Math Rev. Dimensional Analysis

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Math Rev. Dimensional Analysis Powered By Docstoc
					Math Review: Study Guide #2


I.) Standard Units of Measurement:
      Our system of measurement finds its roots in the English System which
came to America with the founding of the first British Colony. Most of the world
along with Britain has since switched to the metric system.
              English System of Weights and Measures
                    Mass                  pound
                    Volume                fluid ounce, quart and gallon
                    Length                Inch, foot and mile
                    Temperature Fahrenheit
                    Energy                calorie
       Starting in 1960, the international Bureau of Weights and Measures, in a
series of international meetings proposed a standard set of units that they
believed all scientists should use. This Systeme International d^Unites,
abbreviated as the SI system, has been largely adopted throughout the scientific

                   Table 1:     SI Units of Physics and Chemistry.
      Quantity             Unit
      Length               meter (m)
      Mass                 kilogram (kg)
                           gram (g)     Typical unit of mass (1000 g = 1 kg)
                                        (Non-SI unit)
      Temperature          Kelvin (K) Defined by absolute zero and the
                               increment of °C.
                           Celsius (°C) Typical unit of temp. (Non-SI unit)
      Time                 second (s)   Minutes, hours and days also used.
      Amount               Mole (mol) # of particles present, will be              of
       Substance                  introduced in chapter 8.
      Volume               m3             Typical unit is the cubic centimeter (cm3)
                                 or milliliter (mL)
                                         106 cm3 = 106 mL =1m3
                          L              Common unit of solution volume is the liter (L).
                         mL              1000 milliliters (mL) = 1 L = 1000 cm3
Study Guide 2                        Math Review

        The SI system is closely related to the metric system, long used in Europe.
However, the SI system represents a further simplification by expressing all
scientific measurements in terms of just seven base units. Despite these good
intentions some units are in such common usage that they are retained. As you
can see on Table 1 the base unit for time is in seconds. But, time in minutes,
hours and days is still accepted. These are commonly used non-SI units. The
standard unit of mass is the kilogram (kg) (1 kg = 1000 g), however, most small
weights are expressed in the non-SI unit of grams (g) as a matter of convenience.
The same can be said for units of volume. The SI unit is meters cubed. The unit
of Liters (L) is most often substituted for m3 (1 m3= 1000 L).

      The SI system permits specification of large units or small units by
choosing the appropriate prefix.

      Example 1:       0.010m (meter) = 1.0 cm (centimeter) = 10. mm (millimeter)
                       10,000 m = 10 km (kilometer) = 1,000,000 cm

                                      SI Prefixes:
   Prefix       Symbol       Power of 10    Prefix        Symbol      Power of 10
    tera          T             10 12        milli          m            10 -3
    giga          G             10 9        micro           µ            10 -6
   mega           M             10 6         nano           n            10 -9
    kilo          k             10 3         pico           p           10 -12
   deci            d             10 -1          fempto       f           10 -15
   centi           c             10 -2            atto       a           10 -18

       The standard unit of temperature is Kelvin (K). However, most temperatures are
collected in the laboratory using the metric unit of degrees Celsius (°C). Fortunately
conversion between degrees Celsius and Kelvin is an easy endeavor.

             K = °C + 273      and       °C = K - 273

      Example 2: Convert 212°C to Kelvin.

                       K = °C + 273 = 212 + 273 = 485 K

       On rare occasions we may want to convert between Celsius and Fahrenheit or
vice versa. The equations below can be used for that purpose.
                   5                       9
              °C = 9 (#°F -32) and °F = 5 #°C + 32
       Units of density, force, velocity, energy and pressure are examples of
algebraic combinations of the S.I. base units. We will visit these individually
during the semester, but density is a quantity we will calculate or apply through
most of this course. So perhaps it will be helpful to define it now and get it out
of the way. Density has units of mass per volume. According to the SI format the

Study Guide 2                       Math Review
units would be kg/m3. However, in practical use the unit is typically g/cm3. The
quantity of density is a physical property of matter and is often used to describe
substances. A given volume of lead for instance weighs more than the same
volume of aluminum due to the difference in density of the two metals.

