UNITS AND UNIT ANALYSIS

Introduction

Quantitative physical properties are recorded as a number with a unit where the unit indicates what instrument was used. In this country, the most common system of units is the English System but in science, the preferred system is the Metric System (SI). The SI system is preferred because all units that are proportional to some basic unit are given in terms of powers of ten of the basic unit. In this chapter, the units for length, mass, temperature, amount and time in both systems are given as well as some compound units. Changing from one unit to another is referred to as unit analysis problems and these are developed along with the introduction of the units. This is preceded by a general strategy for solving unit analysis problems.

General Aspects of Unit Analysis Problems

In either system, there are many units for the same quantitative property. For example, in the English system length can be given in inches, feet, yards or miles. Having different units for the same property may seem objectionable but consider the following. Suppose that the only unit for length was the inch and you are traveling from Waco to Dallas. When leaving Waco you see a sign that said Dallas 5700000 or Dallas 5.7x106 inches being understood. The problem is that very large (or very small) numbers are difficult to interpret. Just how far is 5700000 inches and how long will it take. In actuality, the sign reads Dallas 90 mi. which is much easier to interpret.

Since the choice of unit is somewhat arbitrary, it is necessary to be able to convert from one unit to another. Perhaps some lab uses instruments in the English system but your lab uses the SI system. This leads to a set of problems called unit (dimensional) analysis problems. All problems give information in the form of numbers with units and have as an objective an answer with units. The general form of these problems is given by

[pic] (1)

where CF is a conversion factor that makes the units on both sides of Eq. (1) agree (an equation in science is incorrect if the units on both sides do not agree).

Suppose that the information in the problem is the distance to Dallas of 5700000 in and the objective is to determine the distance in miles. Then Eq. (1) becomes

[pic] . (2)

In Eq. (2), CF must convert in into mi or symbolically

[pic] (3)

where the arrow means that the unit on the left is converted to the unit on the right, i.e. in is converted into mi. The problem is that the number of inches in one mile is not widely known. However, it is widely known that there are 12in in one foot or

[pic] (4)

which is considered to be an exact expression. Equation (4) is used to construct (simple) conversion factors. If rearranged to the form

[pic] (5)

it can convert length in inches to length in ft or in the form of Eq.(1)

[pic] (6)

where yin is the information. When Eq. (4) is expressed as

[pic] (7)

it can convert length in ft to length in inches or from Eq. (1)

[pic] . (8)

Another fairly well known conversion is that

[pic] . (9)

With this knowledge, the CF in Eq. (2) can be written symbolically as

[pic] (10)

which means that one has a product of two conversion factors. The first one converts in into ft and the second converts ft into mi. Using Eqns. (4) and (9) the conversion factor becomes

[pic] (11)

so that Eq. (2) becomes

[pic] (12)

In general, when working with problems in unit analysis, it is helpful to set the problem up with Eq. (1), determine with your known knowledge a symbolic expression for CF as in Eq. (10), then write CF in terms of actual conversion factors as in Eq. (11) and then finally solve the problem as in Eq. (12).

The Basic Units

The basic units to be discussed here are mass, length, time, temperature and the amount.

Length

Metric CF English basic unit meter(m) foot(ft) others 1km=1000m 1mi=5280ft 1m=100cm 1yd=3ft 1m=1000mm 1ft=12in 2.54cm=1in

The above table contains some of the more common units in the two systems. One notices that, in the SI system, all units are related by powers of 10 but that this is hardly the case in the English system. The entry in the CF column (2.54cm=1in) is a somewhat arbitrarily selected conversion between the two systems.

For the previous example, suppose that the distance to Dallas was given is 90.mi and the objective is to determine the distance in km. The Eq. (1) becomes

[pic] . (13)

Knowing the above information, the symbolic form for CF would be

[pic] .

Using the above information gives

[pic]

and using Eq. (13) gives

[pic] .

Chemistry often deals with very small lengths when considering sizes of atoms or molecules and the unit of length is a picometer(pm) where 1m=1012pm. A carbon atom has a radius of 77pm. How many feet would this be? Equation (1) becomes

xft=77pm(CF)

and

[pic]

so that

[pic] [pic]

and

[pic] .

