Arithmetic in Scientific Notation
 Addition/Subtraction with same exponent
3.42 104  5.2 104 
3.42 10 4
5.2 10 4








Light
8.62 10 4
The least significant placeholder is the in the 5.2 number, so we round there to get
the answer of 8.6 10 4 .
Addition/Subtraction with different exponents.
7.89 101 2.45 102 
Takethe numbers out of scientific notation, then subtract.
.7890
.0245
.7645
The least significant placeholder is the in the 4 in the bottom number, we round
there to get the answer .765 or 7.65 101 in scientific notation. (Technically the
answer should round it to .764, because if the end number is a 5 then you round to
the even number.)
Multiplication/Division

1.2 105  1.2 104 
Separate the number in front from the exponent part. Multiply the number then,
add the exponents.
1.2 1.2  105 104 1.44 109
Both these numbers have 2 significant figures so the answer should be reported
with 2 significant figures. 1.4 10 9 . You can do this in the calculator just count
the correct number of significant figures.
c  

 
c  speed of light m s
  wavelength m
  frequency Hz 
What is the wavelength of a particle of light if it’s frequency is   5.10  1014
8 m
c 3.00  10 s
c   , or   
 5.88  10 7 m
14

 5.10  10 Hz
The answer is given with 3 s.f.
Stiochoimetry
 Convert from grams to moles
How many moles are in 32 g NaCl?
58.45 gNaCl  1molNaCl
32 gNaCl   

1molNaCl 
  .55molNaCl
 58.45 gNaCl 
There are 2s.f.’s in the answer.
Convert from moles to grams
How many grams are in 12.6 moles of C8 H 8O6
200 gC8 H 8 O6  1molC8 H 8 O6


12.6molC8 H 8O6    200 gC8 H 8O6   2520 gC8 H 8O6

 1molC8 H 8 O6 
The answer is reported with 3 s.f.’s or in scientific notation the answer is
2.52 103 gC8 H 8O6
Convert from atoms to moles
How many moles of water are there in 4.32  10 22 molecules
6.02  10 23 moleculesH 2 O  1molH 2 O


1molH 2 O
  .0718molH 2 O
moleculesH 2 O  
23
6
.
02

10
moleculesH
O
)
2


The answer has 3 s.f.’s
4.32  10


22
Convert from moles to atoms
How many atoms of Barium are in 1.2 mol?
6.02  10 23 atomBa  1molBa

atomBa 
  7.2  10 23 atomBa
1molBa

1.2molBa    6.02  10


23
Answer has 2 s.f.’s
Convert from liters to moles
How many mol of CO2 are there in 12.7 liters?
22.4 LCO2  1molCO2


  .567molCO2
22
.
4
LCO
2 

There are 3 s.f.
Convert from moles to liters
How many liters is 3.42 mol of CH 4
22.4 LCH 4  1molCH 4
12.7 LCO2    1molCO2



3.42molCH 4    22.4LCH 4   76.6LCH 4
 1molCH 4 

There are 3 s.f.
Percent Composition
What is the % composition for the compound CuSO4
Cu  63.55  1  63.55
S
32  1  32
O
16  4  64
 159.55
63.55
32
 100  39.8%
S:
 100  20.1%
159.55
159.55
64
O:
 100  40.1%
159.55
Convert from grams of substance A to grams of substance B
How many grams of KCl will form from 237.1 g of KClO3 (This is Theortical
Yield below)
2 KClO3  2 KCl  3O2
Cu :

122.55 gKClO3  1molKClO3
74.55 gKCl  1molKCl
2molKCl  2molKClO
 1molKClO3   2molKCl   74.55 gKCl 
  
  
  144.2 gKCl
 122.55 gKClO3   2molKClO3   1molKCl 
The given number,237.1 has 4 s.f.’s so the answer is reported with 4s.f.’s.
Limiting Reagent
What is the limiting reagent of the following reaction if we have 37.1 g of NH 3
and 73.2 g of NO.
Convert both quantities to moles.
17 gNH 3  1molNH 3
237.1gKClO3   



37.1gNH 3    1molNH 3   2.18molNH 3
 17 gNH 3 
73.2 gNO    1molNO   2.44molNO
 30 gNO 
Now use the mole ration to find the limiting reagent
4molNH 3  6molNO
 6molNO 
  3.27molNO
4
molNH
3


In order to use up all of the ammonia we would need 3.27 mol of NO, but
we only have 2.44 mol, so the limiting reagent is NO
Percent Yield
4K  O2  2K 2O
If 43.6 g of K are used in the reaction and only 50.0 g of K 2 O are collected, what
is the percent yield?
2.18molNH 3   

We must first find the theoretical yield (Convert from g of K to g of K 2 O )
39.1gK  1molK PeriodicTa ble 
94.2 gK 2 O  1molK 2 OP.T .
2molK 2 O  4molK BalancedEq uation 
  2molK 2 O   94.2 gK 2 O 
  52.52 gK 2 O
  
  
 39.1gK   4molK   1molK 2 O 
Now we can find the % yield.
Actual
50.0 g
 100 
 95.2%
Theoretica l
52.52 g
Both given numbers have 3 s.f.’s the answer is reported in 3 s.f.’s
43.6 gK    1molK