Download MISE - Physical Basis of Chemistry

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MISE - Physical Basis of Chemistry
First Set of Problems - Due October 16, 2005
Submit electronically (digital drop box) by October 16 – by 6 pm.
Note: When submitting to digital Drop Box label your files with your name first and
then the label - HmwkOne.
• Since this is a Word .doc, you may type in your solution to each problem below the
problem itself – maybe a smaller font size smaller (Times 10 pt. maybe – since this is
Times 12 pt.). Then save the document titled as suggested. Please don’t forget the .doc.
• Some of you may be wondering how to easily submit solutions involving mathematical
expressions. Here are some options.
I. Many recent versions of Microsoft™Word have an “Equation Editor™” option.
To get to this, go to the pull-down ‘Insert” Menu Option ‘ and then to the ‘Object…’
choice that is near the bottom of the list. When you do this, a window will open in
which you can type an equation – using the symbol palette included in the window.
In order to insert the equation back into the document, go to the ‘File’ menu in
Equation Editor™and choose ‘Update’ (keystroke ‘  U ’ on the MAC) inserting
the completed equation into the document. If you want to edit an equation already
inserted, merely double-click on the equation and the Equation Editor™ window will
re-open with the equation, ready for editing
II. You may type the appropriate text merely using your keyboard as effectively as possible
and then include explanatory text to explain any aspects of the calculation that may
seem ambiguous when displayed. If you are typing directly from the keyboard
(not using the “Equation Editor™”), here are some useful keystrokes (if you
are using Word with a MAC).
Desired Symbol
MAC keystroke in Word
∑ (sigma => summation)
Option (or Alt) and w
π (pi)
Option (or Alt) and p
∆ (delta => change)
Option (or Alt) and j
≈ (approx. equal to)
Option (or Alt) and x
√ (check mark or square root) Option (or Alt) and v
∫
(integral sign)
Option (or Alt) and b
µ (mu)
Option (or Alt) and m
≤ (less than or equal to)
Option (or Alt) and , (comma)
≥ (greater than or equal to)
Option (or Alt) and . (period)
∞ (infinity)
Option (or Alt) and 5
º (degree sign)
Option (or Alt) and 0 (zero)
≠ (not equal to)
Option (or Alt) and =
÷ (alternate division symbol)
Option (or Alt) and /
±
(plus-minus)
Option (or Alt) and Shift and =
Of course, switching to Symbol font will allow any Greek alphabetic symbol to be entered.
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Let’s take an example:
When the water is frozen solid (ice), in a refrigerator freezer, its density is 0.917 g/mL.
Suppose that a can is to be constructed in which exactly 355 grams of ice will just fit.
(a) What must be the volume of the can in milliliters?
(b) What must be the height (in inches) of this can – assuming that it is cylindrical with
a circular base of radius 0.75 inches?
Some Info that may be useful for the problem – (See also Supplementary Topic #1 document):
1 mL = 1cm3 ; 2.54 cm = 1 inch (exactly); Volume of cylinder = π(radius)2(height).
Solution to Example:
Method I – Using Equation Editor™ (part of Microsoft™ Word) as needed:
(a) Volume of can = volume of ice that exactly fits.
Volume of ice (or can) =
(insert equation typed in Equation Editor) =
mice
355 g ice


0.917 g ice 
dice


 mL ice 
= 387. mL ice and so the required volume of the can.
(b) We can first convert the volume
 of the can to cubic inches and then use the provided formula
from geometry to find the radius.
3
Vcan(in ) =

1 cm 3  13 in.3 
3
= 23.6 in .
(387. mL)

3
3 
1
mL
2.54
cm



Volume of can
= height of can
 r2

23.6 in 3
Volume of can
height of can (in.) =
=
= 13.4 inches
(0.75 2 )in.2
r 2
Method II – Typing directly from keyboard – not using Equation Editor™:
(This method necessitates careful use of parentheses…)


