Download review notes

Scientific notation, Significant Figures, and Problem Solving
I. Scientific or Exponential Notation
Results of chemical problems often involve very large or very small numbers. The best way to represent
these numbers is NOT to write them all the way out but to use EXPONENTIAL NOTATION.
ex. 432,000 = 4.32  105
.000079 = 7.9  10-5
To convert a large number to exponential notation, move the decimal point to the left until only one digit
remains to the left of the point.
For example, 123,456,789 = 1.23456789  100,000,000 or 1.23456789  108
In exponential notation we moved the decimal point 8 places left: 1.23456789 and multiplied by 108.
To convert a small number to exponential notation, move the decimal to the right until one non-zero digit
is to the left of the decimal point and multiply by 10 raised to a negative power. If you have moved the
decimal m spaces to the right you need to multiply by 10–m. For example:
0.000052 = 5.2  0.00001 or 5.2  10-5
What are the advantages? In addition to being easier to read (you don’t have to count a lot of zeros),
arithmetic using exponents is easy.
Multiplication is easier: To take the product of two numbers, multiply the decimal portion in the usual way
and add the exponents for the tens part.
(30,000)*(15,000) = (3104)(1.5104) = 4.510(4+4) = 4.5108
(320,000)*(.00003) = (3.2105)(310-5) = 9.610(5+-5) = 9.6100 = 9.6
Division is easier: To divide two numbers, divide the pre-exponential part (the decimal) and subtract the
exponents.
30,000/.002 = 3.0104/210-3 = 1.5x10(4 - -3) = 1.510+7
Remember that dividing two numbers is equivalent to multiplying the first number by one over the second
number: a/b = (a)(1/b) and that changing the sign on an exponent is like putting it in the denominator
(or inverting it).
10-5 = 1/105
108 = 1/10-8
Addition and Subtraction: To add and subtract numbers you need to make the exponential parts the same.
In other words you have to convert the numbers so that they are raised to the same power of ten.
1.45106 + 3.4105 = 14.5105 + 3.4105 = 17.9x105 = 1.79106
II. SI Units (Système Internationale)
The seven standard base units are: length — meter (m), mass — kilogram (kg), time — second (s),
temperature — Kelvin (K), number (quantity) — mole (mol), electric current — ampere (A), and
luminosity — candela (cd). All other units are derived from these.
Metric prefixes. Because SI units are metric and based on decimals, larger and smaller units are created easily using
metric prefixes to denote various powers of 10. KNOW THESE!
Prefix
Megakilodecicentimillimicronanopicofemto
symbol
M
k
d
c
m

n
p
f
Meaning
million 106
thousand 103
1/10
1/100
1/1000
10-6
10-9
10-12
10-15
example
one million Hz = 1 MHz (radio frequency)
1000m = 1 km
1 dL (deciliter) = 0.1 L (used in cooking more than chemistry)
1 cm = .01m 100cm = 1 m
1mm = .1 cm = .001 m 1000 mL = 1
1s = 10-6 s 1000 L = 1 mL
1 ng = 10-9 g
1 ps = 10-12s
1 fs = 10-15s
Derived units: Many measurements use units derived from the base units. Let’s take volume as an example. The
volume of a box is its lengthwidthheight. If we have a cube measuring a meter long on each edge, the volume
would be V = (1m)3 = 1 m3. Thus the SI unit for volume is m3. However, most volume measurements in chemistry
are far smaller than that so we will commonly use a unit of a centimeter cubed, cm3, also defined as the milliliter mL,
or the larger unit of a L.
1 cc = 1 mL = 1 cm3 = 10-3 L = 10-6 m3.
also 1 L = 1000 mL and 1000 L = 1 m3.
Temperature: The standard temperature unit is the Kelvin, which has the same degree interval as the Celsius scale,
but the zero on the Kelvin scale is –273.15 C on the Celsius scale. 0 K is the coldest temperature possible and
therefore is called absolute zero.
to convert Celsius to Kelvin use: TK = TC + 273.15
III. Significant Figures
In any measurement you should report as many digits as you are fully confident of plus one more in which there
is some uncertainty. In a length measurement the last digit would be the estimation between ruler lines. The
number of digits reported in a measurement are the number of significant figures, not including placeholder zeros.
Counting significant figures.
