Los Alamos National Laboratory Chemistry Division Periodic Table of the Elements He 7 1.!U, 3A 5 4A SA 6A 6 1.01, 2A 3 Lit Be iFfl^' 2" lit! itim 6.941 i i 9.91 11 Na 2 g .09 ?-4.31 3B 19 20 21 K Ca Sc 39,Ii1 jli,ii^ I1.96 3/ 38 39 Rb Sr Y 26 Fe 55.135 44 Ru I()1.1 76 O s 191).'_ 108 Hs 10 '1I I,OI IT,01 j 13 14 1 15 Al Si P 16.00 16 11 S (OR p). I! i .36.9 31 32 Ga Gc 33 As 3,1 35 36 Se I&m I 11;.19 18 At 49 50 51 52 In Sn Sb Tc 116.7 1>1.. 82 83 Pb l Bi 111., 81 Ti 9.99 53 I 54 ze 1 ' 6.9 85 At 84 Po 413 5B 6B 7B 23 24 25 V Cr Mn 3_On 4.!14 43 42 Moll Tc (() .) 75 Re I'.6.? 107 Bh 2 8B 27 I, 28 Co ) Ni 5S.93 . 5ts.b9 45 46 Rh Pd IU(,.4 78 Pt V6. 1 110 Ds 11B 12B 30 Zn 65.39 AgCd 107.9 112.4 79 80 Au Hg 107.1 201).5 ill 112 Uuu IUub 22 Ti 47.' 40 Zr v;.17 ')1_" 92.91 95.94 55 56 5/ 72 73 74 Cs Ba La* Hf Ta W 13,.`1 _ I^S,3 Inf).9. 1, P) 87 88 89 104 105 106 Fr Ra Ac- Rf Db Sg 1 77 Ir 19!1_ 109 Mt 29 Cu (,.3.55 26 263. 1 (272) 2 ) Lanthanide Series Actinide Seri 58 59 Ce Pr I III.I 111.0 90 91 Th Pa 60 61 Nd Pm 63 Eu Am 64 9b Cm 69 /0 71 Tm Yb Lu I'^.II 103 Lr 1()_'.J 98 1.. I.' 99 Es P- 100 101 102 Fm Md No 68 E1r 66 137 0 Dy Sm 1911.-1, 94 Pu 65 FN Los Alamos NATIONAL LABORATORY 13) CU (2-I7)- 24 4l) 2)4)
Transcript
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Atomic Structure
ATOM COMPOSITION
Atoms are composed of positively charged protons, uncharged neutrons, and negativelycharged electrons. The number protons and neutrons in the nucleus of each element isrepresented using an element's atomic mass (A.K.A. mass number). The number of
Atomic Mass 4 650 4 Atomic number
protons in an atom's nucleus is communicated using an atomic number. In a neutral atomthe atomic number is also equal to the number of electrons orbiting the nucleus in shellsor orbits. The outermost electron shell of an atom is called its valence shell.
IONSAtoms may gain electrons or lose electrons and so become ions. Atoms which lose
electrons have an excess of protons and are positively charged CA+IONS, whereasnegatively-charged ions are called ANIONS.
BOHR DIAGRAMS
There are two main ways to represent atoms: the Bohr model, and the Lewis electron dotdiagram. In the Bohr model, the number of protons and neutrons in the nucleus arerecorded in the centre, and then circular shells are drawn around the nucleus. Electronsare placed into each shell according to the periods of the periodic table. Since the firstperiod contains only two atoms, H and He, the first shell, called the K shell, can holdonly two electrons. Therefore, H and He follow what is known as the duet rule. Thesecond and third periods, representing the L and M shells, respectively, can accommodate8 electrons. Atoms following this rule are said to require an octet of valence electrons inorder to complete their outer shell.
, the atomWhenLEWIS sketching
ELECTRON
Bohr
DOT
diagrams ofDIAGRAMS ions, typically the valence shell is complete with either
must be enclosed in square brackets, and the sign and charge of the ion is included in thetop right corner as a superscript.
two
Lewis
or eight
electron dot
electrons,
diagrams
depending
show ononlythe the valence
structureence of the
electrons
atom.
in an atom.
Furthermore
The electronsare represented as dots around the element symbol in four quadrants-north, south, east,and west. As a rule, there can only be a maximum of two electrons in any quadrant.Generally, the number of valence electrons in an atom is the same as the group number inwhich it is placed on the periodic table, with the exception of He, the only noble gas toobey the duet rule.
1
Periodic Table
Historical DevelopmentThe periodic table was originally published in 1869 by Dimitri Mendeleev who classifiedthe elements according to their regularly repeating physical and chemical properties as afunction of their atomic mass-the number of protons and neutrons in an atom's nucleus.The modern periodic table was introduced in 1911 by A. van den Brock wherebyelements are arranged according to their increasing atomic numbers-the number ofprotons in the nucleus, which, in a neutral atom, is equal to the number of orbitingelectrons.
OrganizationThe periodic table is represented as a grid separated into seven horizontal periods(A.K.A. rows) and eighteen vertical groups (A.K.A. columns or families). Each periodrepresents the filling of an outermost electron shell, called a valence shell. The groupsare chunked into families with the following names.
Group NameAlkali metals
Alkaline earth metals
Transition metalsChalcogens or oxygen family
HalogensNoble gases
Group Number1A2A
B block
6A
7A
8A
Metals, Non-metals, and MetalloidsAll elements on the periodic table can be classified as metals, metalloids, or non-metals.The eight metalloids are found around a "staircase" beginning at boron. Metal atoms arelocated to the left of the staircase and non-metals to the right of the staircase. Hydrogenis a unique non-metal and is often placed with the alkali metals because of its valence ofone.
Properties of Metals and Non-metals
Metals Non-Metals
Compounds and Naming
Compounds are substances containing atoms of more than one element combined in fixedproportions. Compounds can be classified as either ionic or covalent. The followingtable highlights the main characteristics of each type of compound.
Ionic Compounds
a complex ion (negatively charged).
Electrons are completely transferred fromthe cation to the anion
• Can further be classified as Binary IonicCompounds and Ternary Ionic Compounds
Both types of compounds are named simplyby combining the names of both the cationand anion parts
• If an atom has more than one possiblecombining capacity, then the Stock Systemof Naming or the Traditional System mustbe used
Stock System
Traditional System+ uses roman
numerals inbrackets betweenthe cation and
anion names to
indicate the
oxidation state ofthe metal cation.