      Example 3: What is the density of 35.3 mL of a solution that has a mass of
                 38.2 g?
                               mass     38.2 g
                    density = volume = 35.3 mL = 1.08 g/mL

                             1 mL = 1 cm3 by definition
                    density = volume = 1.08 g/mL = 1.08 g/cm3

      Example 4:    What is the density of a metal cube that measures    1.1 cm
                    on each side and weighs 12.31 g.
                               mass               g               g
                    density = volume = length x width x height =

                                      12.31 g              g
                    density =                       = 9.2
                                1.1 x 1.1 x 1.1 cm3       cm3

       The equation for density involves three variables; density, volume and
mass. If any two variables are known the one which is unknown can be

      Example 5: What is the mass of 45.32 mL of a salt water solution which
                  has a density of 1.12 g/cm3?
                  density =                 for    mass = density x volume
                                            45.32 mL = 45.32 cm3
                                    1.12 g
                           mass =          x 45.32 cm3 = 50.8 g

Study Guide 2                      Math Review

      Example 6:     The density of a particular gas sample is 0.0893 g/L. What
                     is the volume of 200. g of the gas?

                               mass                                  mass
                    density =                   for        volume = density

                                                                200. g
                                                    volume = 0.0893 g/L = 2240 L

II.) Dimensional Analysis:
      Dimensional analysis is a large term for a type of mathematics that you have
probably done at some level since you were a child. For example, you have four
friends, two dollars and want to buy candy bars for everyone at 45 cents each. You
intuitively do the math in your head by dividing 50 cents into two dollars and
know that you have enough. Dimensional analysis is a more formal way of doing
the same thing while providing an internal proof regarding the correctness of
your equation.
      Lets look at this simple example to get a solid idea of how dimensional
analysis is applied. We'll find the number of candy bars that can be purchased
with 2 dollars by dimensional analysis. The first step is to pull the pertinent
information out of the text and organize it as below.

      a.) One candy bar requires 45 cents to purchase.
      b.) There are 100 cents in one dollar.
      c.) You have 2 dollars to purchase the candy.

     The ultimate goal of our exercise is to convert dollars to candy bars both
figuratively through calculation and in reality through purchase.

      Statements a and b are true relationships. For every 45 cents we can
purchase one candy bar. This is called a conversion factor. It expresses a true
relationship between two different units (candy bars and cents). At the time of
purchase these two quantities are equivalent.

                                         one candy bar = 45 cents
The equality can be expressed as
                                               45 cents
                                         1 = one candy bar

if we divide both sides by one candy bar.
Or if we divide both sides by 45 cents.
                                              one candy bar
                                         1=     45 cents

Study Guide 2                       Math Review

       Notice that both relations equal one. Therefore, to multiply any term by
these is to multiply it by one. This type of relationship is called a conversion
factor. It is used to convert units from those of the conversion factor's
denominator to that of the numerator.

                            1 candy bar 
       Example 7: 200 cents              = 4 candy bars
                               45cents  

       Conversion factors permit conversion from a given number (200) and
units (cents) to another unit (candy bars) based upon a true relationship(s) (the
conversion factor(s).

       The statement in b.) is a definition indicating                        one dollar
= 100 cents.

                               100 cents            1 dollar
It can also be stated as:   1 = 1 dollar =          100 cents .

       Similar to the above, multiplying any term by either of these is to multiply
it by one. Definitions are also true relationships and can be used for unit
                                            100 cents 
              Example 8a:       2 dollars              = 200 cents
                                             1dollar 

       Notice in both examples how the units cancel to permit you to obtain the
desired units. The chief utility of this technique is that if a problem is correctly
set up, the equation will obtain the correct units.

                                          1dollar                2
              Example 8b:      2 dollars            = 0.02 dollar /cents
                                         100 cents 

       Oops, the units don't work out and the error is apparent. So to obtain the
correct equation for conversion of 2 dollars to candy bars we assemble the
equation so the units correctly cancel.