(a) Volume of can = volume of ice that exactly fits.
Volume of ice (or can) = mice/dice = (355 g ice) ÷ (0.917 g ice/mL ice)
= 387. mL ice and so the required volume of the can.
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(b) We can first convert the volume of the can to cubic inches and then use the provided formula
from geometry to find the radius.
3
Vcan(in ) = (397. mL) (1 cm3 / 1 mL) (13 in.3 / 2.543 cm3) = 23.6 in.3 .
(Volume of can) / (π r2) = height of can
height of can (in.) = (Volume of can) / (π r2) = (23.6 in.3) / [π (0.752 in.2)] = 13.4 inches
Homework Problems
1. An English unit of mass used in pharmaceutical work is the grain. There are
15.43 grains (1/7000 pound) in one gram. A standard aspirin tablet contains 5.00 grains
of aspirin. If a 165 pound person takes two aspirin tablets, calculate the dosage
expressed in milligrams of aspirin per kilogram of body weight.
[Note: 1 pound = 453.6 grams]
10 grains
1lb
1gram
1000 g



165lbs
453.5 grams 15.43grains
1kg

1000mg 10000000mg

 8.659186307mg / kg  8.66mg / kg
1g
1154842.92kg
2. Working with the “triangle” :
(Ratio of combining masses) = (Ratio of subscripts)•(Ratio of Atomic Weights)
Preamble: A useful quantity when working with combining masses, is
mass (weight) proportion or percent. In symbols:
Mass % of “X” in a sample =
mass of X in sample
• 100 %.
mass of sample
The units of mass are up to you, as long as the same units are used in the
numerator and in the denominator. Sometimes this unit is used to express
the (constant) mass proportion of a specific element in a compound containing
this element.
Pretend that you are an assistant to John Dalton and that you have just determined that two
as-yet-unknown elements, X and Z, form four different binary compounds, labeled I, II, III,
and IV in the table below. The elemental mass percent composition of X is next to each
compound.
Compound
Mass % of X
I
61.37
II
51.44
III
44.27
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IV
38.86
(a) Applying the law of simplicity to compound I, what is the relative atomic weight of X to
Z, i.e., what is the ratio of the atomic weight of X to that of Z?
Mass % of Z
38.63
48.56
55.73
61.14
p AW ( X ) 61.37


 AWrelative
q AW ( Z ) 38.63
Compound I:
X ( g ) 61.37

 1.588661662  1.59ratio
Z ( g ) 38.63
(b) Using your result from (a), determine the simplest (empirical) formula for each of the
compounds II through IV. Why is this the best that you can do?
X ( g ) r 1.59
Compound II:
 
Z (g) s
1
51.44 r 1.59
 *
48.56 s
1
51.44
r
6.66 3.33 3
  .6662314921  6.662314921 *10^ 1 

  X 3Z5
(1.59 * 48.56) s
10
10
5
X t 1.59
 *
Compound III:
Z u
1
44.27 t 1.59
 *
55.73 u
1
44.27
t
5.00 1
  .499601064  4.99601064 *10^ 1 
  X 1Z 2
(1.59 * 55.73) u
10
2
Compound IV:
X 38.86 1.59 m
3.997 4.00 2

