2.8: Unit Conversions and Dimensional Analysis

Equivalence Statements

As chemists have different ways of expressing measurements they need to be able to convert between different units. Central to this is the concept of an equivalence statement which says two ways of representing the same thing are equivalent. For example 12 in = 1 foot is an equivalence statement. An equivalence statement allows you to convert from one unit to the other, and this is done by creating a conversion factor. A conversion factor is simply the ration of one part of the equivalence statement to the other, where the numerator has the unit you want to convert to, and the denominator has the unit you want to transform from.

12 in = 1 foot

has two conversion factors

1. to convert from feet to inches multiply by the conversion factor of \[ \left (\frac{12 \; in}{ 1 \; foot} \right)\]
2. to convert from inches to feet multiply by the conversion factor of \[\left (\frac{1 \; foot}{12 \;inches} \right)\]

There are three fundamental types of conversions that you will need to be able to perform. The first involve different ways of expressing numbers within the same scale, which include transforming floating numbers to SI prefixes and scientific notation. The second involves converting from one scale to another, like the metric system to the English [read: American] system, and you are expected to know that 2.54 cm=1 inch [read: memorize that]. The third involves physical constants of specific substance, where you convert from one type of physical measurement to another. Density is such a physical constant, and it allows you to convert mass measurements to volume.

What is consistent with each of these three different types is that they all represent an equivalence statement, and from the equivalence statement you can generate a conversion factor.

Three Basic Types Conversions

A. Different prefixes of same base scale (cm -> Km)
B. Conversions between scales (cm -> in)
C. Physical Constants d=M/V

 

Conversions Involving SI Prefixes

2 Basic Techniques:

1. Basic Technique : Use two equivalence statements
2. Advanced Technique: Use one equivalent statement

Basic Technique

How many micrograms are in a kilogram?

Set up conversion factors based on an equivalence statements.

1 kg = 10³ g and 1 µg = 10^-6 g

so, this gives two conversion factors

\[\left ( \frac{10^{3}g}{1kg} \right )and \left ( \frac{1\mu g}{10^{-6}g} \right )\]

You have two conversion factors. In the first you will convert kilograms to grams, and in the second you will convert the grams to micrograms.

\[\left ( 1kg \right )\left ( \frac{10^{3}g}{1kg} \right )\left ( \frac{1\mu g}{10^{-6}g} \right )= 10^{3-\left ( -6 \right )}\mu g = 10^{9}\mu g \]

In the advanced technique (below) you will solve this problem with only one conversion factor.

Example \(\PageIndex{1}\)

3x10²³fsec = ? Gsec

Solution

Video \(\PageIndex{1}\): Basic technique for unit conversions. https://youtu.be/Jlfk7XjDnwM

(3x10^-1 Gsec)

Exercise \(\PageIndex{1}\)

Convert 32.4 ng to Mg.

Answer

\[\left ( 32.4ng \right )\left ( \frac{10^{-9}g}{ng} \right )\left ( \frac{Mg}{10^{6}g} \right )= 32.4\times 10^{\left ( -9-6 \right )}Mg = 32.4\times 10^{-15}Mg = 3.24\times 10^{-14}Mg \]

Advanced Technique

How many micrograms are in a kilogram?

Set up a single conversion factor based on an equivalence statement.

The trick here is to multiple both SI prefixes by the number that makes them equal to one. That is, if a micro is one millionth, than there are a million micro in one, and and likewise if a kilo is a thousand, than a thousandth of a kilo is one, and so a million micrograms equals a thousandth of a kilogram.

\[1\mu g= 10^{-6}g\therefore 1g= 10^{6}\mu g\]

\[1kg= 10^{3}g\therefore 1g= 10^{-3}kg\]

so, this gives two conversion factors

\[\frac{10^{6}\mu g}{10^{-3}kg}=\underbrace{ \frac{10^{9}\mu g}{1kg}}_{\text{converts kg to }\mu g}\leftarrow or\rightarrow \frac{10^{-3}kg}{10^{6}\mu g}= \underbrace{\frac{10^{-9}kg}{1\mu g}}_{converts \; \mu g \; to \; kg} \]

(Use the ratio which has the starting unit in the denominator and ending unit in the numerator.)

See Video \(\PageIndex{2}\) for the advanced technique combined with scientific notation
(next video below)

Exercise \(\PageIndex{2}\)

Convert 32.4 ng to Mg..

Answer

1 g = 10^-6 Mg = 10⁹ ng

\[\left ( 32.4ng \right )\left ( \frac{10^{-6}Mg}{10^{9}ng} \right )= \left ( 32.4\times 10^{\left ( -6-9 \right )}Mg \right )= \left ( 32.4\times 10^{-15}Mg \right )= \left ( 3.24\times 10^{-14}Mg \right )\]

 

SI Prefix Conversions and Scientific Notation

The following video shows you how to do SI conversions of numbers expressed in scientific notation. The trick is to write the SI prefix in terms of its power of 10.

Example \(\PageIndex{2}\)

Convert: 3 x 10²³ kiloseconds to nanoseconds.

