Chemistry.wlu.edu

Assignment
BEGINNING CHEMICAL CALCULATIONS

The mathematics required for the problems in general chemistry is actually of a very elementary sort, involving arithmetic, a
little algebra, some exponents and some logarithms. The difficulties arising most frequently are those of translating from
chemical terms into simple mathematical operations. The purpose of these assignments is to provide aid in learning to
translate this chemical terminology. Emphasis is placed on "setting up" an entire problem using the dimensional units as a
guide, before performing the numerical computations.
In order to save time on long numerical computations, you are urged to obtain an inexpensive hand-calculator to use with
problem assignments. To accommodate later needs, the calculator should possess exponential and logarithmic capabilities.
A) DIMENSIONS AND DIMENSIONAL ANALYSIS
Placing the proper units (or dimensions) with every physical measurement or number used in a calculation is a topic which
cannot be overemphasized as an aid in performing chemical calculations. If a measurement of length is made, then the
proper dimension must be put with the numerical result of the measurement: for example, a measurement of an object’s
length made with a standard ruler must include a numerical value, such as 18.37, and the appropriate dimensions, in this
case inches, resulting in a reported measurement of 18.37 in. It is rather meaningless to say: "It is 18.37 long". Does this
mean inches, feet, meters or miles? Units, or dimensions, must be used! In mathematical calculation, the arithmetical
operations are performed on the dimensions as well as on the numbers.
Examples:
2 cm x 4 cm = 8 cm2 , 1 g + 2 g = 3 g, etc. Quite frequently a physical quantity will have a dimension which includes the symbol, / . Examples: 3.42 "3.42 grams of the substance per one mL of the substance"
interpreted as "100 centimeters per one meter"
"15.9994 grams of the substance per one mole of the substance"

Any dimension that includes a slash, /, is a relation existing between at least two dimensional units. This relation may be
considered a conversion factor between these dimensional units. It is always possible to invert a conversion factor to the
inverse relation because inversion of this relation is merely a change in emphasis—it does not change the fundamental
relationship between the two dimensions. For example, 2.54 cm and 1 in. are comparable quantities of length (1 inch ≡ 2.54
cm). This reality leads to two conversion factors: Writing 1 in./2.54 cm instead of 2.54 cm/in. merely changes a definition of 1 inch in terms of centimeters to a definition of 2.54 cm in terms of 1 inch. Similarly, the density measurement, 3.42 g/mL, expresses a fundamental relationship that 3.42 g of a particular substance is the same amount of material as 1 mL of that same substance. Thus, the two conversion factors become: 3.42 For calculations involving these quantities, either form may be used as required to obtain the desired result. 01-111-ProbZero-1

Dimensional analysis is the process of deciding what mathematical operations should be performed, involving physical
measurements and conversion factors (each of which has dimensional units), in order to achieve the dimensions desired for
the answer.
Examples:
1)
Convert a measurement of 30.0 inches to the equivalent length in centimeters.
The solving by dimensional analysis proceeds as follows: ✓ What do you want?:
✓ What do you “know”? (Identify given measurement information and necessary conversion factors):

