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New optical microscopes are capable of visualizing objects down to 1 nanometer in size. The photo above shows a conventional micrograph of dispersed oil droplets alongside a simultaneously taken optical image showing features 1 nm across (approximately 5 atoms in width). Advances like this could allow direct visualization of viruses and lead to storage media which hold much more information than present ones.

As you may already realize, we possibly take atoms (and molecules)
for granted. For instance, we possibly don't even think twice
about the conceptual evolution of ideas that led to the notation
H_{2}O. In light of what was said previously, it is perhaps useful
to give a thumbnail sketch of the ideas and struggles that
resulted in the acceptance of the atomic hypothesis as probably
the "way things are", not just as "one of several useful theoretical
frameworks". This will hopefully give us a greater appreciation
for "counting atoms" and the mole concept - which necessitates the
reliable measurement of (relative) atomic weights. An appreciation
of the central role of atomic weight determination in the development
of chemical ideas may allow you to better appreciate the technological
advances that afforded precise measurements of these weights.
.

As recently as 1910, Alexander Smith - a Professor of Chemistry at the
University of Chicago, stated the following
"The atomic hypothesis provides a convenient form of speech, which
successfully describes many of the facts in a metaphorical manner.
But the handy way in which the atomic hypothesis lends itself
to the representation of the characteristic features of chemical
change falls short of constituting proof that atoms have any
real existence."

In 1802, John Dalton proposed the "modern form" of the atomic hypothesis. However, 58 years later the best chemists in the world still thought that the correct molecular formula for water was "HO". Some questions should be stated:

- How do we know the molecular formula of any compound - such as
H
_{2}O? - What gives us the right to say that one atom of carbon (C) is approximately 12 times as heavy as one atom of hydrogen (H)?
- How are atomic numbers for the elements obtained?
- Why do the atoms with the same atomic number but different atomic weights (isotopes) have (nearly) the same chemical properties?
- How do we know that the negative charges in an tom are on the "outside" and that most of the mass of the atom is compressed into a tiny dense nucleus of positive charge?
- What do we mean by the "radius" of an atom? Should not the size of the atom be as difficult to measure as its mass (weight)?
- What laboratory measurements can be related to such "microscopic" parameters?

Aristotle - ironically - gave a decent definition of an
**element**
"Everything is either an element or composed of elements...
An element is that into which bodies can be resolved, and
which exists in them either potentially, or actually, but
which cannot itself be resolved into anything simpler, or
different in kind."

This above statement does not make it clear, however, how to recognize
an element when we "see" it. Even so, Robert Boyle (you will investigate
the results of his J-tube experiments with air for homework) gave a
definition in his book - "The Skeptical Chymist (1661) that is not too
far from our current view
"And, to prevent mistakes, I must advertize you, that I now mean
by Elements, as those chymists that speak plainest do by their
principles, certain primitive and simple, or perfectly unmingled
bodies; which not being made of any other bodies or of one another,
are the ingredients of which all those called perfectly mixed
bodies are immediately compounded, and into which they are
ultimately resolved."

However, during the time of Boyle and his contemporaries, there was no
way to determine if a chemical substance (element or compound) was -
with certainty - pure. Chemical reactions were fairly carefully being
investigated however. For instance, when mercuric oxide (red powder,
modern formula = HgO) is heated, oxygen gas (O_{2}) is evolved and a
silvery mercury liquid (symbol=Hg) remains that weighs less than the
original oxide. [Today, we would write the "balanced" chemical reaction:
2HgO(s) goes to 2 Hg(l) + O_{2}(g).] However, it was not until about
a century later that Lavosier performed careful weighing experiments
on the decomposition in a closed flask to demonstrate that there is
*no overall loss of weight* during the reaction, i.e. mass is
**conserved**. Hence, according to Boyle, mercuric oxide could not
be an element (because it had decomposed into pieces) but mercury
could (unless observation demonstrated that it could be further decomposed).
As the chemist Justus von Liebig stated in 1857
"The elements count as simple substance not because we know
that they are so, but because we do not know that they
are not."