Diatoms, Tetra-atom, and Octa-atomA number of atoms on the periodic table exhibit a unique characteristic in that they areonly found paired with themselves in their free atomic state. That is, these elements arenever found as single atoms. All of these atoms are covalently bonded in compounds,and, therefore, share electrons. Seven compounds are known as "the group of seven" orH-O-F-Br-l-N-Cl diatoms. In order to satisfy the octet or the duet rule, H, 0, F, Br, I, N,and Cl pair with themselves to become H,, Oz, F2, Br2, 12, N2, Cll. Two other atoms onthe periodic table; P, and S, combine to form the P4 tetra-atom, and S8, an octa-atom.
Binary Covalent
Composed of a metal cation (positivelycharged metal ion) and a non-metal anion or
Objective: Your objective is to correctly write the chemical formulas and nameionic compounds.
Materials: 6 ionic dice
Procedure:1) You will be placed in groups of four. Each group will be divided into
teams of two.2) There are 5 different combinations of dice that you will be asked to roll.3) For example, the 1st two dice you will roll are dice #1 and #44) You will roll the dice and then write down the positive and negative ion
on the appropriate section of table 1.5) Your goal is.to determine the correct chemical formula and write the
correct name for the resulting ionic compound.6) 1 person will roll the dice while the other will be responsible for
completing step 5.7) One pair can start at table #1 and the other at table #5.
Table #1Dice
"+"#1 + #4
ion"="
Chemical Formula
Chemical Nameion
Roll 1
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Table #2Dice
#2 + #5Chemical Formula Chemical Name
ion
Roll 1
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Table #3Dice
"+"#3 + #5
ionChemical Formula
Chemical Name
ion
Roll 1
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Table #4Dice#1,+0
Chemical Formula Chemical Name.""+"
ion ion
Roll I
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
Table #5Dice#3+#6Roll I
Roll 2
Roll 3
Roll 4
Roll 5
Roll 6
"+"
I "_"
ion
I ionChemical Formula Chemical Name
Monarch madness Page 1 of 2
MONARCH CONSERVATION
GLIDER THEORY
TOPIC ANTHROPOLOGY
Activeoverwintering
monarchs [aboveright] fly to open
areas to get nectarfrom flowers.Photo by TomTrower. Male
Monarch's menace24 MARCH 1999. Monarch butterflies are the long-ranging kings of insect migration. In spring, they flyfrom winter refuges on a few mountaintops inMexico. In fall, their descendants somehow find thesame isolated mountains, where they huddle by themillions in a massive mingling of insects.
Even though individuals weigh a half-gram each,boughs bend beneath the masses of butterflies. On a
warm day, butterflies cloud the sky. On a coolmorning, you must tread carefully to spare half-frozen butterflies littering the ground.
Sadly, the monarch migration may be in trouble.This winter, there are an estimated 60 millionbutterflies in Mexico. That's an 80 percent drop fromrecent averages, according to monarch authorityLincoln Brower, a biologist from Sweet BriarCollege in Virginia.
For seversdecades,
monarchs havebeen tagged to
the path anddestination of
their migration.Photo by Karen
Oberhauser.
Like much else in Mexico,monarchs are beingpressured by a humanpopulation explosion. Poorfarmers are cutting logs andgathering firewood evercloser to the butterflyreserves. Commercial
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Monarch madness
Page 2 of 2
logging is picking up speed.As trees are thinned out, the butterflies get lessshelter.
But monarchs could be facing problems north of theborder, too. Questions about the effects of changesin their summer range in Canada and the UnitedStates are easy to ask, but difficult to answer.
Until now, there's never been a good way to find outexactly where the Mexican monarchs spend theirsummers and reproduce. Butterfly tagging programsare popular. But only a few tagged butterflies arefound among the millions of monarchs in thewintering grounds. Butterflies are too small to trackwith radios or satellites.
Now, from the science of geology, comes anappealing solution called stable isotope analysis.
So what's an isotope, and how do they tell usabout monarchs' summer habits?
There are J. 2 3 4,5 pages in this feature.Bibliography I Credits I Feedback I Search
The Why Files Staff includes: Terry _Dev,itt, editor; Darrell_Schulte,webmaster; David Tenenbaum, feature writer
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Tracking with isotopes
Page 1 of 4
n; chemistryit'on
Isolating isotopesYou can't see isotopes. You can't feel them. Butthey're here. Think of them as mutant but harmlessatoms of a single element that have different masses.
isotopic t i o otoo ! The common isotope of hydrogen has one protonand one electron. But the hydrogen isotope
M ONARC H toso4deuterium also has a neutron, and is almost twice as
GLIDER TH EORY
heavy as the common hydrogen isotope.
Chemically, these isotopes of hydrogen cannot bedistinguished -- when combined with oxygen, theyboth make water. But because deuterium is moremassive, water containing deuterium (called "heavywater") evaporates more slowly and rains out morequickly.
That difference in behavior gives scientists anopening big enough to drive a mass spectrometerthrough. Isotope-ratio mass spectrometers measurethe ratio of stable isotopes of an element in a sample,and show what relative percentage of each mass ispresent. Isotope-ratio mass spectrometers areparticularly valuable for analyzing light elementslike hydrogen, oxygen, and carbon, which just sohappen to be the most common elements in living
(Stable isotopes are different from the radioactiveisotopes formed during nuclear decay. Radioactiveisotopes change over time into other isotopes.Eventually they finally become stable isotopes,which cannot decay further.)
Geologists have been analyzing stable isotopes fordecades. Lately, climatologists have joined in thefun. Now it's the biologists' turn.
Mind your monarchsThe first study of insectmigration using isotopeswas performed by twoCanadian researchers, whoreported their results in
December, 1998 (see "NatalOrigins of MigratoryMonarch... " in thebibliography). The goal wasto compare isotopes inmonarchs found in Mexicoto the isotopic "signature" ofvarious possible breedinggrounds in Canada and the United States.
The field rearing was done by volunteers andstudents, under the coordination of M0narch_._Watch,at the University of Kansas.
Milkweed -- the host of the monarch caterpillar --and the monarch adults raised at each field site weregathered and analyzed for hydrogen isotopes in amass spectrometer by Leonard Wassenaar and KeithHobson of Environment Canada, Saskatoon, Canada.