Known conversion factor = 2.54 cm/in
✓ How do you get there? (Set-up dimensions to yield the desired dimension):
30.0 in. 2.54
✓ Calculate the result. (Multiply together values in the numerator and divide by values in the denominator
such that a particular dimension listed in both the numerator and the denominator effectively “cancels 6.6 grams of ethane (C2H6) is equivalent to how many moles of ethane? The molecular weight of ethane is
30.0 g/mol.
✓ What do you want?:
? mol ethane =
✓ What do you “know”?:
Physical measurement = mass of ethane is 6.6 grams
Known conversion factor = the molecular weight of
ethane is 30.0g/mol
✓ How do you get there?:
6.6 g ethane 1 mol
? mol ethane =
= mol ethane
30.0 g ethane
✓ Calculate the result:
? mol ethane =
= 0.22 mol ethane
01-111-ProbZero-2
A single tablet of the pain reliever Aleve® contains 0.200 g of naproxen, the active ingredient. If the
maximum recommended dosage is 1 Aleve® tablet every 8 hours, what is the maximum amount of
naproxen, in grams, that would be recommended over a three day period?
✓ What do you want?:
? g naproxen =
✓ What do you “know”?:
Known conversion factors = 0.200 g naproxen/tablet,
tablet/8hours,
hours/day
✓ How do you get there?:
? g naproxen = 0.200
= g naproxen
g naproxen 1
tablet 24
1 tablet 8
✓ Calculate the result:
1.80 g naproxen
Notice: in a problem involving many conversion factors the order of the conversion factors (in terms of first, second, third, etc.) is not significant but the placement of certain dimensions in the numerator and denominator is crucial. The dimensions for which you are solving are placed in the numerator. Also note that “given information” may include conversion factors (e.g. 24 hours/1 day) not explicitly stated in the problem. If you attempted to solve this example without converting hours to days, you would not achieve an answer with the appropriate dimensions. Notice: the above example used only multiplication and division operations. Don’t be fooled into thinking these are the only two operations used in this kind of work--dimensional analysis problems which use addition and subtraction operations (or a mix of mathematical operations) are also possible. To prepare a laboratory reagent, 4.000 g of a white solid is mixed with 135 mg of red powder. What is the
total weight in grams of this mixture?
Note:
For additions and subtractions all factors must have the same dimensional unit.
✓ What do you want?:
✓ What do you “know”?:
Physical measurement = mass1 is 4.000g,
mass2 is 135 mg
Known conversion factor = 1000mg/g
✓ How do you get there?:
✓ Calculate the result:
The bonus of dimensional analysis. Proper use of dimensions can help in avoiding mistakes in the calculations of many
problems, particularly since some of the dimensions used in chemical problems may not be familiar. In fact, dimensional
analysis can often be used to gain fundamental insight into how to solve a problem in the first place.
This emphasis on dimensional analysis for working problems is referred to as the “conversion factor method” of solving.
The placement of all numerical data and dimensions as adjacent conversion factors is referred to as the single line set-up
format. Both will be used extensively in this Problem Manual, throughout this course, and in subsequent chemistry courses.
*
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B) THE SIGNIFICANT FIGURES (SIG.FIGS.) OF A MEASUREMENT
Experimentation often involves making a measurement of some physical quantity such as the length, volume or mass of an
object. Such an operation usually requires reading a scale which is divided into small subdivisions. In reading the scale, it
is only possible to estimate a fraction of the smallest subdivision. Each digit measured is a significant figure (or
significant digit), including the last digit which is estimated. The number of significant figures reported thus contains
information regarding the experimental precision of the instrument used for the measuring and the magnitude (size) of the
quantity being measured.
Examples:
Portions of two different graduated cylinders containing volumes of water are shown below at left and at right. A volume of water in a graduated cylinder forms a downward-curved surface called a meniscus and the volume measurement is read at the bottom point of the meniscus. Reasonable volume measurements would be 27.5 mL for the left cylinder and 4.28 mL for right cylinder. In each case, the recorded measurement includes all certain digits plus the estimated fraction of the smallest division. Notice: the smallest division is milliliters for the cylinder at left and tenths of milliliters for the cylinder at right. If a ruler is subdivided into centimeters (only), then the estimated fraction is tenths of a centimeter. A 5 cm paperclip would be measured, for example, as 5.0 cm by estimating the fraction, 10/10, between the 4 cm and 5 cm marks. There are two significant figures in this number. If the ruler were subdivided into centimeters and tenths of centimeters (millimeters), the paperclip measurement would be, for example, 4.98 cm, and there are three significant figures in the number. The measured quantity, 4.978 cm (four significant figures), would communicate that the instrument used to make the length measurement was subdivided in hundredths of cm and that the measured digits were: 0.008 cm (or .08 mm), or 8/10 of the smallest subdivision. In some cases, the measuring instrument is used to obtain an approximate value, even though the subdivisions Reconsider example (3). If we wanted to measure the paperclip to the nearest tenth of a cm (or nearest mm), the length would be 5.0 cm even though the instrument is capable of producing a more accurate measurement, namely 4.98 cm. There would be only two significant figures wanted in the number. In this course, every number written down for a physical measurement must have the correct number of significant figures ─ all measured numerals plus one estimated numeral. It is wrong to put down more digits than it is possible to measure, or to put down fewer digits than required. The estimated digit is uncertain and is known to be subject to error (an important consideration for further calculations). 01-111-ProbZero-4
The Zero Issue.
All numerals that are non-zero are significant. If a number contains zeros, a question arises as to whether or not the zeros
are significant.
A zero may serve either of two purposes:
(1) as a placer of the decimal point, or
If the zero serves only to place the decimal point, then the zero is not significant. For example, in 0.000123 g the zeros
are not significant, while the digits 1, 2, and 3 are significant.
Otherwise, the zero is significant. For example, in 10.01 grams and also in 10.00 mL all the zeros and each digit 1 are
significant because they represent either measured digits or the first estimated digit.
In a number such as 100. cm, there is ambiguity in deciding whether the zeros are placers or measured digits; a later
device (scientific notation) will be used to resolve the ambiguity.
Similarly,
4037
has four sig. figs.—the zero is a measured value. has seven sig. figs.—all zeros are measured values with the last zero being the estimated digit has five sig. figs.—the zeros to the left of the first non-zero digit are only place-holders has an ambiguous number of sig. figs.—it is unclear whether the zeros after the ‘3’ are measured has six sig. figs.—all zeros are measured values with the last zero being the first estimated digit All zeros between non-zero digits are measured digits and are significant. For a number smaller than 1, any zero to the left of the first non-zero digit is not significant (a placer), and any
zero to the right of the first non-zero digit is significant (a measured digit).
For a number greater than 1, if there are any digits (including zero) to the right of the decimal point, then all digits
(including zeros) are significant.
For a number greater than 1, if there are no digits to the right of the decimal point, then zeros immediately to the
left of the decimal point may be significant (measured) or may not be significant (merely placers).

