Download Essential Math of Chemistry

Essential Math of Chemistry
Scientific Notation
-- used to express very large or very small
numbers, and/or to indicate precision
(i.e., to maintain the correct number
of significant figures)
Form:
(# from 1 to 9.999) x 10exponent
800 = 8 x 10 x 10
= 8 x 102
2531 = 2.531 x 10 x 10 x 10
= 2.531 x 103
0.0014 = 1.4 10 10 10
= 1.4 x 10–3
  
Put in standard form.
1.87 x 10–5 = 0.0000187
3.7 x 108 = 370,000,000
7.88 x 101 = 78.8
2.164 x 10–2 = 0.02164
Change to scientific notation.
12,340 = 1.234 x 104
0.369 = 3.69 x 10–1
0.008 = 8 x 10–3
1,000,000,000 = 1 x 109
6.02 x 1023 = 602,000,000,000,000,000,000,000
Using the Exponent Key
EE
EXP
The EE or EXP or E key means “times 10 to the…”
How
How to
to type
type out
out 6.02
6.02 xx10
102323::
6
0
.
2
EE
2
3
not…
WRONG!
6
0
.
yx
2
2
3
or…
6
WRONG!
.
0
2
x
1
and not…
6
.
0
EE
2
3
TOO MUCH WORK.
0
2
x
1
0
yx
2
3
Also, know when to hit your (–) sign.
(before the number,
after the number,
or either one)
1.2 x 105

2.8 x 1019
Type this calculation in like this:
1
.
2
EE
5
2
.
8
EE
1

9
=
Calculator gives… 4.2857143 –15
or… 4.2857143 E–15
This is NOT written… 4.3–15
But instead is written… 4.3 x 10–9
or
4.3 E –9
7.5 x 10–6 (–8.7 x 10–14) = –6.5 x 10–19
4.35 x 106 (1.23 x 10–3) = 5.35 x 103 or 5350
5.76 x 10–16

9.86 x 10–4 = 5.84 x 10–13
8.8 x 1011 x 3.3 x 1011 = 2.9 x 1023
Essential Math
of Chemistry
Units must be carried into the
answer, unless they cancel.
5.2 kg (2.9 m) = 0.64 kg-m
(18 s)(1.3 s)
s2
4.8 kg (23 s)
(18 s)(37 s)
= 0.57 kg
s
Solve for x.
x+y=z
x and y are connected by
addition. Separate them
using subtraction. In general,
use opposing functions to
separate things.
x+y=z
–y –y
The +y and –y cancel on
the left,
leaving us with…
x=z–y
Numerical Example
Solve for x.
x – 24 = 13
x and 24 are connected by
subtraction. Separate
them using the opposite
function: addition.
x – 24 = 13
+24 +24
The –24 and +24 cancel
on the left,
leaving us with…
x = 37
Solve for x.
x and k are connected by
multiplication. Separate
them using the opposite
function: division.
The two k’s cancel on
the right,
leaving us with…
F=kx
()
()
__
1
__
1
F=kx
k
k
(or)
F=kx
k k
__
F
x=
k
Numerical Example
Solve for x.
x and 7 are connected by
multiplication. Separate
them using the opposite
function: division.
The two 7’s cancel on
the right,
leaving us with…
8=7x
()
()
__
1
__
1
8=7x
7
7
(or)
8=7x
7 7
__
8
x=
7
Solve for x.
One way to solve this
is to cross-multiply.
Then, divide both
sides by TR.
The answer is…
___
BA = ___
TR
x
H
BAH = xTR
( )
( )
___
1 BAH = xTR ___
1
TR
TR
BAH
x = ___
TR
Solve for T2, where…
P
1V1
____
=
P1 = 1.08 atm
T1
P2 = 0.86 atm
____
1 PVT =
V1 = 3.22 L
P1V1 1 1 2
V2 = 1.43 L
P2V2T1
T1 = 373 K
( )
P
2V 2
____
T2
( )
____
1
P1V1
P2V2T1
______
T2 =
P1V1
(0.86
atm)(1.43 L)(373 K)
_____________________
T2 =
= 132 K
(1.08 atm)(3.22 L)