Later in the same migratory cycle, 597 butterfliesthat died naturally were collected at the 13 knownwintering sites in Mexico, and put through a similaranalysis. The researchers analyzed wing membranesbecause their isotopic composition does not changeafter they are formed at the breeding site. Thus theisotopic content of wintering individuals is a goodmarker of the geographic origin of the monarchs.
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Tracking with isotopes
Page 3 of 4
The isotope results showed that about half of allmonarchs originated in the Midwestern corn andsoybean belt.
Reading rainfallIsotopic analysts "read" clues in the form of differentisotope ratios. But why are different ratios present inthe first place? The monarch researchers pin thedifference on rainfall.
Recall that the isotopes of an element are chemicallyidentical, but physically different. When waterevaporates from the ocean, it carries both hydrogenand the heavier isotope, deuterium. The heavierwater, containing deuterium, condenses and formsrain sooner than the light (regular) water. Since mostwater vapor in the atmosphere comes from oceansnear the equator, rain contains more deuterium nearthe equator than near the poles. Although mountainsand wind patterns cause local variations, in general,the further north, the less deuterium.
When plants take up water, they show a similarisotope pattern. Similarly, the butterflies that eatthose plants "inherit" this pattern of isotopes, makingit relatively straightforward to compare plants andbutterflies to determine where the butterflieshatched.
The Canadian research project was the first attemptto probe the migratory patterns of an insect withisotopes, says Wassenaar, an isotope specialist. Heand Hobson have also used the technique to track thebreeding grounds of migratory birds.
A current project looks at scaup, a waterfowl incentral North America. Scaup numbers are decliningin certain regions, but since they breed over a large,remote part of northern Canada, they are difficult totag. "It would be nice to find a way to relateindividuals that are shot to where they are comingfrom without requiring physical tagging," Wassenaarsays. By identifying the isotopic signatures ofdifferent breeding locations, scientists will be betterpositioned to understand and perhaps correct the
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Tracking with isotopes Page 4 of 4
Wry
decline.
The explosion in use of stable isotopes reflectsadvances in technique -- and machines that don't"take a Ph.D. in physics to operate," Wassenaar says.
Tracking Teeth! Desktop-sized stable isotope massspectrometers are handy in anthropology, too.
There are 12 3 4 5 pages in this feature.Bibliography I Credits ^ Feedback ^ Search
Success for All Learners: A Handbook on Differentiating Instruction
ATTACHMENT 21
Fact-Based Article Analysis
Note: The Fact-Based and Issue-Based Article Analysis sheets must be copied back to back.
When you read the article, did it present a certain point of view about an issue under dispute? If so, usethe other side of this sheet. If the article informed you but did not raise any concerns, use this side.
Key concept (written in a sentence). Write an article summary or definition inyour own words. Do not list facts. Givean overview.
Draw a figurative representation.
List at least five key words.
Relevance to today: This is important or not important because ...
Fact-Based Article Analysis: Used by permission of Lynda Matchullis and Bette Mueller, Nellie McClung Collegiate,
Pembina Valley S.D. No. 27.
6.114
Chapter 6: Teaching and Learning Strategies
Fact-Based Article AnalysisChemistry 30S
Write an article summary or definition inyour own words. Do not list facts. Givean overview.
Nylon was the first entirely synthetic fibre andit totally changed industry and everyday life.Nylon was used for many things (luggage,
toothbrushes, carpet, fishing line, surgical
sutures), but perhaps its best known use wasDraw a figurative representation. in fabrics. The popularity of nylon stockings led
to riots in the 1940s. The strength andsheerness of nylon made it the ideal choice andwas also used in parachute cloth ropes andtent fabric.
Nylon also led to other "unnatural" fibres(Teflon, polyester, Spandex) as well as workwith recombinant DNA and the biotechnology oftoday. Work with protein polymers is likely tocreate new methods of curing disease and toserve as very small machines,
List your questions (at least two).
1.What are sub-units?
2. What is recombinant DNA?
Explain the technology presented.
Polymer technology was what created nylon. Itinvolved stringing the sub-units of polymers togetherin a specific order to make a new substance. Polymertechnology led to nanotechnology, which is folding uppolymers into molecular objects which can serve aspieces of extremely small machines and electroniccomponents. With molecular devices, one could makea microscopic computer.
List at least five key words.
• rearrangement• molecular• spin-offs
Relevance to today: This is important or not important because ...
Nylon is very important today because it has created so many things and made so manychanges in our lives. It has provided a far better material for many industries and has allowedthe public to reap the benefits of these changes
no waiting at airports; stronger,sheerer stockings; affordable carpet; etc.). Nylon has also opened up the way for newmaterials and new technologies such as Teflon, po lyester, Spandex, recombinant DNA,biotechnology, and nanotechnoiogy. Nanotechnology will surely be a great device in the future.
Key concept (written in a sentence):
The introduction of nylon 50 years agohas totally revolutionized the way we live.
What are the scientific facts? List at least five.
1. Polymers are molecular chains of sub-units.
2. Nylon can be heat set to make its yarns coil and stretchmuch like telephone cords.
3. Chemists can string some of the polymer sub-units togetherin specific order.
4. Polymers can sometimes be made to fold up into molecularobjects.
5. The simple rearrangement of molecules can transform air,water, and coal into nylon.
• polymer• revolutionized• future
6.71
Name:Date:
Modelling Isotopes Lab
Purpose:
Recent Canadian pennies (after 1978) consist of three different "isotopes." In thislab, you will mass quantities of each of these types of pennies so that you candetermine both the mass of each "isotope" as well as the average "atomic mass"of a penny.
Materials:
A container of Canadian pennies
Balance
Procedure:1.
Take your pennies back to your desk and sort them into three groups-- 1996 andearlier, 1997-1999, and 2000 to the present.
2.
Record the number of pennies in each group, as well as the total number ofpennies.
3.
Determine and record the mass of 10 pennies from each group.
Adapted from: Phillips, John S., Victor S. Strozak, and Cheryl Wistrom. Chemistry: Conceptsand Applications. Ohio: Glencoe ; McGraw-Hill, 2002, p. 102.
Note:
Chem 30S--Rubric For Modelling Isotopes Lab
Total
PurposeObservations
QuantitativeIsotope
1996 and earlier1997-1999
2000-present
umber ofpeennies
(1 mark)
Mass of 10 pennies (g)
Isotope Average mass of the
Percent abundance
Percent abundance xisotope (g)
Average mass of theisotope (a.m.u.)
