# The Mole Lecture by pengtt

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```									Chapter 7

Chemical
Quantities
Standards covered in this Lecture

Standards B, C, and D
23
6.02 x 10
Mole: amount of a substance equal
that       substance.
Ex) mole of molecules
mole of atoms
mole of students, etc.
(exploding mole)
Exploding Mole
# ions,
atoms        solution
known
concentration
grams

mole

gas                unknown
volume                grams
chemical
formulas
converting molecules to moles
Ex) How many moles of CO2 equal
26
3.3 x 10 molecules of CO2?

3.3 x 1026 molecules     mole
6.02 x 1023
(molecules)
= 548 moles CO2
Molar Mass
Atomic mass of an element or
compound expressed in grams.
- used as a custom conversion factor

Ex) molar mass of C = 12.0 g/mol
molar mass of Na = 23.0 g/mol
molar mass of Cl = 35.5 g/mol
(NaCl Demo)
gram atomic mass
gram molecular mass
gram formula mass
}   Molar Mass

Ex) Calculate the molar mass of CaCO3
molar mass
Ca 40.1 x 1= 40.1
C 12.0 x 1= 12.0
O3 16.0 x 3 = 48.0
CaCO3        100.1 g/mol
converting grams to moles
Ex) Calculate the number of moles in
5.6g of NaOH.
step 1) find molar mass
Na     23.0
O     16.0
H       1.0
NaOH 40.0 g/mol
step 2) calculate moles

5.6 g NaOH        mol
[   40.0 g
]= 0.14 mol
NaOH

stop
converting moles to grams
Ex) How many grams are in 3.5 moles
of ammonium sulfate?
molar mass (NH4)2SO4
N2    14.0 x 2 = 28.0
H8    1.0 x 8 =   8.0
S     32.1 x 1 = 32.1
O4 16.0 x 4 = 64.0
(NH4)2SO4       132.1 g/mol
step 2) calculate grams

3.5 moles (NH4)2SO4
[   132.1g
mol
]
= 462.4 g (NH4)2SO4

acetylene balloon demo
acetylene balloon
Gas Volume & the Mole
STP:
Standard Temperature & Pressure

at sea   0o Celsius   1.0 atmosphere
level
at STP:
one mole of any gas    mole
or
22.4 Liters volume    22.4 L
Ex) Convert 0.630 moles of methane
(at STP) to liters.

0.630 mole CH4           22.4 L
[   mole
]
= 14.1L CH4

zinc & sulfur demo
zinc & sulfur reaction
Multiple Step Problems
Ex) How many molecules are in 8.7
grams of glucose, C6H12O6?
(grams       moles       molecules)
molar mass C6H12O6
C 12.0 x 6 = 72.0
H 1.0 x 12 = 12.0
O 16.0 x 6 = 96.0
C6H12O6      180 g/mol
[
8.7g glucose mol
180 g      ][                      ]
6.02 x 1023molecules
mol
22
= 2.9 x 10        glucose molecules
Ex) How many grams are in 4.73 x 1024
atoms of iron?

4.73 x 1024atoms
[ 6.02 x 10
mol
23       ][
atoms        ]
55.8g
mol

= 438 g Fe

ammonium dichromate demo
decomposition of ammonium dichromate
Percent Composition
-relative amount of each element in a
compound

grams element
% mass =               x 100
grams cmpd
Ex) What is the percent of sodium in
sodium chloride? (assume one mole)
% mass Na =    mass Na      x 100
mass NaCl

23.0 g/mole
=                      x 100 = 39.3%
(23.0 + 35.5) g/mole
Ex) Find the percent composition of
aluminum sulfate?
-find % comp of each element
molar mass Al2(SO4)3
Al2   27.0 x 2 = 54.0
S3    32 x 3 = 96
O12   16.0 x 12 = 192
342 g/mol
54 g/mol
% Al =            x 100 = 15.8 % Al
342 g/ mol

96. g/mol x 100 = 28.1 % S
%S=
342 g/ mol

192 g/mol x 100 = 56.1 % O
%O=
342 g/ mol

stop
Empirical Formula
- lowest whole number ratio of atoms in
a compound
Ex) CH2 is the empirical formula for the
compound that has the molecular
formula C6H12.
Ex) H2O is both the empirical formula
and molecular formula for water
Calculating Molecular Formulas
amount       %     empirical   molecular
in grams   comp    formula     formula
Ex) What is the empirical formula of a
sample containing 68.8% C, 4.95% H &
26.6% O?
Step 1) Assume a 100 gram sample so:
68.8% C = 68.8 grams C
4.95% H = 4.95 grams H
26.6% O = 26.6 grams O
Step 2) convert grams to moles

68.8 g C
[   mol
12.0g]= 5.73 mol C

4.95 g H
[    mol
]
1.0g = 4.95 mol H
26.6 g O
[   mol
16.0g]= 1.64 mol O
Step 3) Find the mole ratio by dividing
by the element with the least moles
5.73 mol C   = 3.49 C x 2 = 6.98 ~ 7
1.64 mol
4.95 mol H   = 3.01 H x 2 = 6.02 ~ 6
1.64 mol
1.64 mol O    = 1.00 O x 2 = 2.00 = 2
1.64 mol
empirical formula: C7H6O2
Molecular Formulas
Find the molecular formula of a compound
if its empirical formula is COH2 and its
molecular mass is 120 g.
emp.form. mass
molecular mass
C 12.0           ratio:
emp.form.mass
O 16.0
120 g
H2 2.0           =       =4
30 g
30.0 g
C1x4O1x4H2x4 =

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