Lab 3

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A. Scientific Notation and Significant Figures.

Scientific notation is a convenient way to display very large and small

numbers.  A number in scientific notation will have a base number from 1 to
less than 10 that has a decimal just to the right of the first digit.  See the
examples below.  

Example #1:   3.68 x 103 

Example #2:   5.0 x 10-4

A number is put into scientific notation by moving the decimal to the right or the
left until there is only one non-zero digit to the left of the decimal.

The number of places you moved the decimal is the numerical value of the
exponent. If you do not see the decimal it is to the right of the rightmost digit.

If you moved the decimal to the left the exponent is positive.  If you moved the
decimal to the right the exponent is negative.

Write the following numbers in scientific notation:

Example #1:   3680  =  3.68 x 103 

Note:  this trailing zero is not considered significant because there is no decimal to
indicate that it is.  The trailing zero without a decimal is dropped when
reporting in scientific notation.
Example #2:   0.00050 = 5.0 x 10-4

Note: this trailing zero is significant because there is a decimal to indicate the
precision made in the measurement.  The leading zeros are only for
clarification (to show the position of the decimal point) and are not significant.

Write the following numbers in scientific notation:

          1.        987,000  __9.87 x 105_______________

2. 0.000356 3.56 x 10-4

3. 35.20     3.520 x 101

4. 0.0950 9.50 x 10-2

Write the following in standard numbers:

5. 4.3 x 104 43000

6. 3.690 x102 369.0

7. 1.99 x 10-5 0.0000199

8. 5 x 105 500000
B.  Significant Figures

Significant figures are used in chemistry to reflect the accuracy and precision
of the instrument used to make the measurement.  They allow us to maintain the
accuracy and precision of the data from the instrument as we use the data in
calculations.  

There are two important things to assess about each number we use in
chemistry.  First, the total number of sig figs.  Second, the number of digits to the
right of the decimal point.  Each of these is important for a different type of
calculations and will be discussed below.

How to determine which digits are sig figs in a number:

If a number has a decimal point, the only digits that are not significant are the
leading zeros.  They are only place holders to tell you how small the number is.
 How many sig figs are in the following numbers?

Example #1:   3.6008   has 5 sig figs 

Example #2:   0.0050 has 2 sig figs

Example #3:   0.05012 has 4 sig figs

Example #4:   1.000x103 has 4 sig figs

Example #5:   2.304x10-2 has 4 sig figs

Example #6:   1.00         has 3 sig figs

If a number does not have a decimal point every digit is significant unless it is
a trailing zero. 

Example #1:   36008   has 5 sig figs 

Example #2:   50 has 1 sig fig

Example #3:   5000 has 1 sig fig

Example #4:   5050 has at least 3 sig figs

Example #5:   100456 has 6 sig figs

Example #6:   15    has 2 sig figs

State the number of significant figures in each of the following measurements:

9. 4.5 m 2 13. 305.55 g ____5___

10. 0.0004 L 1 14. 20.00 g 4

11. 50 mL 1 15. 400,000 lbs 1_____

          12. 5.00x103  s___3___ 16. 3.045x10-2 km ______4

How to determine the number of digits to the right of the decimal:

Find the decimal in the number and count the digits to the right of the decimal
 How many digits are to the right of the decimal in the following numbers?

Example #1:   3.6008   has 4 digits to the right of the decimal 

Example #2:   0.005 has 3 digits to the right of the decimal 

Example #3:   1435.6 has 1 digit to the right of the decimal 

Example #4:   0.0400 has 4 digits to the right of the decimal 

Example #5:   345.0 has 1 digit to the right of the decimal 

Example #6:   1.00 has 2 digits to the right of the decimal 

For numbers in scientific notation it is necessary to take them out of scientific

notation to determine the number of digits to the right of the decimal

How many digits are to the right of the decimal in the following numbers?

Example #1:   1.000 x103 = 1000.

has 0 digits to the right of the decimal

Example #2:   1.234 x10-2 = 0.01234

has 5 digits to the right of the decimal 

Example #3:   1.000 x10-1 = 0.1000

has 4 digits to the right of the decimal 

Example #4:   1.58324 x104   = 15832.4

has 1 digit to the right of the decimal

Example #5:   6.49211 x103   = 6492.11

has 2 digits to the right of the decimal

Example #6:   1.7820 x103   = 1782.0

has 1 digit to the right of the decimal

State the number of digits to the right of the decimal in the following:

17. 4.5 m 1___ 21. 305.55 g 2

18. 0.0004 L 4 22. 20.00 g 2____

19. 50 mL___0 23. 400,000 lbs 0___

    20.   5.00x103  s 5000__0 sig fig 24. 3.045x10-2 km
__0.03045 5 sig fig
Calculations with sig figs:

There are two different sig fig rules for calculations.  One for
multiplication/division and one for addition/subtraction.