1996 and earlier1997-1999
2000-presentTotal
(7 marks)Calculations (Show a sample calculation for each step)
1. Determine the average mass of the penny isotopes by dividing the mass of 10pennies by 10.
2. Calculate the percent abundance of each isotope by dividing the number ofpennies from each isotope group by the total number of pennies in the container.
3. Calculate the atomic mass of a penny bya) Multiplying the percent abundance by the average mass of the isotope.b) Adding the values from a).
(4 marks)
Conclusion--Statement of the "atomic mass" of a Canadian penny.
(1 marks)Questions
1. Would the atomic mass be different if you obtained a different container ofpennies containing a different number of pre and post 1997 pennies? Explainyour answer.
2. Why did we use such a large sample size when determining the average atomicmass of each penny "isotope"?
(3 marks)Sources of error (not human errors!)
(1 mark)
TOTAL: 17 marks
Table 4.3 Natural Isotopes of Some Familiar Elements
Mass
Natural
"Average"
Name
Symbol
(amu)
percent abundance
atomic mass
1.0078
99.985
2.0141
0.015
3.0160
negligible
3.0160
0.0001
4.0026
99.9999
12.000
98.89
13.003
1.11
14.003
99.63
15.000
0.37
15.995
99.759
16.995
0.037
17.999
0.204
31.972
95.00
32.971
0.76
33.967
4.22
35.967
0.014
63.929
48.89
65.926
27.81
66.927
4.11
67.925
18.57
69.925
0.62
4.7 Atomic Mass
The mass of an atom is concentrated in the nucleus. You know that the
mass of a proton is 1 amu as is the mass of a neutron. Thus it is reason-
able to expect that the mass of an atom expressed in atomic mass units
should be a whole number. This view, however, ignores the existence
of isotopes. The atomic mass of an element is the weighted average of
the masses of the isotopes of that element. A weighted average reflects
both the mass and the relative abundance of the isotopes as they occur in
nature.The percent abundances of the natural isotopes of some familiar ele-
ments are listed in Table 4.3 along with their atomic masses. How is the
atomic mass of an element calculated? Most elements occur as two or
more isotopes in nature. For example, chlorine has two isotopes, both of
which have 17 protons in their atomic nuclei. One isotope has 18 neu-
trons and an atomic mass of 35 amu. This isotope is chlorine-35. The
Zinc
Helium
Carbon
Nitrogen
Oxygen
Sulfur
Hydrogen `H2H
3H
He
He
12C6
13C6
14N7157N160
1,08
180
8
32S16
33S16
34S16
36S16
3OZn
Zn
,3^'Zn
6'Zn70zn30
The atomic mass of an element isnot always a whole number.
Your grade in a class is often calcu -
lated as a weighted average. Majorexams, quizzes, homework assign-
ments, and laboratory work may all
have a different "weight" (or degree ofimportance) when your grade isaveraged.
1.0079
4.0026
12.011
14.007
15.999
32.064
65.37
78
Chapter 4 Atomic Structure
Ave Atomic Mass worksheet
1) Rubidium has two common isotopes, 85Rb and 87Rb. If the abundance of85Rb is 72.2% and the abundance of 87Rb is 27.8%, what is the averageatomic mass of rubidium?
2)
Uranium has three common isotopes. If the abundance of 2340 is 0.01 %,the abundance of 23SU is 0.71 %, and the abundance of 238U is 99.28%,what is the average atomic mass of uranium?
3)Titanium has five common isotopes: 48Ti (8.0%),47 Ti (7.8%), 48Ti (73.4%),49Ti (5.5%), 50Ti (5.3%). What is the average atomic mass of titanium?
4)
Explain why atoms have different isotopes. In other words, how is it thathelium can exist in three different forms?
Z 2002 Cavalcade Publishing - AN Rights Reserved
GRADE 11 CHE USTRY • Topic 3 Appendices
Appendix 3:3: Isotopes Used in Medicine and Climatology
Isotope Application Use Radiation Half-Life
Sodium-24 Medical •
to detect blood flow Beta 14.8 hradioactive constrictions and emittertracer obstructions
Iodine-131 Medical •
to test the activity of the Beta 8 dradioactive thyroid gland emittertracer
Technetium-99 Medical • to image organs such as Gamma 6 hradioactive heart, liver, and lungs emittertracer • to do 3-phase bone scans
Cobalt-48 Medical •
to determine intake of 71.3 dradioactive vitamin B12 that containstracer non-radioactive cobalt
Iron-59 •
to determine the rate of 45.6 dred blood cell formation(they contain iron)
Chromium-51 • to determine blood 27.8 dvolume and lifespan ofred blood cells
Hydrogen-3 • to determine volume of 12.3 yTritium water in person's body
• to determine the use of(labelled) vitamin D inbody
•
to conduct cellularchemistry research
Strontium-85 • to do bone scans 64 d
Gold-198 •
to do liver scans 2.7 d
Phosphorus-32 • to determine eye 14.3 ddisorders, liver tumours
Radioisotope tracers are used for diagnosis in medicine. One advantage of usingradioactive isotopes is that their particle emissions are straightforward to detect.Photographic imaging techniques or the use of devices known as scintillorneters(counters) can detect their presence even in small amounts.
Topic 3 Appendices -
PERCENTAGE COMPOSITION
Name
Determine the percentage composition of each of the compounds below.
1. KMnO4
K=
0=
2. HC1
H =
=Cl
3. Mg(NO3)2
Mg =
N =
0 =
4. (NH4)3P04N =
H =
P=
5. A12(S04)3
Al =
S-
Solve the following problems.
6. How many grams of oxygen can be produced from the decomposition of 100. g
of KCI03?
7. How much iron can be recovered from 25.0 g of Fe203?
How much silver can be produced from 125 a of Aa.,S? -
COMPOSITION OF HYDRATES
ivame
A hydrate Is an ionic compound with water molecules loosely bonded to its crystal
structure. The water Is in a specific ratio to each formula unit of the salt. For example, the
formula CuSO4.5H20 indicates that there are five water molecules for every one formula
unit of CuSO4. Answer the questions below.
What percentage of water is found in CuSO4.5H20?
2. What percentage of water is found In Na2S•9H2O?
A 5.0 g sample of a hydrate of BaCl2 was heated, and only 4.3 g of the anhydrous
salt remained. What percentage of water was in the hydrate?