One notices the futility of trying to measure this with a yardstick.

As a final example Astronomy deals with very large distances and its unit of length is the light year which is the distance that light travels in one year. Use of this unit gives how far in the past the observation of the light from the star corresponds to. The speed of light (c) is 3.00x108m/s. Given that a star is 5.0 light years from earth, determine its distance in mi.

Since distance is equal to the speed times the time and one light year is cx1yr, Eq. (1) becomes

[pic] .

One notices that two units need conversion, m must be converted to mi and yr must be converted to s to cancel time units. This can be written as two CF’s in the form

[pic] .
For distance

[pic] so that [pic] .

For the time

[pic] so that [pic] .
Then
[pic] .

It should be noted here that multiple conversions can be easily re-expressed in terms of simple conversions.

EXERCISES

Determine how many miles are in 77.5m. (ans. 0.0482mi)

The diameter of a hydrogen atom is 6.57x10-14mi. Determine the diameter in picometers. (ans. 106pm)
The distance from the earth to the moon is 3.84x105km. Determine how long in seconds it would take for light to travel from the moon to the earth. (ans. 1.28s)

Mass

Metric CF English basic unit kg(kilogram) slug others 1kg=1000g 1lb(weight) 1g=1000mg 1lb=16oz 1ton=2000lb 454g=1lb(sea level)

If a person weighs 180.lb, determine the mass in kg.
Equation (1) becomes

[pic] and [pic]

so that with the above information

[pic]

and

[pic] .

As a second example, suppose that one wanted to determine how many kilograms are in one slug. To do this suppose that a person weighed 184lbs so that they would have a mass of 83.5kg. From the definition of weight one has that w=mg where g is the gravitational constant which in the English system is 32ft/s2. Then

[pic] or [pic] .

Using the SI mass above gives

[pic] .

EXERCISES

Determine how much an object would weigh in pounds if its mass was 195kg.
(ans. 430. lb)
Determine the mass in slugs of the above object. (ans. 13.4 slugs)

Time

Both systems use the standard unit of time as one second. The conversions, 1min=60s, 1hr=60min, etc. are assumed to be well known.

Temperature

Metric CF English basic unit oC oF Centigrade Fahrenheit or Celsius [pic]

The CF between the two systems can be easily derived. Since the degree marks on both thermometers are equally spaced, there must exists a linear relation between them of the form

[pic] (14)

where a and b are constants. In the metric system, the normal freezing and boiling points of water are 0oC and 100oC respectively and the corresponding points in the English system are 32oF and 212oF. Using these sets of points in Eq. (14) gives

[pic] (15) and [pic] . (16)

Subtracting Eq. (15) from Eq. (16) gives

[pic]

and using this in Eq. (15) gives

[pic] .

Substituting these into Eq. (14) and rearranging leads to

[pic] . (17)

This equation can be rearranged to give the conversion between t(oC) and t(oF).

As an example, if the temperature in Paris is 33oC, determine the temperature in oF. Rearranging Eq. (17) gives

[pic] (18) so that [pic] .

Another temperature scale that is very important in science is the absolute temperature scale in degrees Kelvin (K). The lowest temperature on this scale is 0K. It is related to the Centigrade scale by

[pic] (19)

and from Eq. (17) to the Fahrenheit scale by

[pic] . (20)

One notices the lowest oC temperature is -273.15oC and the lowest oF temperature is -459.67oF. Rearrangement of Eq. (19) and Eq. (20) gives an equation for t(oC) and t(oF) in terms of t(K) respectively.

EXERCISE
Determine the temperature of an object in oC and oF if its temperature is 125K.
(ans. -148oC, -235oF)

Amount of matter

Both systems use the mole (mol) as the standard unit for amount. This unit will be dealt with at length later.

Compound Units

Volume

Metric CF English basic unit m3 ft3 others 1m3=103L gal 1L=1000mL 1gal=4qt 1mL=1cm3 1qt=2pt 1L=1.0567qt

A lot of laboratory chemistry deals with liquid solutions and lab drawers contain many pieces of volumetric glassware (graduated cylinders, volumetric flasks, etc.). The metric basic unit for volume is the cubic meter which is roughly the size of a moving box. More appropriate units for laboratories are the liter(L) and milliliter(mL) or cubic centimeter (cc or cm3).