*
  .3997424202  3.997424202 *10^ 1 

  X 2Z5
Z 61.14 1
n
10
10
5
(2c) Through painstaking labor - and some luck - you determine that the atomic weight of Z
(relative to unity for hydrogen) is about 32. Using this value for the atomic weight of Z,
determine the atomic weight of X. Referring to a modern periodic table of the elements
(which also lists hydrogen with a relative atomic weight of about 1), determine the identity
of elements X and Z and specify the chemical symbol of each.
AWx
 1.59
AWz
AWx
 1.59
32
AWx  50.88  Vanadium,V
AWz  32  Sulfur , S
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(2d) Given the chemical symbol of X and Z determined in (d), rewrite the chemical formulas
determined in (c) in terms of these symbols. Does knowing the identity of elements X
and Z allow you to determine the “true” chemical formula of compounds I, II, III,
and/or IV? Explain why or why not.
I X1Z1 = VS
II X3Z5 =V3S5
III X1Z2 =VS2
IV X2Z5 =V2S5
I think that knowing the identity helps to determine the chemical formula, although the sample X, is an
approximate identity since our answer is 50.88 and Vanadium is 50.94. But it helps with knowing the
subscripts and being able to write the chemical formulas we determined.
3. Deducing the mole … the chemist’s “dozen”…
Up to now, we’ve been talking about relative atomic weights and we have been
working in ratio - using the “triangle”. Since individual weights appear in the
periodic table, there has to be a mass standard, i.e., a reference mass - so that
the ratio of atomic weights can become individual values. Since hydrogen was
believed to be the lightest element , H was assigned the weight of “1” and all
other atomic weights were determined relative to the ratio with hydrogen. A lot
of history intervened - such as isotopes, i.e., atoms of the same element could
have different masses, etc. The bottom line is that the reference atomic weight
was changed. For our purposes, only the following is relevant:
The reference atom (isotope) was a particular isotope of carbon (C), i.e., “carbon-12”.
It was symbolized as: 12C. The mass of one atom of this particular carbon atom
was defined as exactly 12.0000…. atomic mass units (amu). So, the conversion
factor is: 1 atom of 12C = 12.00… amu.
This means that the atomic mass unit (amu) is equal to 1/12 of this mass.
In grams, the amu is: 1 amu = 1.661 x 10-24 g.
With this standard and the appropriate atomic weight ratios from experiment, the
remainder of the atomic weights were determined. The atomic weights listed
in the periodic table are individual atom weights in amu. [Actually, each
listed atomic weight is an average over the various isotopes, but we don’t have
to worry about that for the time being. This is why the atomic weight of carbon
in the periodic table is not 12.00… but 12.011.] Back to the story.
This means that any atomic weight listed in the periodic table (O = 16.00,
Mg = 24.31, S = 32.06, and Cu = 63.45 are the individual atom masses on the amu scale:
Atomic Weight of O = 16.00 amu = mass of 1 atom of O.
Atomic Weight of Mg = 24.31 amu = mass of 1 atom of Mg.
Atomic Weight of S = 32.06 amu = mass of 1 atom of S.
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Atomic Weight of Cu = 63.54 amu = mass of 1 atom of Cu.
The problem is that these masses are so small. [Using dimensional
analysis, determine the mass of 1 Mg atom in grams.] The chemist’s
next mission was to “super-size” the atomic weight scale so that the
listed atomic weights would prove useful for weighing things out on
laboratory balances.
Question: Can an atomic weight scale be developed - related in a simple way to the
existing one - such that the listed numerical values can still be used?
The answer is yes and it is simply an exercise in dimensional analysis! The question
can be re-stated as follows:
If one atom of 12C weighs (exactly) 12.00 amu, then how many atoms of
C will weigh exactly 12.00 grams?
12