4. A 2.5 g sample of a hydrate of Ca(NO3)2 was heated, and only 1.7 g of the
anhydrous salt remained. What percentage of water was in the hydrate?
5. A 3.0 g sample of Na2CO3. H2O Is heated to constant mass. How much anhydrous
salt remains?
6. A 5.0 g sample of Cu(NO3)2•nH2O Is heated, and 3.9 g of the anhydrous salt remains.
What is the value of n?
molesex4 Page I ot, I
SCH3A Grade 11 Chemistry Marvin Da Mole Strikes Again!
1. How many particles are there in one mole?
2. It is estimated that a sample of matter contains 1.38 X 102' atoms. How many moles are present in thesample?
3. How many moles of barium are present in a sample having a mass of 22.3 grams?
4. A chemical reaction requires 3.7 moles of boron. What mass, in grams, of boron must be used in the reaction?
5. A sample of naturally occurring carbon has a mass o f 1.732 grams. Calculate the number of moles of carbonin this sample.
6. A chemical reaction results in 57.2 grams of the gas carbon dioxide, CO2. How many molecules of gas were
produced?
7. Calculate the mass of one trillion molecules of oxygen, 02.
8. Calculate the number of moles in:a) 25 grams of oxygen, 02
b) 0.27 g of ammonia, NH3
c) 10.5 g of sodium, Nad) 347 g of ammonium nitrate, NH4NO3
9. Calculate the mass, in grams, of.a) 1.24 moles of water, H2O
b) 0.269 moles of amonium chloride, NH4CI
c) 5.62 moles of sodium hydroxide, NaOHd) 2.35 moles of sodium sulphate, Na2SO4
10. Calculate the number of molecules in:a) 3.00 moles of chlorine, Cl2
b) 3.00 moles of uranium hexafluoride, UF6
c) 3.00 moles of hydrogen chloride gas, HCId) 3.00 moles of any kind of molecule
11. Calculate the number of atoms in:a) 3.00 moles of chlorine, C12
b) 3.00 moles of uranium hexafluoride, UF6
c) 3.00 moles of hydrogen chloride gas, HCId) 3.00 moles of ammonium sulphate, (NH4)2SO4
Return to the Notes
molesex3
SCH3AI Grade I1 Chemistry Marvin Da Mole! Our Hero
1. Calculate the mass of
a) 2.00 moles of water, H2O
b) 4.38 moles of chlorine, Cl2
c) 0.025 moles of ammonia, NH3
d) 1.8 moles of oxygen, O2
2. Calculate the number of moles in
a) 25 g of helium, He
b) 12.5 g of methane, CH4
c) 0.364 g of iodine, 12
d) 40.0 g of sodium, Na
3. Calculate the number of particles in
a) 2.50 moles of Neon, Ne
b) 0.050 moles of iron, Fe
4. Calculate the number of moles in
a) 9.03 X 1023 atoms of Cu
b) 3.76 X 1025 molecules of SO2
c) 8.6 X 1018 electrons
5. Calculate the number of molecules in
a) 12.5.g of nitrogen, N2
b) 0.76 g of ammonia, NH3
c) 0.60 g of hydrogen, H2
6. Calculate the mass of
a) 4.25 X 1024 atoms of C
b) 6.02 X 1021 molecules of H 20
c) one trillion atoms of Zn
d) one atom of U
Return to the Notes
3/22/00
Chemistry 30S Mole Calculations Lab
ObjectiveSolve problems requiring interconversions between moles, mass, and number of particles
Procedure1. In groups, students will answer one of the following questions showing all their
work, including units.
2. Once each group member has calculated and recorded their answers to thequestion they must have their calculations inspected by the teacher who will thenpermit them to weigh out the appropriate amount of substance.
3. After weighing, each group must have their substance inspected by the instructor.
4. The group must then return their massed substance to the original container andreplace any equipment in their lab cupboard.
Conversions & Calculations1. Mass out 2.01 x 1022 atoms of zinc.
Student observesStudent does not Student is nagged
All aspects of theproper safety
wear their safety once or more aboutlab are conducted in
precautionsgoggles or has a lab not following proper
a safe manneraccident safety precautions
Student followsEvidence that the
Evidence that thewritten and student has not
student has not Student follows all
verbalfollowed more than
followed one writtenwritten or verbal
instructions one written or verbalor verbal instruction instructions
instruction
Evidence ofLittle or no evidence Some evidence that a Evidence that all
equitableof equitable group member group members
participation participation by a participated more participatedgroup member than another equitably
More than one pieceEvidence of Entire station is
Lab station is of evidence remainschemicals or a piece cleaned up with no
cleaned up after of chemicals orof equipment remains evidence remaining
lab equipment afterafter cleanupp
of equipment orcleanup chemicals
5:6 Empirical Formulas
87
EMPIRICAL FORMULA OF A COMPOUND40.0% Carbon, 6.71% Hydrogen, 53.3% Oxygen
Percent l Mole ratioEmpirical
formula
C 40.0%
6.71%CH Ow^
VH
0 53.5%
FIGURE 5-4. This picture repre-sentation shows the steps Incalculating an empirical formulafrom percentage composition.
34. Cei3
EXAMPLE: Empirical Eor puiaWhat is the empirical formula of a compound which is 66.0% Caand 34.0% P?Solving process:Assume a 100-g sample so that we have 66.0 g Ca and 34.0 g P. Con-vert these quantities to moles of atoms.
.66.0
1 mol Ca = 1.65 mol Ca40.1 g .Ga
34.0 g.P 1 1 mol P = 1.10 mol P( 31.0$P
Dividing both results by 1.10, we obtain 1.50 to 1. This result isnot close to a whole number. Substituting the fractional form of1.5, we get 3/2. That ratio is 3 to 2. Thus, the ratio of Ca atoms toP atoms is 3 to 2 and the empirical formula is Ca3P2. Supposewe have an empirical formula problem which produces a ratio of1 to 2.33. What is the correct whole number ratio? We can say2.33 = 21/3. Since 21/3 is 7/3, the ratio is 7 to 3.
PROBLEMSFind the empirical formulas of the following compounds.34. 1 .67 g Ce, 4.54 g I35. 31 .9 g Mg, 27.1 g P
36. 4.04 g Cs, 1.08 g Cl
37. 9.11 g Ni, 5.89 g F
38. 6.27 g Ca, 1.46 g N
DETERMINING
Name
EMPIRICAL FORMULAS
`,4hat is the empirical formula (lowest whole number ratio) of the compounds below?