C) CALCULATIONS AND SIGNIFICANT FIGURES

A beginning chemistry student should realize that uncertainty in measurement and calculation is a crucial issue. This
issue is more rigorously handled later (i.e., in Chem 112) using statistical methods that are beyond the scope of this
course. However, the use of significant figures allows us to estimate the results obtained by a more rigorous treatment of
uncertainty. The following rules govern the use of significant figures in calculations.
Addition and Subtraction. Any column which contains an estimated figure (digit) will have a result which is estimated, or
uncertain, in that column. Determine, reading from left to right, which column is the first to contain an estimated
(uncertain) digit. The digit in that column will be the last significant figure in the answer.
Examples:
14.327 g + 162.1 g = 176.4 g
01-111-ProbZero-5
0.537 mL + 0.14 mL = 0.68 mL
(The digit in the 0.01's column in the final answer is rounded up because the 0.001's column is known to have a value greater than 5. The answer has two 83.62 m − 9.1228 m = 74.50 g
(The zero in the final answer is the uncertain digit and is absolutely necessary.)
173.2711 g − 173.22 g = 0.05 g
(The final answer has only 1 significant figure!)
Notice: in addition calculations the number of significant digits in the answer can be the same as or greater than the number
of significant digits in the starting numbers. In subtraction calculations the number of significant figures in the answer can
be the same as or less than the number of significant figures in the starting numbers.
Multiplication and Division. Determine which factor in the calculation contains the fewest number of significant digits.
The number of significant digits in the answer is the same (regardless of where the decimal point is) as in that factor with
the fewest number of significant digits.
Examples:
0.018 cm3
g/cm3 x 0.018 cm3 =
(4 sig.fig. in 17.32, 2 sig.fig. in 0.018, thus 2 sig.fig. in answer) 41.835 cm
0.02 cm x 41.835 cm x 8.0 cm
(1 sig.fig. in 0.02, 5 sig.fig. in 41.835, 2 sig.fig. in 8.0, thus 1 sig.fig. in answer) 0.0632 hr
hr/12.640 hr =
12.640 hr
(3 sig.fig. in 0.0632, 5 sig.fig. in 12.640, thus 3 sig.fig. in answer; note that the digits 0.00 in the
answer show the location of the decimal point only.) (The dimensions cancel out, yielding no dimensions
for the final answer.)