       The following is a map of the conversions we'll use.

           dollars --->cents ---> candy bars

The equation then is:

                                100 cents   onecandybar 
                      2 dollars                          = 4 candy bars
                                 1dollar   45 cents 

Study Guide 2                       Math Review
        Well you are now saying to yourself why bother with all this hassle. From
the above you can see that you obtained the correct answer based on the fact that
all, the units correctly cancel. This probably is not enough for you. So lets prove
the point by doing a problem that is just a little more challenging.

        Could a stack of 1 dollar bills representing the national debt of   5
trillion dollars reach the moon. Assume each 20 dollar is 0.3 mm thick and the
average distance to the moon is 250, 000 miles.

1.) Lets write down what we have.
        a.) One bill = 1 dollar.
        b.) One dollar bill is 0.3 mm thick.
        c.) One stack represents 5 x 1012 dollars (5 trillion dollars).
        d.) The distance to the moon is approximately 250,000 miles.

2.) What we want to determine: Is the stack of dollar bills higher
   than 250,000 miles?

3.) Since the thickness of an individual bill is in mm we will need to
   convert at some point from millimeters (1 mm = 0.001 meters) to
   miles. Some additional definitions of length will be required.

       a.) 1 meter (m) = 1000 millimeters (mm)
       b.) 1000 meters (m) = 1 kilometer (km)
       c.) 1 kilometer (km) = 0.621 miles (mi)

4.) One possible solution map for this problem is:
5.0 x 1012 dollars ---> bills ---> mm ---> m       ---> km ---> mi

                    1bill   0.3mm   1m   1km   0.621 mi 
5.0 x 1012 dollars                                         = 931,500 mi
                   1dollars   1bill  1000 mm  1000 m   km 

       Inverting any of these conversion factors would lead to an error in the final
calculation. In the absence of some expert to check your work the error might go
unnoticed. Using dimensional analysis permits calculation at an advanced level by mere
amateurs. Like a cross word puzzle you can continue to peck away at the problem until
the correct result is obtained and confirmed. You can prove your result and easily
communicate how and why you obtained it. This is the style of mathematics that will
be applied to all calculations during this course. Resist the temptation to use
alternative methods. The wrong answers on the multiple choice exams are often the
answers obtained by inverting conversion factors.

Study Guide 2                      Math Review

III.) Applied Unit Conversions:
       Mathematical operations in chemistry involve unit conversions. For
instance, a volume of water can be expressed in fluid ounces, cups, quarts,
gallons, milliliters or liters or in units of weight such as ounces, pounds, tons,
grams, kilograms to name but a few. Inevitably this leads to the necessity of
converting from one unit to another. Dimensional analysis provides the
mathematical where-with-all for the unit conversion and a proof that we set up
the mathematics correctly. Before using this technique it is important to
remember that units are treated mathematically in the same way we treat
       Example 9: When like units appear in the numerator and denominator of
                  a fraction they cancel.
          19 m        19        1m     19
       3.5 kg x m = 3.5 kg    x 1m = 3.5 kg       x 1 = 5.4 kg -1

      Units may be treated like numbers. They can be raised to a power,
multiplied or divided.
      Fundamental to unit conversions is the use of conversion factors. As
mentioned earlier these are obtained from any true relationship that can be
expressed as an equality.

       Example 10: 1 cup contains 8 fluid ounces: which implies
                                                                 1 cup
                                  1 cup = 8 fl. ounces or                 = 1.
                                                              8 fl.ounces
                          We could just as easily have divided both sides by         1 cup
                          and obtained
                                       8 fl. ounces
                                           1 cup       = 1.