1. 75% carbon, 25% hydrogen
2. 52.7% potassium, 47.3% chlorine
3. 22.1% aluminum, 25.4% phosphorus, 52.5% oxygen
. 13% magnesium, 87% bromine
5. 32.4% sodium, 22.5% sulfur, 45.1% oxygen
6. 25.3% copper, 12.9% sulfur, 25.7% oxygen, 36.1 % water
5:8 Hydrates
Tim Courlas
m experi-we need
ne of thealculated)mpoundiolecularnolecule.the ratioila. Thenit in oneunit has
e molec-
FIGURE 5-5. The difference incolor between the anhydrousand hydrated forms of a com-pound are shown. Heatingdrives off water molecules whichcauses the color change.
formula
d, if the29?
e, if thes 184.5?
mpirical
-centage
2.1.
a waters of theintain aL to dryound tot showsIla unit.s of theiply the' NiS03
EXAMPLE: Hydrate CalculationWe have a 10.407 gram sample of hydrated barium iodide. The
sample is heated to drive off the water. The dry sample has a
mass of 9.520 grams. What is the ratio between barium iodide,
Ba12, and water, H20? What is the formula of the hydrate?
Solving process:
The difference between the initial mass and that of the dry
sample is the mass of water that was driven off.
10.407 - 9.520 = 0.887 g
The mass of water and mass of dry Ba12 are converted to moles.
9.520
( 1 mol BaI2 =391.2
0.024 34 mot BaI2
0.887 Z_%9-
1mol H2O = 0.0492 mol H2O
18.0 gJi &
The ratio between Ba12 and H2O is seen to be 1 to 2. The
mental data. In order to calculate a molecular formula, we needone additional piece of data, the molecular mass. In one of theexamples in the previous section, the empirical formula calculatedwas CH2O. If we know that the molecular mass of the compoundis 180, how can we find the molecular formula? The molecularformula shows the number of atoms of each element in a molecule.Knowing that the elements will always be present in the ratio1:2:1, we can calculate the mass of the empirical formula. Thenwe can find the number of these empirical units present in onemolecular formula. In the substance CH2O, the empirical unit hasa mass of
12 + 2(1) + 16, or 30.
It will, therefore, take six of these units to equal 180 or one molec-ular formula. Thus, the molecular formula is C6H1206.
PROBLEMS
39. The molecular mass of benzene is 78 and its empirical formulais CH. What is the molecular formula for benzene?
40. What is the molecular formula of dichloroacetic acid, if theempirical formula is CHOCl and the molecular mass is 129?
41. What is the molecular formula of cyanuric chloride, if theempirical formula is CC1N and the molecular mass is 184.5?
42. What is the molecular formula of a substance with empiricalformula TIC2H2O3 and molecular mass 557?
43. Find the molecular formula for a compound with percentagecomposition 85.6% C, 14.4% H, and molecular mass 42.1.
5:8 HYDRATESThere are many compounds which crystallize from a water
solution with water molecules adhering to the particles of the
Hydrates are crystals which con-
crystal. These hydrates, as they are called, usually contain a
tain water molecules, specific ratio of water to compound. Chemists use heat to drythese compounds and then calculate the ratio of compound towater. An example of a hydrate is NiS03.6H2O. The dot showsthat 6 molecules of water adhere to 1 molecule of formula unit.To calculate the formula mass, we add the formula mass of thecompound and water. For NiSO3 we obtain 139. We multiply the18 for water by 6 and add to the 139. The formula mass of NiS036H2 0 is then 139 + 6(18), or 247.
Molecular mass is a whole num-
ber multiple of the empiricalformula mass.
39. C6H6,
40. C2H2O2C12
41. C3C13N3
42. T12C4H4O643. C3 H6
E
fo
P
Fi
44
4E
4E
47
4E
DETERMINING- MOLECULAR
Name
FORMULAS (TRUE FORMULAS)
Solve the problems below.
1. The empirical formula of a compound is NO2, Its molecular mass Is 92 g/mol.What is its molecular formula?
2. The empirical formula of a compound is CH2. Its molecular mass is 70 g/mol,What is its molecular formula?
A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen.Its molecular mass is 60. g/mol. What is its molecular formula?
4. A compound is 64.9% carbon, 13.5% hydrogen and 21.6% oxygen. Its molecularmass is 74 g/mol. What is its molecular formula?
5. A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molecularmass is 88 g/mol. What is its molecular formula?
FORMULA OF A HYDRATE LAB
OBJECTIVES1. Experimentally determine the percent composition by mass of water in a hydrate.2. Calculate the formula of a hydrate.3. Verify the Law of Definite Composition.
CHEMICALCopper (II) sulfate hydrate (A.K.A. bluestone)
SAFETY• Wear eye protection over eyes• Keep a safe distance from hydrate when heating• Copper sulfate is toxic, rinse affected areas with copious quantities of water• Observe appropriate fire precautions• Clean up all spills immediately• Wash hands after performing the lab
EXPERIMENTAL SETUP
DATA
1. mass of clean dry crucible2. mass of blue hydrate3. mass of dried salt and cool crucible4. mass of dried salt5. mass of water lost
OBSERVATIONSSight, sound, and other sensory data (minimum of 3)
FORMULA OF A HYDRATE LAB
CALCULATIONSYour calculations must follow a logical and correct mathematical sequence, showing all stepsand units. When you are finished, click here to check your calculations. If you missed the lab,click here for sample data.1. Percent composition of water by mass.
2. Moles of water.
3. Moles of dried salt.
4. Mole ratio of water to dried salt.
5. Experimental error of percent composition of water by mass.
GRAPHING1. Construct a graph of moles of water versus moles of CuSO4 in the hydrate from the data
sheet provided by the instructor.2. Calculate the slope of the line of best fit. Include complete calculations including units on
the graph.
CONCLUSIONSummarize the results of the experiment by answering the 3 objectives. Use the hydrate labrubric to evaluate the soundness of your conclusion.
HAND IN title page, prelab, calculations, graph, and conclusion.