Multiplication/Division:  These calculations involve determining the total

number of sig figs in each number.  Answers to calculations involving
multiplication/division are limited to the number of sig figs in the number that
has the least number of total sig figs.

Example #1:  (2.97 x 102) x (6.0 x 10-7)

This can be entered into your calculator directly, be sure to ask your instructor
how to enter numbers in scientific notation into your calculation.

The answer your calculator gives is 1.782 x 10-4

Applying the rules for sig figs, there can only be 2 sig figs in the final answer
due to 2 sig figs in the 6.0 x 10-7.  Therefore, rounding up is required.  The
value reported is 1.8 x 10-4.

Addition/Subtraction:  These calculations involve determining the number of

digits to the right of the decimal in each number.  Answers to calculations
involving addition/subtraction are limited to the number of digits after the
decimal in the number that has the least number of digits after the decimal.

Example #2:      6.02 + 5.119 + 0.04218  

(Hint:  line up the numbers to be added so that the decimals are in line)
6.02
5.119
0.04218
       11.18118 Report:  11.18

You can see that when adding and subtracting it is the minimum number of
digits past the decimal point that determines the sig figs in the final answer. 
6.02 is the value with the least number of digits past the decimal point, so our
answer must be reported to 2 decimal places as 11.18.

Another way to think about this problem is as follows.

6.02???
5.119??
 0.04218
       11.18118 Report:  11.18

The top two numbers were taken with instruments that were not as accurate
as the one used for the bottom number.  Since we do not know the other digits
in the top two numbers we are not able to add them.  After all, what is ?+?+8. 
Since we can’t tell we cannot report those digits in the answer.

Do the following calculations after looking at the examples and express the
answer using correct scientific notation and significant figures:

25.  (4.4 x 10-2)  / (3.409 x 10-7) = 1.3e5

26.  (8.2 x10 -3) x (1.10 x10-2) =9.0e-5

27. (139.0 + 42.6 + 1.39) = 183.0

28.           (6.4 x 10-5)       =0.00011

(8.8 x 103) x (1.90 x10-4)

29. (6.00 x 1023) x 3.00  =7e21

      284
 30. (2.32 x 102) + (8.6 x 10 -5) = 232
Hint: Write out the numbers in standard notation, not scientific notation!
 

Units are an important part of calculations.  Be sure to always include them.

Example #3:

Example #4: 

Try these:  Watch units, sig figs and parentheses!

31. (38 feet) 2

  /   12.5 feet 

= 38 feet x 38 feet /12.5 feet = 115.52 feet

32.    100.0 m3 /   (10.0 m x 100.0 m)

=100 m x m x m / 10 m x 100 m
= 0.1 m

33.  42.31 cm x 32.5 cm x 12 cm =16500.9 cm3

 C. Unit Factor Method

The unit factor method for problem solving is one of the most important skills in your
math tool box.  It is important to be very detailed and careful in every step.  DO NOT
SKIP STEPS!  If you skip steps then you risk missing an important calculation.

Example:  How many minutes are in 2.75 weeks? 

Step 1:  Derive the conversion factors between weeks and minutes.

Answer:    week ↔   days ↔  hours  ↔  minutes  

7 days  = 1 week 
 
If you divide both sides of this equation by 1 week you get a conversion factor that will  = 1:

7 days1 week=1 week1 week=1

You can also divide both sides by 7 days to get the other conversion factor:
1=7 days7 days=1 week7 days

Therefore, every equality has 2 conversion factors. You will use the factor that allows you to
cancel out the units to leave only the unit you desire.

Step 2:  Use the conversion factors to create one math problem. Doing all the
math at once (rather than a series of consecutive conversions) reduces the
chances you will make a mistake by entering numbers incorrectly into your
calculator. 

data x  factor   x  factor  x  factor    =    answer (include units)

2.75 weeks17 days1 week24 hours1 day60 min1 hour= ?min

Make certain each of the units cancel out.  For instance, in this case weeks in the
numerator cancel with weeks in the denominator, days in the numerator
cancel with days in the denominator and hours in the numerator cancel with
hours in the denominator.  If the units do not cancel out in this fashion try to
invert the factor!

Answer    =    27720  minutes   

However, we must determine the correct number of sig figs for the answer.  This was a
problem involving only multiplication and division so we would use total
number of sig figs.  Even though a few of the factors seem to have 1 sig fig
leading one to believe that the answer should have 1 sig figs this is not the
 case.  Factors that are exact such as 7 days in one week do not influence the
sig figs.  (There are exactly 7 days in 1 week.) Since all the factors are exact, we
use the original number to determine sig figs.  Since it had 3 sig figs and only
multiplication and division were used so the answer should have 3 sig figs. 

Answer    =    2.77 x 104   minutes   

Every answer must have the correct significant figures and be in scientific
notation. Remember that conversion factors are assumed to have an
infinite number of significant figures. 