       As we observed before, multiplying a number by a conversion factor is
essentially the same as multiplying by 1. To convert from cups to fluid ounces
we need a conversion factor that contains both the units we want to eliminate
and the desired units.
        Example 11: Convert 33.65 cups to fluid ounces.
                                         By correctly setting up the conversion
           8 fl. oz.                   factor the units of cups cancel, leaving
33.65 cups            269.2 fl. oz.   the desired units of fluid ounces.
           1 cup 
If the conversion factor had been set up incorrectly the result would have had
units of cups2/ounces and our error would be apparent. (see below)

Study Guide 2                      Math Review
        Example 12: Convert 33.65 cups to fluid ounces.
                                                The error is readily apparent
            1 cup                 2           when the conversion factor is
33.65 cups             4.206 cups            inverted.
            8 fl. oz.               fl. oz.

       On many occasions a single conversion factor for the desired operation is
not available, so we must choose a series of conversions that will take us
ultimately to our goal. A solution map is helpful in that it allows us to try out
ideas that might lead to a solution without writing out the entire solution and
checking units.

       Example 13: The distance from Fresno to San Francisco is
                   180 miles what is that distance in meters?

       A proposed solution map: 180 mile --->ft --->in --->m (desired)

The solution map always starts with the unit of measurement that is given in the
problem and ends in the units desired. The individual steps are conversion
factors or definitions.

      Units of ft result when a conversion factor is selected that has the
undesired unit(miles) in the denominator and the desired unit of feet (ft) in the
numerator. The concept is continued for conversion to inches (in) and then
meters (m), which was the desired unit.

The result based upon the solution map is:

                              5280 ft  12 in   1 m              5
                    180 mi                            = 2.9 x 10 m
                              1 mi   1 ft   39.37 in 
       Conversions involving area or volume have units such as m2, m3 or cm3,
that require some special handling. SINCE THE UNITS HAVE A POWER THE

Study Guide 2                     Math Review

      Example 14: How many cubic meters are present in 123,500 cm3?

                    The definition necessary for relating meters to centimeters is
             1 m = 100 cm. However the conversion won't work unless the
             conversion factor is also cubed as below.

             Solution map: cm3 ---> m3

                           1 m  1 m  1 m 
              123,500 cm3 100 cm100 cm100 cm =
                                            
                           1m 
              123,500 cm3 100 cm =
                                
                       1m  3                1 m3 
         123,500 cm3  3 3 = 123500 cm3  6 3 = 0.1235 m3
                      100 cm               10 cm 


Math Review: Study Guide #2

Problem set 1: Do the temperature conversions indicated.

a.) Convert 33.5 °C to Kelvin.

b.) Convert 45.8 K to degrees Celsius.

c.) Convert 75°F to degrees Celsius.

d.) Convert 295 K to degrees Fahrenheit.

Problem set 2: Do the density calculations as follows.

a.) What is the density of an unknown metal if 45.36 g of metal occupies a
    volume of 3.62 cm3?

b.) A 55.3 mL sample of a salt solution has a mass of 56.213 g. What is the
    density of the solution?

Study Guide 2                      Math Review

c.) What is the mass of 35.2 mL of a solution that has a density of 1.12 g/cm3?

d.) A rock sample has a density of 10.3 g/cm3 and weighs 243.12 g. What is the
     volume of the rock?

e.) A particular gas sample has a density of 1.12 g/cm3. What is the mass in kg of
     4000 L of gas?

Problem set 3: Use your conversion factor handout to make the necessary unit
conversions and report the answer with correct sig. figs and units. Use the form
presented in the study guide and show all work.

a.) 746 miles --> ft

b.) 15.3 yr. --> day

c.) 8.31 x 105 in --> ft

d.) 9.6 x 104 yd --> m

e.) 4.671 x 106 L --> Gal

f.) 2.37 x 104 g --> lb

g.) 895 miles --> km

Study Guide 2               Math Review
h.) 15.3 yr --> hr

i.) 0.4837 km --> mm

k.) 56 cm2 -->m2

l.) 94.72 km to cm

m.) 7.1 x 103 L --> fl oz

n.) 5.76 kg --> oz

o.) 8.7 Gal --> mL

p.) 184.3 in2 --> m2

q.) 7.71 x 104 µm --> km

r.) 2.6 tons -->kg

s.) 5.22 qt --> L


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