At first sight, it may seem that unit analysis problems are more complicated for volumes than for lengths. However, consider the simple problem. Determine how many cm3 are in one m3. For this, Eq. (1) becomes

[pic]

which can be expressed as

[pic] .
Since
[pic]

then [pic]

i.e. there are 106cm3 in one m3. One notices that the only conversion factor needed was the one from m to cm.

As another example, determine how many m3 are in 8.0ft3. Equation (1) becomes

[pic] where [pic] so that [pic] and [pic] .

As another example, determine how many gallons are in 1.0ft3.
Equation (1) becomes

[pic] .

Consider the following symbolic map for CF given by

[pic]

which can be also written as

[pic] so that [pic]

[pic]
Then

[pic]

i.e., there are about 7.5 gallons in one cubic foot.

EXERCISES

Determine how many cubic feet are in 5.45L. (ans. 0.192 ft3)

Determine the volume in gal of the carbon atom used previously. (V=4πr3/3)
(ans. 5.05x10-28 gal)
Determine how many pints are in 6.57 cubic yards. (ans. 1.06x104 pts)

Density

The density of a system (d) is defined as the total mass of the system (m) divided by the total volume occupied by the system (V) so that [pic] . (21)

One notices that the mass of a system does not depend on temperature or pressure but that the volume does so that the density depends on both temperature and pressure. This dependence is relatively small for liquids and solids but large for gases. The units for density are usually g/cm3 or g/mL although, since gases have much lower densities than liquids or solids, the unit used for gases is often g/L.

Densities can be quite helpful in doing laboratory work. Tables list the densities of many common substances at usually 25oC and one atmosphere pressure. Suppose that one needed 15.0g of water for an experiment. This could be accomplished by weighing a dry container and then by adding water until the weight increased by 15.0g. Although this is a straightforward procedure, it is somewhat tedious to carry out. An easier method would be to locate the density of water in a table (1.00g/mL) and use Eq. (21) in the form

[pic] (22)

to determine the volume that this mass of water occupies. In this case one has

[pic]

which could be easily obtained using a graduated cylinder or 15mL pipette.

In unit analysis problems involving density, one should look for two properties. If the mass and volume are given, Eq. (21) can be used to determine the density. If the mass and density are given, Eq. (22) can be used to determine the volume. Finally, if the volume and density are given, Eq. (21) can be solved for m,

[pic] (23)

in order to determine the mass.

For example, a person claimed that a metal was gold. To verify this claim, you measured the mass of the metal and found it to be 97.5g. You also found that when placed into a graduated cylinder, the water level rose from 25.0mL to 30.6mL. From this information, the density of the metal is

[pic]

The table value for the density of gold is 19.3g/mL so that the person’s metal can not be pure gold.

As another example, given the above density for gold, determine how much, in lb, 1.0 cubic yard of gold weighs. From Eqns. (1) and (23) one has

[pic] which can be also written as

[pic] .

For the g CF one has

[pic]

and for the yd3 CF one has

[pic] or [pic]

so that [pic]

and [pic] [pic]

EXERCISES

An object weighs 0.486lb and occupies a volume of 0.266gal. Determine the density with units of g per mL. (ans. 0.219 g/mL)

Determine the weight in lbs of gold that has a volume of 1.25gal. (ans. 201 lb)

Percent mass

Consider a system that is made up of substances (components) A, B, C, ...... Let the total mass of the system be m and the mass of one of the components, say X, be mX. The percent mass of component X (%X) is defined to be

[pic] . (24)

Notice that percent masses are unitless numbers and that the sum over all components of the percent masses for the components must be 100%. One can directly show that percent mass is an intensive property.

As in the case of density, in problems dealing with percent masses, one expects two properties. If given %X and mX, rearranging Eq. (24) determines m. If given m and mx, Eq. (24) gives %X or if given m and %X, rearranging Eq. (24) gives mX. Furthermore, for only two components, say A and B, only %A needs to be given since %B=100-%A.