Solution

Video \(\PageIndex{2}\): Advanced technique with SI notation. https://youtu.be/43wWTLB7GS0

(3 x 10³⁵ nsec)

Exercise \(\PageIndex{3}\)

Convert 3.57x10¹⁴ ml to ml

Answer

\[\left ( 3.57\times 10^{14}\mu l \right )\left ( \frac{10^{3}ml}{10^{6}\mu l} \right )= \left ( 3.57\times 10^{\left ( 14+3-6 \right )}ml \right )= \left ( 3.57\times 10^{11}ml \right )\]

 

Conversions Between Different Scales

There are typically two different types of conversions between scales; those between different systems (like inches to meters), and those between SI base units and derived units (like ml to m³).  America still uses the English or "British Imperial System" for common measurements like mass, length and volume, even though the British don't, and thus we often need to convert between the two different systems.  What is required, as in any conversion, is an equivalence statement.  In this class you are only responsible to memorize two conversion factors of this nature

You need equivalence statements between different units. These are the same techniques as the previous, just that the conversions are between different types of units.

Note

UALR students are required to memorize the following conversion factors

• 2.54 cm=1 in (exact) (Metric length to English Length)
• 1 ml = 1 cm³ (Metric volume to Metric length cubed)

Exercise \(\PageIndex{4}\)

How many dm are in an object which is 2.450 feet long?

given 12 in = 1 ft,

Answer

\[\left ( 2.450\,ft \right )\left ( \frac{12\in}{1\,ft} \right )\left ( \frac{2.54\,cm}{1\,in} \right )\left ( \frac{10^{1}dm}{10^{2}cm} \right )= \left ( 7.4676\,dm \right )= \left ( 7.468\,dm \right )\]

Example \(\PageIndex{3}\)

What is the volume in Gallons of a 1.00 cubic foot object?

(Try this first, the answer is at the bottom of the video. )

Solution

Note, you are required to know a 1 ml =1 cm³, 1 in = 2.54 cm, and of course the SI prefix conversions.

Video \(\PageIndex{3}\): Conversions between scales

(7.48 gal)

Exercise \(\PageIndex{5}\)

How many gallons are in a cuboid can that is 2.05 ft wide, 12.4 inches long and 1.30 yards deep?

Given: 4 qt = 1 gal, 1.057qt=1L, 12in = 1 ft, 3 ft = 1 yd:

Answer

\[\left ( 2.05ft \right )\left ( 12.4in \right )\left ( 1.3yd \right )\left ( \frac{3ft}{1yd} \right )\left ( \frac{12in}{1ft} \right )^{2}\left ( \frac{2.54cm}{1in} \right )^{3}\left ( \frac{1mL}{1cm^{3}} \right )\left ( \frac{1L}{1000mL} \right )\left ( \frac{1.057qt}{1L} \right )\left ( \frac{1gal}{4qt} \right )= 61.8gal\]

 

Conversions Involving Physical Constants

Many substances have physical properties showing a linear relationship that can be described by the equation of a straight line Y=mX+b. If these are extensive properties (section 1.6.1), then the Y intercept (b) equals zero and this takes the form of Y=mX, which is an equivalence statement, and allows one to convert measurements in the unit of X to those of the unit of Y. These are often tabulated by the constant m \( \left (m=\frac{Y}{X} \right )\), which allows conversions of measurements between the units of X and Y.

This can be understood by looking at one such constant, density, which is the linear proportionality constant for the extensive properties of mass and volume, where m = dv (the y intercept is zero because if you have nothing, its mass and volume both equal zero - review extensive properties). So if you know the density of something, you can convert its mass to volume, or its volume to mass. Therefor it is common to compile tables of densities and other physical constants.

It should be stated that although these generate conversion factors, they are really describing relationships, that is, mass and volume describe completely different aspects of matter and density allows you to convert from one to another. This is completely different than definind 12 inches as a foot, where inches and feet are different units that describe the same thing.

Conversions Involving Density

Density is defined as

\[d=\frac{m}{v} \]

This gives the equivalence statement

\[m=dv\]

and there are thus two conversion factors

\[\underbrace{ \frac{m}{d}}_{\text{converts mass to volume }}\leftarrow or\rightarrow \underbrace{dv}_{\text{converts volume to mass }} \]

That is, dividing the mass by the density converts it to its volume, and multiply the volume by the density converts its volume to its mass

Example \(\PageIndex{4}\)

How many liters is the volume of 12.140 kg of Iridium? Use 22.65 g/ml as the density of Iridium

(Please do this before looking up the answer at the bottom of the video.)

Solution

Video \(\PageIndex{4}\): Conversion Factors involving physical scales.

0.5360 L, note, the number of sig figs was determined by the density and not the starting mass.

Exercise \(\PageIndex{6}\)

How many pounds would 61.8 gallons of water weight if at a given temperature the water has a density of 0.98g/mL?

Answer

\[\left ( 61.8gal \right )\left ( \frac{4qt}{1gal} \right )\left ( \frac{1L}{1.057qt} \right )\left ( \frac{1000ml}{1L} \right )\left ( \frac{0.98g}{mL} \right )\left ( \frac{1lb}{453.6g} \right )= 505.27lb= 510lb\]

Note the answer has 2 significant digits because of the density.

 

Note

If you look at the answer of exercise \(\PageIndex{6}\) you will see that each equivalence statement is in a set of parenthesis.  You will also note that all three types of conversion factors are used.  (\(\frac{4qt}{1gal}\)) is between two different units (4 quarts =  1 gallon),  (\(\frac{1000mL}{L}\))  is between different SI prefixes (1000mL=1L) and (\(\frac{0.98g}{ml}\)) is a physical constant, the density of water (0.98g H₂O = 1mL H₂O, with two sig figs).

 

Test Yourself

Homework: Section 2.8

Graded Assignment: Section 2.8

Query \(\PageIndex{1}\)