12 * 1.661*10^-24g= 1.9932*10^-23g
1
 6.02 *10^ 23, Avagrado' snumber
1 mol = 12g * 1.9932 *10^ 23
In other words, how many atoms (of any element) are required such that
the numerical value of the listed atomic weight in amu can be replaced
the same number - but in grams? Using dimensional analysis, and the
above conversion factors as needed, determine how many atoms of
12C will weight 12.00 g. This number - labeled the mole is the
chemist’s “dozen”. It is a convenient counting number so that any
atomic weight (AW) can be interpreted in two (2) ways.
For atom X, the listed atomic weight of atom X (AWX) can be interpreted as:
•
mass of 1 atom of X in amu
•
mass of 1 mol (of atoms of) of X in grams.
OR
This will be a convenient pair of conversion factors for future use.
Just remember, the mole (abbreviated as mol) is not a magical
number - any more than a dozen. It is just convenient - allowing us
to use the listed number for each atomic weight in two ways.
Since we believe in conservation of mass, it is hopefully easy to see that:
Molecular Weight (MW) of a compound = sum of AW’s of constituent atoms.
Then, it is also hopefully evident that the MW can also be interpreted in two equivalent
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ways: MW = mass of 1 molecule of a compound in amu = mass of 1 mol (of molecules).
With these relationships between grams of an element or compound
and the number of atoms or molecules present, we are finally at the point
that Dalton and his colleagues could have only dreamed!
4. Consider the following four (4) molecules:
• Water (H2O, a liquid at room temperature with a density of about 1.00 g/mL).
• Octane - major component of gasoline
(C8H18 , a liquid at room temperature with a density of 0.703 g/mL).
• Nitrous oxide – a.k.a. “laughing gas”
(N2O , a gas at room temperature with a density of 1.80 g/L ; note, per Liter).
• Nitrogen dioxide – a highly poisonous gas
(NO2 , a gas at room temperature with a density of 1.88 g/L ; note, per Liter).
(a) Imagine a 2.40 L sample of each of the compounds. Determine what mass (in g) of
each compound would be present.
Water
1000ml
 2400ml *1.00 g / ml  2400 g  themassofwater
2.40L *
1L
Octane
m
2400ml * 0.703g/ml=
*2400ml=1687.2g=mass of Octane
2400ml
Nitrous oxide
m
* 2.40 L  4.32 g  massofNitrousoxide
2.40L * 1.80g/L=
2.40 L
Nitrogen dioxide
m
* 2.40 L  4.512 g  massofNitrogendioxide
2.40l * 1.88g/L = 2.40 L
(b)
How many molecules of each of the compounds would be present in the 2.40 L
samples?
Water
2 H, 1 O= 3 atoms 2amu=H
16amu=O total= 18amu of H2O
1.661 *10^ 24
 2.9898 *10^ 23
18amu *
1amu
2400 g
 8.0272928 *10^ 21 molecules of H2O
2.9898 *10^ 23
Octane
8 C, 18 H= 26 atoms 96 amu=C
18amu=H total= 114amu of C8H18
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1.661810^ 24
 1.89354 *10^ 22
1amu
1687.2 g
 8.910295003 *10^ 24molecules
1.89354 *10^ 22
114amu *
Nitrous oxide
2 N, 1 O=3 atoms
28amu=N
16amu=O total= 44amu of N2O
1.661 *10^ 24
 7.3084 *10^ 23
44amu *
1amu
4.32 g
 5.911006513 *10^ 22molecules
7.3084 *10^ 23
Nitrogen dioxide
1 N, 2 O=3 atoms
14amu=N
32amu=O total= 46amu of NO2
1.661 *10^ 24
 7.6406 *10^ 23
46amu *
1amu
4.512 g
 5.905295396 *10^ 22molecules
7.6406 *10^ 23
(c) Determine the mass percent of each element in each of the above compounds.
Water
16 gO
*100%  88.88888889%  88.9%ofO
18 gwater
2 gH
*100%  11.11111111%  11.1%ofH
18 gwater
Octane
96 gC
*100%  84.21052632%  84.2%ofC
114 g
18 gH
*100%  15.78947368%  15.8%ofH
114 g
Nitrous oxide
28 gN
*100%  63.636364  63.6%ofN
44 g
16 gO
*100%  36.36363636  36.4%ofO
44 g
Nitrogen dioxide
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14 gN
*100%  30.43478261  30.4%ofN
46 g
32 gO
*100%  69.56521739  69.6%ofO
46 g
(d)
Apply the Law of Multiple Proportions to the nitrous oxide – nitrogen dioxide
pair. That is, show that the ratio of the number of grams of nitrogen per gram
of oxygen for NO2 : N2O is a ratio of whole numbers.
14 gN inNO2  14 g 1