Counts and Definitions. There are two ways to have an infinite number of significant digits associated with a number:
definitions and counts. In a definition (or a conversion factor based on a definition) the numbers involved are infinitely
significant.
Examples of definitions:
01-111-ProbZero-6
In the examples above, 1000 m, 1 km, 16 ounces, 1 pound, 2.54 cm and 1 in. all have an infinite number of significant digits
and will not limit the number of significant digits reported in a calculation.
In contrast, a conversion factor based on measurement is not infinitely significant. A density reported as 3.42 g/ml has three
significant figures. The implicit “1” in the denominator of 3.42 g/1 ml does not affect the significant figures.
Sample calculations:
If 47.03 pounds are measured, how many ounces is this?
(The 16 ounces/pound conversion factor does not limit the number of significant digits in the calculation because 16 ounces is exactly equal to 1 pound by definition and the 16 ounces has an infinite number of significant figures). If 5473.2 mL methanol (density = 0.7914 g/mL) is measured, how many g of
methanol

(The 0.7914 g/mL conversion factor does limit the sig. figs. of the final answer because the density conversion factor is
based on measurement)
If something has been counted the integer value of the count is infinitely significant.
Examples of counts: 47 students,
In the examples above, 47 students, 538 grains of sand, and 2 history books are all infinitely significant because they have been counted. These numbers will not limit the number of significant digits reported in a calculation. Sample calculation:
The average student has $37.21 in their wallet. What amount of money would (The 47 students does not limit the number of significant digits reported in the final answer because the 47 is a counted integer which is infinitely significant.
D) SCIENTIFIC NOTATION (the writing of numbers in exponential form)
Many problems in scientific work involve calculations with very large numbers or with very small numbers. For example,
the total amount of sulfuric acid produced in the U.S. annually is 42,900,000,000,000 g whereas the mass of a single proton
is 0.000000000000000000000001672623 g. Writing numbers like this over and over again would be tiresome and lead to
lots of mistakes. It is the purpose of this section to show:
• how to express very large or very small numbers in exponential form, • how to perform calculations with these numbers, and • how to use significant figures with numbers in exponential form. 01-111-ProbZero-7

When a number is expressed exponentially, the number is usually given as a single digit (1 to 9) followed by the decimal
point (only one digit to the left of the decimal point), then all remaining significant figures, and finally multiplied by the
number 10 raised to some power. Expressing numerical values in such exponential form is known as scientific notation.
Thus, 123. is given as 1.23 x 102. The power to which 10 is raised is determined by the number of places the decimal point
has to be moved to give the proper exponential form (i.e., +1 for each place moved to the left, and −1 for each place moved
to the right). Note that the original number (123. x 100) and the number in exponential form (1.23 x 102) have exactly equal
value.
Examples:
1)
1,234,500,000 (ambiguous number of significant zeros)
Since the decimal must be moved to the left 9 places to get 1.2345, x = +9, and the number is: 1.2345 x 109 if none of the zeros were significant. To illustrate the inclusion of significant zeros: Note that the ambiguity of whether zero is a significant figure or not is resolved by including all significant zeros in the number which is multiplied by a power of 10. 0.000123 (this kind of zero is not significant; each is a placer)
The decimal must be moved to the right four places to get 1.23; y = -4 , and the number is: If further aid is desired in understanding the exponents, it may be well to recall the meaning of the powers of 10: 10o ≡ 1
E) CALCULATIONS AND SCIENTIFIC NOTATION

Addition and subtraction. When numbers expressed in the exponential form are being added or subtracted, it is important
that all numbers have the same power of 10 as well as the same dimensional units. This is required in order to have all
numbers lined up correctly with respect to a decimal point for the arithmetic operations.
01-111-ProbZero-8

Examples:
1.437 x 103 + 6.263 x 103 = 7.700 x 103
8.3 x 103 m + 2.7 x 102 m = (8.3 x 103) + (0.27 x 103) = 8.6 x 103 m
The answer is certainly not 11.0 x 103, nor 11.0 x 102 (from addition of 8.3 + 2.7 to get 11.0). 1.387 x 10−4 g − 1.42 x 10−6 g = (1.387 x 10−4) − (0.0142 x 10−4) = 1.373 x 10−4 g

Multiplication. When numbers in exponential form are being multiplied, multiply the measured numbers (the significant
digits), add the powers of 10, and multiply the dimensions.
Examples:
(2 x 103) x (4 x 106) = (2 x 4) x 103+6 = 8 x 109
(2.0 x 10−8 m) x (4.0 x 103 m) = (2.0 x 4.0) x 10-8+3 = 8.0 x 10−5 m2
(3.0 x 10−3 N) x (1.0 x 10−4 m) = (3.0 x 1.0) x 10−3+(−4) = 3.0 x 10−7 N•m