As an example, suppose that a homogeneous liquid solution of two components A and B has a density of 1.35g/mL and that %A is 36.7%. Determine the volume in gal of a solution that contains 2.50kg of B. From Eqns. (1) and (22) one has

[pic] but m is the mass of the solution. Since %B=100-36.7=63.3%, rearranging Eq. (24) gives

[pic]

so that

[pic] .

For the kg CF one has

[pic]

and for the mL CF one has

[pic]

which gives

[pic].

Using these results gives

[pic] .

EXERCISE

For the system above, determine the weight in pounds if the volume of the solution is 1.75gal. (ans. 19.7 lb)

SIGNIFICANT FIGURES-REPORTING ANSWERS

Suppose that the density of an object is to be determined.

Using a scale the mass of the object is 1.1250g.

Placed into a graduated cylinder initially containing 5.00mL of water the volume becomes 8.25mL.

Then

[pic]

and your calculator gives the answer of 0.384615384615g/mL.

Clearly neither measurement was as accurate as this number so what should the answer be.

The mass measurement has four figures of accuracy but the volume measurement has only three figures of accuracy.

Since the overall accuracy can not be greater than the least accurate measurement the answer should only have three figures of accuracy so that the answer is

d = 0.385g/mL .

A number is made up of digits (integers from zero to nine) and it may or may not include a decimal point.

In order to determine the number of significant figures (digits) to have in the answer it is necessary to determine the number of significant figures that each number used to obtain the answer has.

Rules for Determining the Number of Significant Figures for a Number

1) All nonzero digits in a number are significant. 2) All zeros between nonzero digits are significant. 3) All zeros to the left of the first nonzero digit are not significant. 4) If the number contains a decimal point, all zeros to the right of the last nonzero digit are significant. 5) If the number does not contain a decimal point, all zeros to the right of the last nonzero digit are not significant.

The meaning of rule 1 is that nonzero digits must have been read or estimated from the instrument and therefore must be significant.

The meaning of rule 2 is that zero digits between nonzero digits must have also been read from the instrument and therefore must also be significant.

The meaning of rule 3 can be seen as follows; Suppose that a meterstick is used to measure the length of an object and it gives 1.0cm. Expressing this result in meters gives 0.010m. The result in cm has two significant figures and simply changing units can not change the number of significant figures. Therefore zeros to the left of the first nonzero digit can never be significant.

The meaning of rule 4 is that zero digits to the right of the last nonzero digit in a number with a decimal point must have been read from the instrument and must therefore be significant.

For rule 5, the lack of a decimal point means that the zeros to the right of the last nonzero digit may or may not be significant and are therefore considered to be not significant.

Examples;

[pic]

Reporting Answers

Multiplication and division;

The number of significant figures in the answer is the same as the number with the fewest number of significant figures.

The final answer is obtained by rounding off the last significant figure.

In rounding off the last significant figure if the digit immediately following is

i) equal to or greater than five, increase the last significant figure by one. ii) Four or less, leave the last significant figure unchanged.

Examples;

[pic]

[pic]

[pic]

Problem with rounding off;

Suppose the calculator gave the number 74976.

If the answer should have four significant figures this would become

74980 .

If the answer should have three significant figures the number would become 75000

which has only two significant figures.

This problem can be avoided by using Scientific Notation;

In scientific notation a number is written as some number usually between one and ten that includes all significant figures times ten to some power. For the previous number 74976 in scientific notation this number would be

7.4976x104

With four significant figures

7.498x104

With three significant figures

7.50x104

With two significant figures

7.5x104

With one significant figure

8x104

Avogadro’s number 602,213,670,000,000,000,000,000=6.0221367x1023 1pm=0.000000000001m=1x10-12m

Addition and Subtraction;

In order to add or subtract all numbers must have a common decimal point. The precision of a number is defined to be the number of significant figures to the right of the decimal point. The precision of the answer is the same as the least precise number.

Examples;

Add the numbers 17.55, 5689.2 and 4.315.

17 .55 5689 .2 4 .315 5711 .065 = 5711.1

25.04 - 23.235 = 1.805 = 1.81

5.04x104 +1.56x102 –3.67x103 = (504. +1.56 –36.7)x102 =468.86x102

=469x102 = 4.69x104