Appendix D | Units and Conversion Factors
|
Factor |
Name |
Abbreviation |
|
Factor |
Name |
Abbreviation |
|
101 |
deca |
da |
|
10-1 |
deci |
d |
|
102 |
hecto |
h |
|
10-2 |
centi |
c |
|
103 |
kilo |
k |
|
10-3 |
milli |
m |
|
106 |
mega |
M |
|
10-6 |
micro |
µ |
|
109 |
giga |
G |
|
10-9 |
nano |
n |
|
1012 |
tera |
T |
|
10-12 |
pico |
p |
|
1015 |
peta |
P |
|
10-15 |
femto |
f |
|
1018 |
exa |
E |
|
10-18 |
atto |
a |
|
1021 |
zetta |
Z |
|
10-21 |
zepto |
z |
|
1024 |
yotta |
Y |
|
10-24 |
yocto |
y |
Table D.1 SI prefixes
|
SI unit: metre (m) |
||
|
1 metre (m) |
≈ |
39.37 inches (in.) |
|
1 centimetre (cm) |
= |
0.01 m |
|
1 millimetre (mm) |
= |
0.001 m |
|
1 kilometre (km) |
= |
1000 m |
|
1 angstrom (Å) |
= |
10-8 cm |
|
|
= |
10-10 m |
|
1 inch (in.) |
= |
2.54 cm (exact, definition) |
Table D.2 Units of Length
|
SI unit: cubic metre (m3) |
||
|
1 litre (L) |
= |
0.001 m3 |
|
|
= |
1000 cm3 |
|
1 millilitre (mL) |
= |
0.001 L |
|
|
= |
1 cm3 |
|
1 microlitre (µL) |
= |
10-6 L |
|
|
= |
10-3 cm3 |
Table D.3 Units of Volume
|
SI unit: kilogram (kg) |
||
|
1 gram (g) |
= |
0.001 kg |
|
1 milligram (mg) |
= |
0.001 g |
|
1 kilogram (kg) |
= |
1000 g |
|
|
≈ |
2.205 lb |
|
1 ton (metric) |
= |
1000 kg |
|
1 pound (lb) |
≈ |
0.4535924 kg |
|
|
= |
16 ounces |
|
1 atomic mass unit (amu) |
≈ |
1.66054 × 10-27 kg |
Table D.4 Units of Mass
|
SI unit: joule (J) |
||
|
1 joule (J) |
= |
1 kg • m2/s2 |
|
|
≈ |
9.4778 × 10-4 BTU1 |
|
1 thermochemical calorie (cal) |
≈ |
4.184 J |
|
|
≈ |
4.184 × 107 erg |
|
1 erg |
= |
10-7 J |
|
1 electron-volt (eV) |
≈ |
1.60218 × 10-19 J |
|
|
≈ |
23.061 kcal mol−1 |
|
1 nutritional calorie (Cal) |
= |
1000 cal |
|
|
≈ |
4184 J |
Table D.5 Units of Energy
|
SI unit: kelvin (K) |
||
|
0 kelvin (K) |
= |
-273.15°C |
|
|
= |
-459.67°F |
|
K |
= |
°C + 273.15 |
|
°C |
= |
59
(°F – 32) |
|
°F |
= |
95
(°C) + 32 |
Table D.6 Units of Temperature
|
SI unit: pascal (Pa) |
||
|
1 pascal (Pa) |
= |
N m-2 |
|
|
= |
kg m-1 s-2 |
|
1 Torr |
= |
1 mm Hg |
|
1 atmosphere (atm) |
= |
760 mm Hg |
|
|
= |
760 Torr |
|
|
= |
101 325 N m-2 |
|
|
= |
101 325 Pa |
|
|
= |
1.01325 bar |
|
1 bar |
= |
105 Pa |
|
|
= |
105 kg m–1 s–2 |
Table D.7 Units of Pressure
Dimensional Analysis
Dimensional analysis is a form of proportional reasoning. It uses conversion factors to convert a quantity from one unit to another.
|
Quantity with desired unit |
= |
Quantity with given unit |
× |
Conversion factor |
In general, this method starts with the given value that will then be multiplied or divided by a known ratio or proportion. When setting up the ratios, the unit in the denominator must match that of the numerator of the given value. Continuing with the unit of the numerator in the next ratio, it has to match the denominator of the following ratio or of the units necessary for the answer.
For example, say you were trying to convert 3.41 grams of He to a number of atoms of He. You would identify 3.41 grams as the given quantity with grams as the given unit. The first step is always to place the given quantity in front of your equation. Then find a ratio that will help you convert the units of grams to atoms. As you probably have already guessed, you need to use a couple of ratios to help you in this problem. The ratio that 4.002 g of He = 1 mole (molar mass) will help you in this problem. Avogadro’s number, 6.022 x 1023 atoms = 1 mole, will also help you in this problem. Then you set up your ratios so that your units will cancel successfully (the same unit must be in the numerator of the equation and also in the denominator of the equation). Lastly, multiply through to get your final answer. As always, your final answer should contain the correct number of sig figs and the correct units.
3.41g×1mole4.002g×6.022×1023atoms1mole=5.13×1023atoms
Flipping the Conversion Factor
Note that a conversion factor can be flipped. For example, days are converted to hours by multiplying the days by the conversion factor of 24. The conversion can be reversed by dividing the hours by 24 to get days. The reciprocal 1/24 could be considered the reverse conversion factor for an hours-to-days conversion. The term “conversion factor” is the multiplier, not divisor, which yields the result.
Consider the following relationship
1kg1000g=1000g1kg
Both fractions are equal to 1 when the units are ignored. As the quotients are
both equal to 1, it does not change the equation, just the relative numerical values with
various units.
Solving Dimensional Analysis Problems
When doing dimensional analysis problems, follow this list of steps:
Identify the given amount with the given units (see previous concept for additional information).
Identify conversion factors that will help you get from your original units to your desired unit.
Set up your equation so that your undesired units cancel out to give you your desired units. A unit will cancel out if it appears in both the numerator and the denominator during the equation.
Multiply through to get your final answer. Don’t forget the units and sig figs!
Example Problems
How many hours are in 3 days?
Solution:
1. Identify the given: 3 days
2. Identify conversion factors that will help you get from your original units to your desired unit:
3. Set up your equation so that your undesired units cancel out to give you your desired units: 3 days ×
4. Multiply through to get your final answer: 72 hours
Converting between moles and grams
Find the amount of moles in 22.34 g of water.
Solution
22.34gH2O×1molH2O18gH2O=1.24molesH2O
