Download - Chapter 3. Units and Calculations
Chapter 3.Units and Calculations
All measurements have three parts:
1. Number (value, quantity)
2. Uncertainty (error, shown by sig figs)
3. Unit (nature of quantity, label)
Units must always be shown with numbers!
The Metric System
The metric system is a decimal system of weights and measures based on the meter as a unit of length, the kilogram as a unit of mass, the second as a unit of time, and the kelvin as a unit of temperature.
Decimal: Unit conversions are factors of 10.
The Metric SystemBasic (fundamental, defined) Units:
Length; meter (m) (about 1 yard)in lab, centimeter and millimeter
Mass; kilogram (kg) (about 2.2 pounds)in lab, gram (g)
Time; second (s) (same as English)minutes and hours, not decimal
Temperature; kelvin (K) (no negative values)also Celcius, centigrade (C)
The Metric System
The Metric System
The fundamental units in the metric system are too large to be convenient in chemical labs. How do we get smaller units?
Some of the derived units in the metric system are very small. How do we get larger units?
The Metric System
We multiply the unit by some power of ten, for example 103 (1000) or 10-2 (0.01).
These multipliers relate to prefixes. The prefixes are combined with names of fundamental units to obtain larger or smaller units:
kilogram = 1000 gramscentimeter = 0.01 meter
The Metric SystemMetric system prefixes (multipliers) to know:
My king died chewing M & M's
Prefix Symbol Meaning Value Exp.mega m million 1,000,000 106
kilo k thousand 1,000 103
deci d tenth 0.1 101
centi c hundredth 0.01 102
milli m thousandth 0.001 103
micro millionth 0.000001 106
The Metric System
How to use prefixes and multipliers:
Name of unit Value of unitprefix unit multiplier x unitmilligram 0.001 x 1 gram
one thousandth of a gram
The Metric SystemExamples:
One centimeter = 1 cm = 0.01 meter
One kilogram = 1 kg = 1000 gram
One millisecond = 1 ms = 0.001 second
One megahertz = 1 MHz = 1,000,000 Hz
One microfarad = 1F = 0.000001 F
The Metric SystemConversions within the metric system, e.g. convert
75833 meters to kilometers
1. Set up equality: prefix unit = multiplier x unit 1 kilometer = 1000 meters
2. Convert to ratio with desired unit in numerator: 1 km
1000 m3. Multiply ratio by given units:
1 km x 75833 m = 75.833 km 1000 m
The Metric System
Convert:
0.0285 kilograms to grams (kg to g)
27935 meters to kilometers (m to km)
53.8 milliseconds to seconds (ms to s)
0.084 meters to millimeters (m to mm)
The Metric System
Convert:
0.000850 meters to micrometers (m to m)
250 micrograms to milligrams (g to mg)
The Metric SystemDerived units are obtained by mathematical
operations on one or more basic units.
Area = length squared 1 square meter = 1 m2
Volume = length cubed (space occupied)1 cubic meter = 1 m3
The basic unit of volume in chemistry isthe liter (L). 1 L = 1 dm3 = 1000 cm3
The Metric System
The Metric System
Other derived units:
Speed = distance/time, m/s
Acceleration = distance/time2, m/s2
Force = mass x acceleration, kgm/s2 newton, N
Pressure = force/area, kg/ms2 pascal, P
Energy = force x distance, kgm2/s2 joule, J
Units in MathUnits can be multiplied, divided, squared,
canceled, etc. -- just like numbers!
102 x 10 = 103 m2 x m = m3
Pressure = force/area = kgm x 1 = kg sec2 m2 msec2
Energy = force x distance = kgm x m = kgm2 sec2 sec2
Units in Math
Conversion factors are ratios that specify how one unit of measurement is related to another unit of measurement. They can also be expressed as equalities.
2.54 cm = 1.00 inch (exact)
1.00 in 2.54 cm2.54 cm 1.00 in
Units in Math
Example:
How many centimeters are there in 12.0 inches?
2.54 cm x 12.0 in = 30.48 cm = 30.5 cm 1.00 in
Units in MathDimensional Analysis is a method for setting up
calculations in which the units associated with numbers are used as a guide.
Set up the calculation so that desired units remain in the answer, and all others cancel.
Dimensions are quantitative properties such as length, time, mass.
Units are defined measurements of dimensions, such as meters, seconds, and grams.
Dimensional Analysis
How to do it:1a. Figure out what quantity is to be deter-
mined, and what are the desired units.
1b. Identify given quantities in the problem.
Dimensional Analysis
2a. Choose a given quantity or a conversion factor that has the desired units.
2b. Start an equation with this quantity. If it’s a ratio, the desired units should be in the numerator.
2c. Multiply this quantity by other given val-ues and conversion factors to make un-wanted units cancel and retain desired units.
Dimensional Analysis
3a. Perform mathematical operations as indicated in the equation you created.
3b. Reality check: Does the result make sense?
3c. Clean up: Round to correct number of sig figs.
Dimensional Analysis
Example:
A premature infant weighs 1703 grams. What is its weight in pounds?
454 g = 1.00 lb (inexact)
Dimensional Analysis
Example:
At room temperature, 1.00 L of water has a mass of 1.00 kilograms. What is its mass in grams?
Dimensional Analysis
Example:
I can ride my bicycle at 9.6 miles per hour. How long will it take me to go 23 miles?
Dimensional AnalysisTypes of conversion factors:
Equality conversion factors are ratios that interconvert different units of the same dimension.
0.454 kg = 454 g = 1.00 lb
1.00 lb 0.454 kg 454 g 1 lb
Dimensional AnalysisTypes of conversion factors:
Equivalence conversion factors are ratios that interconvert units of differ-ent dimensions.
Speed = distance miles time hour
Density = mass grams volume cm3
Dimensional AnalysisExample:
An investigator found that 50.3 cm3 of bovine fat had a mass of 45.1 gram. What is the density of the fat?
The investigator also found that 49.8 cm3 of bovine lean muscle had a mass of 55.0 g. What is the density of the muscle?
Which is more dense?
Percentage Problems
Percent is the number of items of a specified type in a group of 100 total items.
Parts per hundred
Percent = number of items of interest x 100% total items
Percentage Problems
A student answered 19 items correctly on a 23 point test. What was his score as a percentage?
Percentage
23
19
0 5 10 15 20 25
1
Points on a test
Percentage ProblemsRange as a percent of the average is a way
to express precision. % of average = (highest – lowest) x 100%
average = (20.50 – 19.25) units x 100 % =
6.32% 19.78 units
Measurements and the Average
19.7819.2519.60
20.50
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
1Run #
Measurement Units
Percentage Problems
A technician measured the breaking strength of three samples of plastic. His results were:
Run 1: 65.8 MPaRun 2: 72.4 MPaRun 3: 68.3 MPa
What was the range of his measurements as a percent of the average?
Note: 1 MPa = 145 pounds/in2
Percentage Problems
Percent difference is a way to express accuracy.
% difference = (measured – actual) x 100% actual
= (19.78 – 20.00) units x 100% = –1.1% 20.00 units
Measured and True values
19.7820.00
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Measurement Units
Percentage Problems
A student determined the density of aluminum metal to be 2.64 g/cm3. The accepted value is 2.70 g/cm3. What is the percent differ-ence between her result and the accepted value?
Did she do a good job?
Percentage Problems
A student did three experiments to determine the density of rubbing alcohol. Her results were: 0.778 g/mL; 0.795 g/mL; 0.789 g/mL. What is her precision as % of average?
The true value is 0.785 g/mL. What is her accuracy?