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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 (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.
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 = 1cm^3 ; 2.54 cm = 1 inch (exactly); Volume of cylinder = (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 ) =
m (^) ice d (^) ice
355 g ice 0.917 g ice 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.
Vcan(in^3 ) = (387. mL ) 1 cm^3 1 mL
13 in.^3 2.54 3 cm^3
= 23.6 in (^3).
Volume of can r^2
= height of can
height of can (in.) = Volume of can r^2
= 23.6 in^3 (0.75 2 ) in.^2
= 13.4 inches
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/d (^) ice = (355 g ice) ÷ (0.917 g ice/mL ice)
= 387. mL ice and so the required volume of the can.
= X 8 Z 5
(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?
II. X .5144 Z.4856 X .5144/.4856 Z .4856/.4856 X 1.1 Z 1 X subscripts by 10 X 11 Z 10
III. X. 4427 Z .5573 X .4427/.4427 Z .5573/.4427 X 1 Z 1.3 X subscripts by 10 X 11 Z 13
IV. X. 3886 Z. 6114 X .3886/.3886 Z .6114/.3886 X 1 Z1.6 X subscripts by 5 X 5 Z 8
Because we do not know the actual atomic weights(masses) for X or Z
(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.
- 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: 12 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 12 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 10-24^ 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.
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 12 C weighs (exactly) 12.00 amu, then how many atoms of (^12) C 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 (^12) C 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.
Answer 12 g ( 1 Carbon atom/12 amu)( 1 amu/1.661E-24g) = 6.02E23 Carbon Atoms
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.
NO 2 N = ( 14.0067)= 14.0067g O 2 = 2(16) = 32g molar mass = 46 g N= 14.0067g/46g x 100% = 30.44% O = 32g/46g x 100% = 69.56%
(d) N 2 O:NO 2 N 2 O = 28.01/16 = 1.75 NO 2 = 14/32 =. The ratio between N 2 O and NO 2 = 1.75/.4375 which equals 4 which is a whole number
- 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)••(radius) 3 ]
- What is the density of the nucleus of this hydrogen atom in g/cm 3? V= (4/3)(3.14)(5 E-14)^3 = 5.23 E-40cm^3 d=m/v d= 1.661 E-24g/5.23 E-40cm^3 = 3.174 E15g/cm^
- What is the density of the entire hydrogen atom in g/cm^3? V = (4/3)(3.14)(5 E-9)^3= 5.23 E-25cm^3 d = 1.661 E-24g/5.23 E-25 = 3.175 g/cm^
- If the nucleus of this hydrogen atom were scaled-up to the size of a typical garden pea (a sphere of diameter equal to ”), estimate the weight of this garden pea in tons? [2.54 cm = 1 in. ; 1 lb. = 453.6 g ; 1 ton = 2000 lb.] Mass = V x D diameter of the H atom = 2.54cm/4 = .635 in .635/2 = .3175 radius =. V= (4/3)(3.14)(.3175)^3 = .1339cm^3 3.175 E15 g/cm^3 x .1339cm^3 = 4.25 E14g 4.25 E14g (1 lb/4.53.6 g ) = 9.372 E11 lb. 9.372E11 lb. ( 1 ton/ 2000 lb) = 468,000,000 or 470,000,000 tons
(b) The size of a mole … a “feeling” for 6.022 x 10^23 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 ]
197,000,000 square miles (197,000,000 square miles) = 14,035.66885 miles (one side) 14,035.66885 miles ( 5280 feet/ 1 mile) = 74,108,331.5 feet ( 12 in./ 1 foot) = 889,299,978 inches 889,299,978 inches/2 inch blocks = 444,649,989 (444,649,989)^2 = 1.977E17 blocks on base 6.022E23/1.977E17 = 3046029.337 blocks high 3,046,029.337/ 6 (2 inch blocks x 6 = 12 inches) = 507,671.5562 blocks 507,671.5562/5,280 miles = 96.15 miles
- If we could see the molecules and count them: Imagine the following “chemical reaction”:
3 +^24
Imagine the following initial situation:
