MISE - Physical Basis of Chemistry

MISE - Physical Basis of Chemistry First Set of Problems - Due October 1, 2006 Bill Wagenborg MISEP

Submit electronically (digital drop box) by October 1 – 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 = Page 2

Of course, switching to Symbol font will allow any Greek alphabetic symbol to be entered.

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) = m 3 5 5 g i c e i c e  (insert equation typed in Equation Editor) = d 0 . 9 1 7 g i c e  i c e    m L i c e 

= 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.

1 c m 3  1 3 in .3  V (in3) = ( 3 8 7 . m L )   = 23.6 in3 . can 1 m L 2 .5 4 3 c m 3 

V o l u m e o f c a n = height of can  r 2  V o lu m e o f c a n 2 3 . 6 i n 3 height of can (in.) = = = 13.4 inches r 2 ( 0 . 7 5 2 ) i n . 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. Page 3

(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

Bill Wagenborg October 1,2006 Misep Chemistry 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]

 5grains  2asp  1gram   1000miligram         = 6.4808814 * 10 4  1asp     15.43grains  1gram   453.6grams  1kg  165lbs.     74,844,000kg  1lb   1000g  6.4808814 *104 mg     8.66mg / kg  74,844,00kg 

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 = • 100 %.

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 Page 4 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 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? (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?

(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. (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.

A) z  100  61.37  gX   p  AW        x   gZ   q   AWz   61.37g  1  AW        x   1.59  38.63g  1  AWz 

 51.44  48.56 2 b)II     .66   X Z  1.59  3 2 3  44.27  48.56 1 III     .499   XZ  1.59  2 2  38.86  61.14 2 IV     .399   X Z  1.59  5 2 5 This is the best we can do since we do not know the true identity of the elements or the actual mass. We assume it to be 100 grams. Page 5

 AW  c) x   1.59  AWz   AW   x   1.59  32 

AWx  1.59  32

AWx  50.88 X  Vanadium Z  Sulfur d) I  VS

II  V2 S3

III  VS2

IV  V2 S5 We are still assuming there to be 100 grams of each element, so this will not give us the “true” Chemical formula. The actual amounts of each element would change the ratio and thus the formula.

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: C. 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 C = 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 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, Page 6

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.

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 12C will weigh exactly 12.00 grams?

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 OR

• mass of 1 mol (of atoms of) of X in grams.

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. Page 7

Then, it is also hopefully evident that the MW can also be interpreted in two equivalent 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!

Answer:  12amu  19.932  1024        1atom   1carbon   19.93  1024   12grams       6.02  1023 atoms  1atom   xatoms 

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. (b) How many molecules of each of the compounds would be present in the 2.40 L samples? (c) Determine the mass percent of each element in each of the above compounds. (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. Page 8

 mass  a)density     volume  mass  1.00g / ml     2400  1  2400grams:water  2400ml   mass  .703g / ml     2400 .703  1687.2grams:oc tane  2400ml   mass 1.8g / L     2.4  1.8  4.338grams:nitrousoxide  2.4L   mass  1.88g / L     2.41 1.88  4.512grams:nitrogendioxide  2.41L b)water:H  1  O  16  2  18   masscompound  sample    moles  in  compound    masscompound   6.02  1023 molecules moles  in  compond   molecules  in  sample  1mole   2400    6.02  1023  8.03  1025 molecules  18  oc tane:C  12  8  H  1 18  114  1687.2    6.02  1023  8.91 1024 molecules  114  nitrous  oxide: N  14  2  O  16  44  4.338    6.02  1023  5.94  1022 molecules  44  nitrogen  dioxide: N  14  O  16  2  46  4.512    6.02  1023  5.91 1022 molecules  46  Page 9

 mass  c) Mass ~ percent   element   massmole  water:  2  hydron   11%  18  16 oxygen   89%  18 oc tane:  96  Carbon   84%  114  18  Hydrogen   16%  114 nitrous ~ oxide:  28 Nitrogen   64%  44  16 Oxygen   36%  44

Nitrogen ~ dioxide:  14 Nitrogen   30%  46  32 Oxygen   70%  46 d) Using 100 grams: Nitrous oxide :Nitrogen 64% = 64 grams and Oxygen 36% =36 grams Nitrogen dioxide: Nitrogen 30% = 30 grams and Oxygen 70% = 70 grams  64    1.78:nitrous ~ oxide  36  30    .429: Nitrogen ~ dioxide  70

Using whole numbers the ratio would be 1 : 2 Page 10

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

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? • What is the density of the entire hydrogen atom in g/cm3? • 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? Page 11

[2.54 cm = 1 in. ; 1 lb. = 453.6 g ; 1 ton = 2000 lb.]

Density= Mass/Volume  1.661 1024  a)   3.17  1015 g / cm3   5.24  1040   1.661 10241 b)   3.17g / cm3   5.24  1025   2.54cm c).25in   .635cm / 2  .3175  1in  volume  4 / 3.31753   .1340663538cm3 Volume  density  mass  1lb   1ton  .1340663538  3.17  1015  4.25  1014 grams     4.68  108 tons  453.6grams  2000lb

(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]

3 volumeblock  8in 3 3  1foot   1mile  volume  8  6.02  1023  4.816  1024 in3      1.893393895  1010 miles3 total  12in   5280 feet  volume  height  width  length 1.89  1010  height  197000000 height  1.89  1010 / 197000000 height  96.1miles Page 12

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 all species. [This is what a “Limiting Reagent” problem would boil down to if we could only see the molecules!]

There are15 circles and 12 squares in the initial. This means that 15 of the circles will react with 10 of the square to produce 20 triangles. That means in the final box only 2 squares will be left.