  1 : 2ratio
28gN inN 2 O  28g 2
5. A feeling of the “size” of things – chemically speaking …using
dimensional analysis…
(a) An individual atom …
At the beginning of the 1900’s, technology and the “right frame of mind”
allowed scientists to investigate the sub-structure and size of atoms. The
atom was not homogeneous, i.e., not of uniform density. Most of the mass
of an atom was contained in a very small volume – termed the nucleus.
This nucleus had a net positive charge. The remainder of the atom was
mostly “empty space” populated with negative electric charge equal in
magnitude (and opposite in sign) to the positive charge of the nucleus. The
particles of negative charge were termed electrons and the low density
region of the atom surrounding the positive nucleus containing these electrons
tremendously “spread out” was termed the electron cloud. The particles of
positive charge constituting the nucleus were termed protons and found to
have a mass about 1833 times the mass of the electron. A cartoon of this
atom is below:
Source: http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/What
_Atoms_Look_Like/920424670__atom2.jpg
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If this cartoon were to scale…. and represented a typical hydrogen atom…
The diameter of the hydrogen atom’s nucleus would about 1 x 10-13 cm and have a mass of
about 1.00 amu.
The diameter of the entire hydrogen is about 1 x 10-8 cm and its mass is about the same
(1.00 amu).
Please answer the following. Assume that the nucleus and the entire atom can be
represented as spheres of the appropriate diameter.
[Recall: Volume of a sphere = (4/3)•π•(radius)3]
• What is the density of the nucleus of this hydrogen atom in g/cm3?
V=(4/3)*π*(5.00*10^-14cm)^3
V=(4/3)* π*(1.25*10^-40cm^3)
V=5.23598776*10^-40cm^3
D=1.661*10^-24g/5.23598776*10^-40cm^3
D=3.172276323*10^15g/cm^3 of Hydrogen nucleus
• What is the density of the entire hydrogen atom in g/cm3?
V=(4/3)* π*(5.00*10^-9cm)^3
V=(4/3)* π*(1.25*10^-25cm^3)
V=5.23598776*10^-25cm^3
D=1.661*10^-24/5.23598776*10^-25cm^3
D=3.172276323g/cm^3 of entire Hydrogen atom
• If the nucleus of this hydrogen atom were scaled-up to the size of a typical garden pea
(a sphere of diameter equal to ¼”), estimate the weight of this garden pea in tons?
[2.54 cm = 1 in. ; 1 lb. = 453.6 g ; 1 ton = 2000 lb.]
R=1/8inches
1/8in * 2.54cm/1in= 2.54cm/8= .3175cm= radius
V=(4/3)* π*(.3175cm)^3
V=.1340663538cm^3
D*V=M
(3.17*10^15g/cm^3)*(.1340663538cm^3)=M
4.2478*10^14g = M
(4.247*10^14g) * (1lb/453.6g) * (1ton/2000lbs)= 468,231,922.4 tons
(b) The size of a mole … a “feeling” for 6.022 x 1023 things …
Imagine that you are assigned the task of building a skyscraper with a square base equal to
the surface area of the earth (about 197,000,000 square miles). How high would this
skyscraper be if its height were dictated by the requirement that you had to use every last
one of a mole of cubical toy building blocks that are 2” on an edge?
[1 mile = 5280 feet ; area of a square = (edge)2 ;
volume of a rectangular solid = length•width•height ; thus the
volume of a cube = (length of any edge)3]
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√197,000,000sq. mi. = 14035.66885 miles
(length of side of the base in miles)
14035.66885 miles * 5280 ft/1mile = 74,108,331.53 ft
74,108,331.53 ft * 6 = 444649989.2 ft(number of blocks on one side)
(444649989.2ft)^2= 1.977*10^17 (number of blocks in base layer)
(6.022*10^23)/(1.977*10^17)= 3,046,029.337 (number of layers)
3,046,029.337 * 2in = 6,092,058.675 in (height)
6,092,058.675in * 1ft/12in = 507,671.5562ft
507,671.5562ft * 1mi/5289ft = 96.14991595= 96.14 miles high
6. If we could see the molecules and count them:
Imagine the following “chemical reaction”:
3
+ 2
4
Imagine the following initial situation:
INITIAL
FINAL(??)
Your mission is to fill in the “Final” box with the correct number of
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all species. [This is what a “Limiting Reagent” problem would boil down to if we
could only see the molecules!]