1. All digits written in exponential notation are significant. 1.23450 x 107 has 6 significant figures.
2. In regular notation all digits are significant except:
a. All zeros at the beginning of the number. 0.0003 has one sig.fig.
b. Terminal zeros without an explicit decimal place are not significant. Ex. 30,000.00 has 7 sig.figs, but
30,000 has 1 sig. fig. If written in exponential notation the number of zeros given would show us how many
sig.figs were intended: 3104, 3.0104, 3.00104, etc.
In other words, zeros that are merely placeholders are insignificant.
3. Exact numbers are defined to be infinitely precise, and therefore the concept of sig.fig doesn’t apply. (Consider
them to have an infinite number of sig.figs.)
a. Exact numbers include integers – a countable item ex. There are 10 plastic balls in my test tube. 10 is exact
(you know its not possible to have 9.9 or 10.1 balls)
b. Some defined relationships are exact. 1 inch = 2.54 cm (exactly)
c. Numbers like  -use as many digits as you need.
Arithmetic with Sig. Figs.
1. Multiplication and division: keep the lowest number of significant figures.
Ex. (1602.0 mL)(2.1 g/mL) = 3.4103 g
5 s.f .  2.s.f.  2 s.f.
2. Addition and subtraction: Keep only the fewest decimal places.
Ex. 42.0172 mL + 1.2 mL = 43.2 mL
4 dec. pl.
+ 1 dec. pl. 1 dec. pl.
3. Because exact numbers have as many sig.figs. as you like, they should never change the number of
sig.figs reported in a calculation.
Ex. Volume occupied by 10 balls = (0.952 mL/ball)(10 balls) =9.52 mL
IV. Problem solving, Dimensional Analysis or the Factor Label Method.
When you do a calculation always KEEP THE UNITS WITH THE NUMBERS throughout the
calculation because:
1. The units for the answer will come out of the calculations.
2. Mistakes in using a formula will become apparent.
3. Sometimes it can help you figure out the formula to use.
Calculating from an equation.
In most science problems you are given some information and asked to calculate something from it.
Sometimes you will know what equation you need and you can solve for the unknown variable in terms of
the variables you do know. Write down the equation and rearrange it to solve for what is wanted. Write
down all the values of all the known variables and include the units. Then solve the equation. Remember:
When you substitute a number for a variable in an equation, PUT IN THE UNITS AT THE SAME
TIME.
Example: If you drive 90 kilometers per hour, how long will it take to travel 150 km?
The equation we can use is d=rt where d=distance, r=rate (speed), and t=time.
The value of the variables: d = 150 km,
r = 90 km/hr t = ?
Rearrange the equation to solve for t.
t = d/r = (150 km)
(90 km/hr)
= 1.7 (km)
(km/hr)
= 1.7 hr
We got hours for units because the kilometers cancelled and 1/1/hr. is the same thing as hr. So you
see we do the math on both the value of the numbers and on their units.
Unit Conversions
Often you will be given information in one unit and want to express the answer to a calculation in
another unit. Converting is easy if we know the relationship between the two units and many of these are
found in the inside our book.
Example. In the previous problem we calculated that we traveled 150 km. How many miles is that?
1. Write down the equality relation between the two units: 1 mi = 1.6093 km
2. Make a conversion factor.
3. Since you can multiply anything by 1 without changing it, you can multiply 150 km by the conversion
factor to get the equivalent distance in miles.
(km units cancel)
Using Dimensional Analysis to solve problems without an equation.
Sometimes you don’t have an explicit equation to use to solve a problem. What you do have is a series
of relations between different variables. You can very often use these in a way that is exactly the same as
unit conversions above. Lets look at an example.
In the country of Ferdovia the Ferds thrive on potatoes. The average Ferd earns $142 per week and
spends 1/14 of his income on potatoes. If potatoes cost $2 .00 per pound, how many pounds of
potatoes does the average Ferd consume each year?
1. What do we need to calculate? Pounds of potatoes per year.
2. Write down the equivalency relations that the problem gave you.
For example, if the average weekly wage is $142 then $142 income = 1 week
Also, $2.00 = 1 lb potatoes and
$14 income = $1 of potatoes.
3. Make conversion factors from these equivalencies and multiply them together to give you the quantity
(and units) that you want.
lbs of potatoes per year =
In this example we had to add the conversion factor of 52 weeks in a year to cancel out weeks and end up
with “per year”.
4. Carry out the multiplication and be sure to cancel out units appropriately.
= 264 lb. potatoes/year.
$ potatoes cancel, $ income cancels, and weeks cancel leaving lb. potatoes/year.