Chemistry 11 – Mole Concept Study Guide
4
Gram ! Mole & Mole ! Gram Conversions:
•
The use of the MOLAR MASS allows us to calculate the mass of a given number of
moles of a substance, and the calculation of the number of moles in a given mass of a
substance.
•
The unit for moles is “_____”
Mol
•
conversion factor
In order to calculate from grams to moles, you need to use a ______________________!
•
Conversion factors are used to relate the number of moles to the mass of material present.
•
Since 1 mole of a given substance “X” has a mass of (molar mass of X) g…
x
g
1
mol
Examples:
or
(
use
)
lowercase
1
____________________s
MH
C
kg
1×12.0=12.0
02×16.0=32.0401
What is the mass of 3.25 mol of CO2?
[molar mass of CO2 = 44.0g/mol]
-7
g)
What you want = (What you have) X (Conversion Factor)
=4<.0gI
Mass of CO2
For
X&÷
Use
least #
of
SF
IMOI
3. mol CO X molar mass of CO
= 3.25
2
2
1 mol of CO2
1
3SF
25401402×44.0=9*2
= 3.25 mol
CO2 X 44.0 g CO2
3.25×44.0902
1 mol CO2
3sF
= 143 g CO
2
143
802
How many moles of N2 are there in 50.0 g of N2?
N
2×14
.O=
[molar mass of N2 = 28.0g/mol]
mol )
→
( g
What you want = (What you have) X (Conversion Factor)
.
Moles of N2
xf.fm#P
.o#
so g N2 X
= 50.0
1 mol N2 s
Molar mass N2
T.79molN2]
1.785/4442/81/6
530%6=238
= 50.0 g N2 X 1 mol N2
28.0 g N2
= 1.79 mol N2
.%mo1N2
mowa
-
9
upbk
Round
of
•
When you are calculating molar mass, the units for molar mass is ____________!
glmol
5
Is
g
mol
molecule
→
atom
-
Chemistry 11 – Mole Concept Study Guide
pam.de
{
Molecule
=
A group
multiple
of
bonded
atoms
5
) 02
eg
Moles ! Molecules & Molecules ! Moles Conversions:
compound
•
eg HZO
There are times when chemists would like to know the number of particles or molecules
involved in a reaction.
6.02×1023
=
A
elements
multiple
with
molecule
)
=
•
Thus, we need another set of Conversion Factors! =Av= Avogadro 's #
mol
_____________________
particles
or
1
6.02×1023
6.02×1023 particles
______________________s
particles
mol
1
Examples:
particles
Av
molecule
( mol
)
→
How many molecules are there in 0.125 mol of molecules?
What you want = (What you have) X (Conversion Factor)
0.125*4
Fide
# of molecules = 0.125 mol X
0.125×6.02×10
1
23
6.02 x 10236.02×1023
molecules
1 mol
I
Molecule
s
= 7.53
1022 molecules
7. x5251×1022
molecule
[email protected]
How many moles are in 1.00 x 1022 molecules of H2O?
Hzo )
20
mol
¥7
What you want = (What you have) X (Conversion Factor)
Moles of H2O = 1.00 x 1022 molecules X
1 mol
s
6.02 x 2×1023
1023 molecules
[email protected]
O.co/66H2/9/5mo1Hz0#
X 10moles
@ 22
= 0.0166
1
÷
6.02×10
1
23
=
Try these problems in Hebden: pg 82 Q’s #8-10; pg 84 Q’s #15 bcdeg, 17 bd
g
mol
→
molecule
→
€
0mm
molecule
=
media
atoms
Exl How many
atom
->
^
of
HZO
3
atoms
are
?
there in
1
•fQ@
Chemistry 11 – Mole Concept Study Guide
•
•
8
So far we’ve learned how to calculate g! mol, mol ! mol, and g ! molecules
calculations.
We can also learn how to calculate molecules ! atoms as well!
From Mole Map:
Molecules
(Particles)
X # of atoms
1 molecule
X 1 molecule
# of atoms
r
Count
#
of atom
Atoms
mmoKcuW#m#,1mogcmaetoHm#
Example 1:
How many atoms of oxygen are found in 1.00 x 1023 molecules of ozone (O3)?
Road Map: molecules O3 ! atoms O
3 atoms of oxygen for every molecule of ozone.
Note: There are ____
Calculate:
35T
xi023atomT)
FOO
1.00 x 1023 molecules O3 x
Example 2:
3SF
ASF
0$ 3¥50
Mayone
= 3.00 x 1023 atoms O
3 atoms O
1 molecule O3
•
031
SF
How many atoms of hydrogen are there in 1.00 g of H2O?
MM
Av
formula
Road Map: g H2O ! mol H2O ! molecules H2O ! atoms H
-
MM(H20)= 18.09401
AV
Formula
a atoms of hydrogen for every molecule of H2O.
Note: There are ____
6.023¥
*
Calculate:
3 ato
ms O
←
= 3.00
[email protected]
6.02×10^023
1 ÷ 18 × (
)X 2
=
=
cost
#
6$98/9/×/O"atomsH3=
1.00g H2O x
6.02×1023
=
HZO
?
Q2l )
atom
At3WsH
a)
c) 10
CHDZCO
a)
Q23
FtItT=6atomI
f)
e) 46
d) 15
Ntkcl
1. oomol
23
2+6+1+1
⇒
CzH6cO
=
Ntkcl
b)
gate
=
,
molecule
vinegar
=
Tttttitt
'zI
9-
in
?
→
#
moke.my#
A
BF
3sF
atoms
Formula
SSF
town
(6.02×1023)×6
.tk#nFEIx6fmEMwiM0lNH4Cl
3.612×41024
=
1
×
atoms
=
351=3.61×1024910^57
|
-
Q23l )
9.5×10
Q23tQTf
"
Omit
STP
"
'
¥01
MM
9 ( Formula )
)
NHS
3gNH3→?atom(
NH
molecule
→
}
(
( 3)
Au )
g.
5×10
.
-
t.MX
(6.02×103)×4
msn.IT#nxtiIiIx4TFI=/.3/45H4Y705HIa0m9.5*03
MMNH } )
=
14.0+3.0=17.094014
QSF
3SF
¥
IF