Division. When exponential numbers are being divided, divide the measured numbers, subtract the powers of 10 (the
dividend minus the divisor), and divide the dimensions.
Examples:
(8.0 x 106)/(4.0 x 103) = (8.0/4.0) x 106−3 = 2.0 x 103
(9 x 10−2 g)/(3 x 10−6 mL) = (9/3) x 10−2−(-6) = 3 x 10−2+6 = 3 x 104 g/mL
(6 x 104 J)/(3 x 10−6 s) = (6/3) x 104−(-6) = 2 x 104+6 = 2 x 1010 J•s
(6.0 x 10−8 kg)/(2.0 x 104 hr) = (6.0/2.0) x 10−8−4 = 3.0 x 10−12 kg/hr

Powers and roots. When taking powers or taking roots of exponential numbers, perform the operation on the number and
multiply the power of 10 by the power or root desired. Also perform the operation on the dimensions.
Examples:
(2.0 x 104 m)2 = (2.0)2 x 104x2 = 4.0 x 108 m2
(4.0 x 10−6 m2)1/2 = (4.0)1/2 x 10-6x1/2 = 2.0 x 10−3 m
The 1/2 power is the same as square root; 1/3 is cube root; 2/3 is the square of the cube root; etc. If any answer calculated is greater than 10 or less than 1, the decimal should be moved, with a consequent change in the power of 10, to make exactly one digit to the left of the decimal point. Examples:
Decimal being moved to left (+1 per place): 01-111-ProbZero-9
Decimal being moved to right (−1 per place): All final results should show one and only one digit to the left of the decimal point. Recall that expressing numerical values
in such exponential form is known as scientific notation. Caution: never do what some calculators do and express a
number like 2.5 x 103 as “2.5 E 3” ! An answer expressed in this way would be marked wrong.

F) DENSITY
Some examples of problems involving density are provided below as illustrations of the use of dimensional analysis and
reporting results with the correct number of significant figures.
Density is defined as the mass per unit volume, and may be calculated by dividing the mass of the object by the volume of
the object. Therefore, the dimensional units of mass (grams, g) are placed in the numerator and the dimensional units of
volume (milliliters, mL) are placed in the denominator.
Note: Mass is measured in the laboratory by weighing the object, thus the terms “mass” and “weight” will be used
interchangeably with the dimensional unit, grams. (In some courses, a subtle distinction is drawn between these
two words, with the word 'weight" having a special meaning.)