FORMULA OF A HYDRATE LAB
RELATIONSHIP BETWEEN THE MOLES OF WATER AND DRIED SALT IN A COPPER (II) SULFATEHYDRATE ILLUSTRATING THE LAW OF DEFINITE COMPOSITION OF MASS
LJ I I^T I I I I 111 11 I-F
FORMULA OF A HYDRATE LAB
NAME & PARTNERS: LAB GROUP NO.:
Pre Lab, Calculations, and Graphing Rubric (6 marks)
Criteria Novice Intermediate ExpertPrelab has no additions Prelab contains all parts of the labfrom the lab outline (0)
Prelab contains 3 of 4 expert outline, including labelled sketch ofPrelab or contains less than 3
level criteria (1.5)experimental setup, all recorded masses,
of the expert level and a minimum of 3 relevantcriteria (1) observations (2)
No calculationsCalculations section contains a Correct answers to all 5 calculations
Calculationssubmitted (0) or section
minimum of 3 of five criteriasubmitted on a separate page, follow a
contains I or 2 criteriafrom expert level (1.5)
correct mathematical sequence, showingfrom expert level (1) all units (2)
Graph contains a descriptive title,Graph is missing (0) or
Graph contains at least 4 of six labelled axes, including units andGraphing less than 4 expert level
expert level criteria (1.5) appropriate scale, correctly plottedcriteria are present (1) coordinates, a ruled trend line, and a
correctly calculated slope (2)
Lab Report Conclusion Rubric (4 marks)
Criteria Novice Intermediate Ex pert
Conclusion isAspects of the
originalconclusion are the same
(no other criteriaas those of anotherstudent or the entire
Intermediate level does not apply Conclusion is written in the student'sare evaluated if this
conclusion is the sameto this assessment criteria own words
criteria is assessed at as that of anotherthe Novice level)
studentNot all objectives are
ObjectivesObjectives are not summarized, or objectives are
Objective(s) is/are summarized orsummarized
summarized or clarified simply recopied using the sameclarified in the student's own words (1)
(0) wording as the statedobjective(s) (0.5)
Each objective is correctly andNo conclusion present concisely answered in paragraph format
Objectives answered (0) or conclusion isConclusion contains all butut two results that are correct, including
by referencing missing more than twoof the expert level criteria (1.5)
units and associated percentresults of the expert level experimental error, and reference
1. In the reaction shown here. what weight of iron is needed to react com-pletely with 32.0 g of sulfur? Fe
FeS
2. When zinc reacts with sulfuric acid, as shown here. what weight of hydro-gen is produced from 31.8,9 of zinc? Zn. + H2SOa -- ZnSO 4 -+- H2,1
3. How much sulfurous acid can be pr-oduced.when 128 gof sulfur.dioxidecombines with water? SO2 4- H2O H2S03
4. Silver bromide can be precipitated by the reaction_ of silver nitrate withsodium bromide. What weight of-precipitate can be produced startingwith 34.3 of sodium bromide?-. NaBr + AgNO, - NaNO3 + AgBr I
5. Hydrochloric acid is added to 50.0 g of iron (II) sulfide. What - weight of,hydrogen sulfide is produced? FeS 2 HCI - FeCI2 + H2S t
6. How much nitric acid is'needed to react completely with 25.Og:of magne-sium in the following reaction? Mg -r 2 HNO3 - Mg((NOs)2 +=H2 f
7. How much copper (I) chloride can be produced beginning with 75-09 ofcopper (I) oxide?' Cu20 + 2 HCI - 2 CuCl + H20.
8. What volume of oxygen gas is produced by the decomposition of 1001.0 g32.4 L. of sodium nitrate? 2 NaNO3 - 2 NaN02 + 02 1
9. What volume of oxygen is produced when 75. O g of water is decomposedby electrolysis? 2 H2O -- 2 H2 + 02. t.
10. What volume of carbon dioxide is required to produce 50.0 L of carbonmonoxide according to the following reaction? CO2 + C -- 2 CO t
Copyright 1984. Instructional Horizons Inc.Published by J. Weston Watch Publisher. Portland Maine 041040658
StoichiometryPractice Problems (Level 2)
1. When aluminum is heated in oxygen, aluminum oxide is formed. Whatweight of the oxide can be obtained from 25.0 g of the metal?
2. When steam (hot water) is passed over iron, hydrogen gas and iron (111)oxide are formed. What weight of steam would be needed to react com -pletely with 100.0 g of iron?
3. How much ammonium hydroxide is needed to react completely with 75.0g of copper (II) nitrite in a double replacement reaction?
4. When ammonia is burned in oxygen, free nitrogen gas and water areproduced. What volume of ammonia will react, completely with 25.0 L ofoxygen? What volume of nitrogen gas is formed?
5. When sodium carbonate reacts with hydrochloric acid, the carbonic acidthat is formed immediately breaks down into carbon dioxide and water.What weight of sodium carbonate would have been present originally it5.0 L of carbon dioxide were obtained in this way?
6. How much copper metal can be obtained by the single replacementreaction between copper (I) nitrate and 30.0 g of iron metal? (Iron (IIInitrate is formed.)
7. What weight of $utfuric acid will be needed to react completely with 35.5 gof ammonia, in the production of ammonium sulfate?
What weight of chlorine gas will be needed to react completely with 8544of potassium iodide in a single replacement reaction?
9. In the neutralization reaction between sulfuric acid and potassium.hydroxide, how much potassium sulfate can be produced if you have150.0 g of sulfuric acid to begin with?