NOTE: Factors are exact when derived through counting, when converting from
metric units to metric units or when converting from English units to English
units.  Those derived from converting from metric units to English units or vice
versa are not exact and do influence the sig figs.  The only exception to this
rule is the factor 1 in = 2.54 cm.  This is considered exact.

Try these:  

34.  Tall cans of juice contain 46.0 fl oz.  How many mL is this? 

Derive both factors from this equality:  (1mL = 0.0338 fl oz)

 =46.0x 29.574 =1360.38 mL

35.  If 1 inch = 2.54 centimeters, how many cm are in 15 in?

=15 x 2.54 =38.1 cm

36.  Knowing that 1 g = 0.0353 oz. and 16 ounces = 1 lb (this relationship is

exact).  Calculate the number of grams in 10.0 pounds. 

10 Ib =10 x (16 /1 ) x ( 1/0.035) =4571.43 g

 37. If one mile is 5280 feet, how many inches are in 3.94 km? (Remember
1in=2.54 cm is an exact conversion.)

1 km=39370.0 in
=3.94 x 39370 = 155118.1 in

38.  Suppose the speedometer in your car reads 55.0 mph (also written as
mile/hour, be sure to write this as mile in the numerator and hour in the
denominator).   milehour What is your speed in m/sec?  (1 km = 0.621 mi.)

=55 x (1000/(0.621 x 3600))= 24.6 m/s

39.  Carl Lewis, a sprinter in the 1988 Olympic Games, ran the 100.0-m dash
in 9.92 s.   What was his speed in feet per second  (ft/s) ?  

1 m = 3.28 foot

100 m = 328 foot

Speed = 328/9.92 = 33.07 ft/s

40. Knowing that 1 in = 2.54 cm, calculate how many in2 are equal to 521 cm2. 
(Remember 1in=2.54 cm is an exact conversion.)

=521 x (1/2.54) x (1/2.54) = 80.75 in2

 D.  Metric Conversions

Use the unit factor method for metric conversions. Answers must be in
scientific notation with the correct number of significant figures. Convert to the
base unit (grams, liters, or meters) as the first step in all metric conversions. Then
convert from the base unit to the desired final unit. 

example: Convert 325 mg to kg.

41.  Convert: 0.471 mg to g

=0.471 mg x (1 g/103 mg) = 4.71 x 10-4 g

42. Convert: 2030.0 cm to km. 

=2030.0 x (1/100000 km)=0.02030 km

43. Convert: 210.0 ng to mg. 

=210.0 ng x (1/1x106 )=0.00021 mg

 44. Convert 502.0 mL to cL. 

=502.0 ml x (1/10cL)=50.2 cL

E.  Density Problems

Density is a physical property. Every pure substance has a unique density. The
formal definition of density is mass per unit volume. Usually the density is
expressed in grams per mL. Remember, “per” means division. When you use a
density as a conversion factor, the mass will be on one side of the fraction and
the volume will be on the other.  For example, the density of gold is 19.3 grams
per mL. In fractional form this is:

Density of gold=19.3 g/ml=19.3 g1 mL

Example #1:  A sample of zinc weighs 253.9 g has a volume of 35.60 cm3. 
What is the density of zinc?

           

Answer:

When measuring the densities of solid substances, the volume unit of cm 3 is often
used. 1 cm3 = 1 mL. 

Example #2:  A cube of zinc with a mass of 332 g would have a volume of how
many cm3? The Density of Zinc is 7.132 g/cm3.

Using the unit factor method with density as the factor:

7.132 g/cm3 means   

          
The inverse is also true:

    

45. A cube of metal is 1.42 cm on an edge. It has a mass of 16.3 g. 

Calculate    the density of this metal cube. (Remember the volume of a cube is
determined by cubing on edge length. Ex: (3.00 cm)3 = 27.0 cm3 

density = 16.3 / (1.42)3

=5.69 g/cm3

46. The density of ethanol is 0.7902 g/mL.  Calculate the mass of 15.0 mL of
ethanol.  

mass = density x volume

= 0.7902 x 15

=11.85 g

47. What is the volume occupied by 27g of mercury? (The density of mercury is
13.6 g/mL) 

volume = mass / density

=27/13.6 = 1.99 mL
 F.  Temperature Conversions

°C = (5/9)(°F – 32)
°F = 1.8(°C) + 32
 K = °C + 273.15

The answers don’t need to be in scientific notation.

48. Convert: 25°C to Kelvin.

25°C + 273.15 = 298.15K

49. Convert: 39°C to Fahrenheit.

(39°C × 9/5) + 32 = 102.2°F

50. Convert: 76°F to Celsius.

(76°F − 32) × 5/9 = 24.444°C

51. Convert: 325 K to Celsius. 

325K − 273.15 = 51.85°C