Examples:
Find the density of an object with a mass of 32 grams and a volume of 16 mL.
✓ What do you want?:
✓ What do you “know”?:
Physical measurements: mass = 32 grams,
volume = 16 mL
Known conversion factor: Density =mass/volume
✓ How do you get there?:
✓ Calculate the result:
= 2.0 g/mL (two sig.figs. throughout)
Determine the density of an object whose weight is 850 grams (may be two or three significant figures
─ ambiguous) and whose size is 3.0 cm by 4.0 cm by 10.0 cm (two sig.fig. in 3.0 and 4.0).
✓ What do you want?:
✓ What do you “know”?:
Physical measurements: mass = 850 grams,
size = 3.0 cm by 4.0 cm by 10.0 cm
Known conversion factor: Density =mass/volume
Necessary additional conversion factor: 1mL/cm3
(see Appendix ─ this is an exact relation)
01-111-ProbZero-10
✓ How do you get there?:
3.0 cm x 4.0 cm x 10.0 cm
✓ Calculate the result:
= 7.1 g/mL
(Answer is limited to two sig.figs., thus sig.fig. of 850 does not matter.) Find the density of an object given the following information:
wt. container = 5.818 g (tare); wt.
container + liquid = 14.0 g; volume of liquid in container = 8.204 mL.
✓ What do you want?:
✓ What do you “know”?:
Physical measurements: wt. container = 5.818 g;
wt. container + liquid = 14.0 g; volume of liquid =
Known conversion factor: Density =mass/volume
✓ How do you get there?:
(14.0 g – 5.818 g)
✓ Calculate the result: (14.0 g – 5.818 g)
= 1.0 g/mL
(sig.fig. of net weight of liquid limits answer to two sig. figs.)
Calculations involving density include finding the volume of a certain weight of a substance of known density or finding the
weight of a certain volume of known density.
Examples:
1)
What is the weight, in grams, of 73. 5 mL of a liquid whose density is 0.30 g/mL?
What is the volume, in mL, of a sample whose weight is 36 g and whose density = 2.4 g/mL?
= 15 mL (two sig.fig. throughout)
Find the moles of lead (Pb) metal in a 2.0 cm3 sample, given the following information: Atomic wt. of Pb
metal = 207.21 g/mol; density = 11.3 g/mL. (Necessary conversion factor: 1 mL/cm3)
= 0.11 mol Pb
mL 207.21
(sig. figs. of volume limits sig. figs. in final answer)
Note that the density conversion factor, 11.3 g/mL is equal to 11.3 g/1 mL where the numeral 1 is considered an exact
numeral with a value of exactly 1.
01-111-ProbZero-11
What is the density of ice, in g/mL, if a 6.33 x 1012 ton iceberg covers an area 10.0 miles wide by 30.0 miles
long by 0.50 miles deep?
✓ What do you want?:
✓ What do you “know”?:
Physical measurements: weight = 1.13 x 1012 tons
volume = 10.0 mi x 30.0 mi x 0.50 mi
Necessary conversion factors: 1 mL/cm3, 2000 lbs/ton,
454 g/lb, 5280 feet/mile, 12 inches/foot, 2.54 cm/inch
✓ How do you get there?:
lb 10.0 mi x 30.0 mi x 0.50 mi
(5280 ft)3
(2.54 cm)3 1
✓ Calculate the result:
= 0.92 g/mL
(two sig. figs. in the 0.50 mi measurement
Note that when a factor is raised to a power the operation must be performed on both the number and the
dimension. Also note that a single computation of the numerical values yields the final answer. In this course, the
complete single line set-up (including all numerical data and dimensions) should be written out before any
computation is done. The complete set-up has several important advantages:
1) it displays the entire logical framework of the problem, 2) it reveals the appropriate final number of sig.figs., and 3) it arrays all numerical data in a manner particularly suitable for using a calculator. 01-111-ProbZero-12
G) MOLES
The atomic weights, shown on the Periodic Table and elsewhere in tables, are relative weights based on the isotope of
carbon of mass number 12 (represented as 12C), which is defined as precisely 12, an exact number with no limit (infinite) to
the number of significant figures. These atomic weights have no dimensional units of mass, since they are only relative
weights (compared to the 12C isotope). Thus, the atomic weight of hydrogen is 1.008 (not 1.008 g or 1.008 lbs). The
number 1.008 means that the average mass of one atom of hydrogen is 1.008/12.000 times the mass of one atom of 12C.
If one collects enough atoms of 12C to have a combined mass of exactly 12.000 lbs, then the same number of atoms of
hydrogen would weigh 1.008 lbs, because of relative atomic weights. Or, 12.000 yahoos of 12C will contain the same
number of atoms as 1.008 yahoos of hydrogen. Any mass unit applies similarly. In the chemistry laboratory, we work
primarily with quantities of materials on the order of grams in mass. Therefore, we need to refer to the number of atoms in
12.000 grams of 12C or in 1.008 grams of hydrogen. The number of atoms in 12.000 grams of 12C has been shown by
several experiments to be 6.0221367 x 1023 (known as Avogadro’s number). For the purposes of this course, you should
memorize 6.02 x 1023 as Avogadro’s number and treat it as having three significant figures. 6.02 x 1023 particles is a mole
of particles, just as 6.02 x 1023 sugar molecules is a mole of sugar molecules or 6.02 x 1023 bumblebees is a mole of
bumblebees.
The word, mole, may be applied to any specified type of particles ─ atoms, molecules, ions, or formula groups. Note that
(by convention) the word, mole, is abbreviated, not to "m" which stands for meter, but to "mol" ─ thus 12.000 grams of
(Note: Several other particular words, synonymous with mole, may be encountered particularly in the
older chemical literature: gram-atom (when referring to atoms), gram-mole (when referring to
molecules), gram-ion (when referring to ions), etc. We, however, shall use the word mole exclusively.)