10. What volume of nitrogen gas is needed to react completely with 150.0 L ofhydrogen in the production of ammonia?
Copyright t 984. Instructional Horizons. Inc.Published by J. Weston Watch. Publisher. Portland. Maine 04104-0658
GRADE 11 CHEMISTRY ' Topic 3 Appendices
Appendix 3.10: Gas Density Table (Student Resource Material)
Density of Gases at 25°C and 101.3 kPa (760 mmHg or 1.0 atm) Pressure
Name
Ammonia
Argon
Butane
Carbon dioxide
Carbon monoxide
Dichlorine
Ethane
Ethene
Ethyne (acetylene)
Helium
Dihydrogen
Hydrogen chloride
Hydrogen iodide
Krypton
Methane
Neon
Nitrogen monoxide
Dinitrogen
Dinitrogen monoxide
Nitrogen dioxide
Dioxygen
Ozone
Propane
Sulphur dioxide
Xenon
Formula
NH3
Ar
C4H 10
CO2
CO
C12
C2H6
C2H4
C2H2
He
H2
HCI
HI
Kr
CH4
Ne
NO
N2
N2O
NO2
02
03
C3H8
SO2
Xe
Molar Mass (g/mol)
17.03
39.944
58.12
44.01
28.01
70.91
30.07
28.05
26.04
4.003
2.016
36.47
127.93
83.70
16.04
20.18
30.01
28.02
44.02
46.01
32.00
48.00
44.09
64.07
131.30
Density (g/L)
0.696
1.633
2.376
1.799
1.145
2.898
1.229
1.147
1.064
0.164
0.082
1.490
5.228
3.425
0.656
0.825
1.226
1.145
1.799
1.880
1.308
1.962
1.802
2.618
5.367
Topic 3 Appendices - 35
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Business Proposal to Mr. Mole
Introduction: Now that you have learned conversions and stoichiometry inchemistry, you will be able to apply these concepts in a business proposal. Yourtask is to create a business proposal to Mr. Mole who is a venture capitalist.There are some things about Mr. Mole you should know before you begin. Mr.Mole does not like waste (money, products etc.). Mr. Mole understandsstoichiometric terms but does not like to talk about chemicals. Mr. Mole isimpressed by stoichiometric calculations, and most of all Mr. Mole likes to makemoney.
Your task:Come up with a product and a business model that can be expressed as anequation. The components of your product are the reactants and your product iswhat is produced when all reactants combine in the appropriate ratios. At leastone of the components in your equation must be a number other than one. Besure to include profits and expenses in your equation.
Your detailed business proposal must include:3 A persuasive cover outlining your product and business model
3 Supporting argument(s) to convince Mr. mole to finance your proposal3 A business equation describing all reactants and products3 Supporting data tables (include potential profits and expenses)3 Sample calculations of formulas used in the data table
Business ideas for your proposal:â Manufacturing of a productâ Cooking/ food serviceâ Agricultureâ Miningâ Other (Your choice)
ChecklistUnique or novel product is presented in the cover letter using effective
persuasive languageSupporting arguments in the cover letter make sense and are highly
influentialBusiness equation is present and clearly outlines all the variables present
in the cover letterA business equation with ratios more complex than 1:1 is present and all
variables are present in the cover letterSupporting data table contains all variables stated in the equation
Business proposal is concise, organized so it is easy to read, and wordprocessed
Business Proposal RubricCriteria Novice (0-1) Intermediate (2) Exert (3)
Unique product ispresented in the
Cover letter clearly None (0) or one of Only two of threecover letter using
effective persuasiveoutlines the product the expert levelexpert criteria met language withand the business criteria met (1)
influentialsupportingarguments
Business equation A business equationBusiness equation
is present andwith ratios more
Business equation is missing (0) orcontains more or
less variables thancomplex than 1:1 is
contains relevant business equation isstated in the cover
present and allinformation not referred to in
letter of businessvariables are
cover letter (1)equation contains present in the cover
only 1:1 ratiosletter
No supporting datatable (0) or data Supporting data
Supporting datatable has little or no table contains moretable contains allSupporting data connection to the or less variables
variables stated intables have correct variables in the than stated in thethe equation or theinformation business equation equation or thecover letter (if theor cover letter (if cover letter (if the
equation is missing)equation missing) equation is missing)
(1)Sample calculations
Only one criteria show all work
Sample calculationsmet (1) or sample Only two of three according to the
data x relationship =calculations are criteria are metanswer formulamissing (0) ,include units and
are correctProposal evokes
Business proposalfeelings of
Aestheticsembarrassment in Only two criteria is concise,
organized so it isthe reader (0) or meteasy to read, andonly one criteria isword processedmet (1)
/15
Perhaps you should consider
T
a company logo and letter head?
222-2222 •
1204} 222-32
My Really Exciting & Creative Company's Address(do not use this format if you have incorporated an address into a logo or letter head)
Here is where you sell your product to Mr. Mole. A neat, word processed document isyour best bet. You have to make it really desirable for him to give you money for yourbusiness proposal. The best way to do this is to develop a creative product and use yourbest persuasive language in this cover letter.
You only have a page to convince him in writing that your product is exactly what he islooking for. At about this point in the cover letter you have to inform him of each of thecomponents in your business equation. Another decision that you must make is whetherto ask Mr. Mole for a specific sum of money. Be careful in your assumptions! The bestdecision would probably be to wait and only provide your ideas about specific dollars inthe page(s) that follow because you don't know what amount Mr. Mole is willing to fundfor your venture and you don't want to insult him or turn him off by suggesting a specificsum of money that he has no intention of lending to you.
Likely at this point you should be wrapping up your cover letter by making a fewflattering statements to Mr. Mole and then inviting him to contact you to further discussyour proposal. If you follow this advice, you will have the best chance of success in yournew business venture. Good luck and happy composing!
Select a salutation and insert it here,
Insert your signature above this line (real or otherwise)Write your matching name here (and include your real name if you have used apseudonym)
Perhaps you should considerl 92a company logo and letter head?
1234 222-2222 •
234; 222-3333 „Ydeal
Your super creative genius product in the form of a business equation:
2 g fuel + fabric bunny shell + fuse + 3 kg stuffing + $14 3 1 bunny blast plush toy + $28
Now I have selected to show my overhead, cost, selling price, and profit information:
BunnyBlasts
FuelCost
ShellCost
FuseCost
StuffingCost
TotalCost
SellingPrice
Profit
10 $40 $20 $20 $60 $140 $280 $140
10,000 $40 k $20 k $20 k $60 k $140 k $280 k $140 k250,000 $1 M $500 k $500 k $1.5 M $3.5 M $7 M $3.5 M
*Note: prefix k represents thousands of dollars; M represents millions of dollars.
Here I have selected to show my sample calculations for one row, including each part ofthe business equation, using the Data x Relationship = Answer method. Notice that unitsare included in every step.
Costs for 10 Bunny Blasts:
Item Unit Cost Unit AmountFuel $2/g x
2g/BB x
10 BB
= $40Shell $2/shell x
1 shell/BB x
10 BB
= $20Fuse $2/fuse x
1 fuse/BB x
10 BB
= $20Stuffing $2/kg x
3 kg/BB x
10 BB
= $60TOTAL $140
*Note: BB means bunny blasts
Now that I have presented my equation and numbers to Mr. Mole, it may be fitting topresent something to conclude the proposal. Whatever you chose to write or show,remember that it should leave a lasting impression on Mr. Mole so that he will want tofollow up with your proposal and contact you. Good luck!
