 1:2:2
CO2F2
empirical formula
12.00 + 2(16.00) + 2(19.00) = 82
molar mass = 82 so this is also the molecular
formula!
CHEM131 - Fall 11 - September 14
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One More Example
Magnesium citrate (MgC6H6O7 molecular
weight = 214.41) is used as a nutritional
Mg supplement.
What mass of magnesium citrate is
needed to supply 400. mg of Mg?
% Mg = (24.31/214.41)x100 = 11.34% Mg
Mass of magnesium citrate =
0.400 g Mg (214.41 g citrate/24.31 g Mg)
= 0.400/.1134 = 3.53 g citrate
CHEM131 - Fall 11 - September 14
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Quiz of the Day
What is the empirical formula of a compound
with the following % composition
37.48% C
12.58% H
49.94% O
Assume 100.00 g
# moles C = 37.48/12.01 = 3.12
# moles H = 12.58/1.008 = 12.48
# moles O = 49.94/16.00 = 3.12
C:H:O = 1:4:1
CH4O
CHEM131 - Fall 11 - September 14
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