It is important to distinguish clearly the identity of the substance under consideration. When chemists speak of substances
such as "hydrogen" or "oxygen" or "sulfur", they often intend to mean the element as it occurs in nature (i.e., H2
molecules, O2 molecules, S8 molecules). When one wants to refer to the individual atoms, it is necessary to state specifically
the word, atom; thus, "hydrogen" means H2 molecules, whereas "hydrogen atoms" or "atoms of hydrogen" means H
atoms. The chemical shorthand of formulas is designed to specify exactly what is being considered ─ H2 for hydrogen
molecules and H for hydrogen atoms.
Use of the word, mole, is of central importance in chemical calculations, and the Avogadro number needs to be
memorized. Observe how the idea of mole is used in the following table of data.
Regarding H atoms:

The number of grams per mole of any atom is the same numerically as its atomic weight (by definition of the word,
mole), and it is accepted practice to append the units "g/mol" to atomic weights.
Examples:
The atomic weight of arsenic = 74.91 g/mol As The atomic weight of oxygen = 15.999 g/mol O Several types of problems arise from these considerations. In the present group of examples, we are considering atoms only, and not the molecules of elements as they occur in nature. (The other examples will appear in subsequent assignments.) 01-111-ProbZero-13
Examples:

1)
Determination of the mass of a mole of atoms:
Al = 26.98 (atomic weight from a Periodic Table) F = 19.00 (atomic weight from a Periodic Table) In any chemical problem, the word, mol, must always be followed by an appropriate chemical formula.
Weight of a given number of moles:
3.00 mol C
mol Pt 195.1
= 97.6 g Pt
Number of moles of substance in a given mass:
= 3.0 mol O
= 0.14 mol Mn
Number of atoms in a specified sample
(using the Avogadro number with proper dimensions -- 6.02 x 1023 atoms/mol of specified substance):
0.60 mol N 6.02 x 1023 atoms N
= 3.6 x 1023 atoms N
(Note two sig.figs. in 0.60 moles) 6.02 x 1023 atoms U
= 72 x 1023 atoms U = 7.2 x 1024 atoms U
01-111-ProbZero-14
6.02 x 1023 atoms C
= 3 x 1023 atoms C
(At.wt. of C = 12.011; but fewer sig.figs. are sufficient for given data.) 2.01 x 1019 atoms Ge
72.6 g Ge 1
Weight of a given number of atoms.
= 5.33 x 10-23 g S
6.02 x 1023 atoms S 1
one atom is a count, and infinitely significant!) 1.00 x 10100 atoms of H = ? kilograms of H 1.00 x 10100 atoms
kg H = 1.67 x 1073 kg H
6.02 x 1023 atoms H 1

F) DIMENSIONAL ANALYSIS INVOLVING MATHEMATICAL FORMULAS

Problems involving multiple mathematical formulas should be solved by appropriately manipulating the formulas,
substituting values in for the variables, and then performing the dimensional analysis. For example, later in this course (not
now!) you will memorize the following equations:
where E is energy in Joules (J), h is Planck’s constant (6.626 x 10-34 J•s), ν is frequency in reciprocal seconds (s-1), c is the speed of light (3.00 x 108 m/s), λ is wavelength in length dimensions. 01-111-ProbZero-15
Examples:
1) What is the frequency of a photon that has 4.37 x 10-19 kJ of energy?

✓ What do you want?:
? ν (s-1) =
✓ What do you “know”?:
Photon energy = 4.37 x 10-19 kJ
Known equation that relates energy to frequency = E = hν
Known constant: h = 6.626 x 10-34 J•s, 1000 J = 1 kJ
✓ How do you get there?:
E = hν ➜ (rearrange formula to solve for frequency) ➜ ν = E/h 4.37 x 10-19 kJ
6.626 x 10-34 J•s 1 kJ
✓ Calculate the result:
= 6.60 x 1017 s-1

2) What is the frequency of light that has a wavelength of 23 inches?

✓ What do you want?:
? ν (s-1) =
✓ What do you “know”?:
= 23 inches
Known equation that relates wavelength to frequency: c = λν
Known constant: c = 3.00 x 108 m
100 cm = 1 m, 1 cm = 2.54 in.
✓ How do you get there?:
c = λν ➜ (rearrange formula to solve for frequency) ➜ ν = c/λ 3.00 x 108 m
1 inch 100
23 inches
✓ Calculate the result:
5.1 x 108 s-1
(sig. figs. in final answer limited by two sig. figs. in 23 inches measurement)
Note again, in dimensional analysis problems:
The entire set-up should be expressed before the (one and only) numerical computation is performed. Avoid the
temptation of making numerous partial calculations. With a single line set-up of the data and conversions, all pertinent
information is in view, and only one decision needs to be made regarding appropriate sig.figs. for the final answer.
01-111-ProbZero-16

Source: http://chemistry.wlu.edu/WebPDF-allaccess/111%20Problem%20Sets/assignments/PZero-2001.pdf