Some of the first elements to be recognized were metals (c.f. Gold (Au),
Copper (Cu), Tin (Sn), Iron (Fe), Platinum (Pt), Lead (Pb), Zinc (Zn),
Mercury (Hg), Nickel (Ni), Tungsten (W), and Cobalt (Co)). Of the
non-metals: Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), and
Xenon (Xe), were discovered in the mixture of gases that remained
after air was depleted of all its nitrogen (N_{2}) and
Oxygen (O_{2}).

Thus, many chemists in the 1700's spent much of their investigations
preparing and describing pure compounds - and subsequently decomposing
them into their pure elements. Henry Cavendish discovered hydrogen
(H_{2}) in 1772 and Joseph Priestly invented carbonated water
(CO_{2}(aq))
and identified nitrous oxide (N_{2}O - "laughing gas"), nitric oxide
(NO), carbon monoxide (CO), sulfur dioxide (SO_{2}), hydrogen chloride
(HCl), ammonia (NH_{3}), and oxygen (O_{2}). In 1781,
Cavendish proved
that water consists only of hydrogen and oxygen [after he had
witnessed Priestly explode the two gases in a "random experiment
to entertain a few philosophical friends."] After oxygen was discovered,
Antoine Lavosier was able -through careful quantitative measurements -
to overthrow "phlogiston theory".

- 1 gram of hydrogen combines with 8 grams of oxygen to form water.
- 1 gram of hydrogen combines with 3 grams of carbon to form methane.
- 1 gram of hydrogen combines with 35.5 grams of chlorine to form hydrogen chloride.

- All matter is made up of atoms. These particles are indivisible and indestructible.
- All atoms of a given element are identical, both in weight and in chemical properties.
- Atoms of different elements have different weights and different chemical properties.
- Atoms of different elements can combine in simple whole numbers to form compounds.
- When a compound is decomposed, the recovered atoms are unchanged and can form the same or new compounds.

Dalton - as had Lavosier - paid careful attention to the quantitative apsects (weights). Hence, he developed the chemical reaction symbolism and chemical formula notation - except that he used "pictures" for the elements rather than letter symbols. Nevertheless, his symbols did not represent arbitrary amounts of a given element - it represented either one atom or a standard specified number of atoms.

Dalton then tried to understand why the ratios of elements in
compounds were fixed. His hypothesis was: **A compound consists
of a large number of identical molecules - each one of which
is built up of the same small number of atoms - arranged in the
same way.** This was termed his
**Law of Constant Composition or Fixed Composition**.
Dalton still had the problem of determining what that number
proportion was. For example, in binary (two-element) compounds
such as the oxide(s) of carbon and water - what was the simplest
formula. Dalton thus established a guide - "the rule of
simplicity". He reasoned that the simplest binary compound
between two elements "X" and "Y" should be XY. Next would be
XY_{2} or X_{2}Y, etc. Unfortunately his first attempt at writing
down a formula (for water) proved a mistake. He presumed (relative
to 1 for hydrogen) that the combining weight of oxygen was equal
to its atomic weight. Hence, Dalton postulated that oxygen had
an atomic weight of 8 (relative to 1 for hydrogen). Thus, Dalton's
formula for water was "HO". Bad data collection by Dalton
further exacerbated the problem. Anyhow, Dalton had developed a
way to relate numbers of atoms with their masses and hence could
determine (relative) atomic weights!! The table below exemplifies
the application of these ideas to the oxides of carbon and also to
the oxides of nitrogen.

C/O | |||
---|---|---|---|

Mass | Possible | Possible | |

Ratio | Formula #1 | Formula #2 | |

Oxide "A" | 3/4 | CO | C_{2}O |

Oxide "B" | 3/8 | CO_{2} |
CO |

Atomic weight of C (if O=8) | 6 | 3 | |

Atomic weight of C (if O=16) | 12 | 6 |

N/O | ||||
---|---|---|---|---|

Mass | Possible | Possible | Possible | |

Ratio | Formula #1 | Formula #2 | Formula #3 | |

Oxide "A" | 3.5/8 | NO | NO_{2} |
NO_{4} |

Oxide "B" | 7/8 | N_{2}O |
NO | NO_{2} |

Oxide "C" | 14/8 | N_{4}O |
N_{2}O |
NO |

Atomic weight of N (if O=8) | 3.5 | 7 | 14 | |

Atomic weight of N (if O=16) | 7 | 14 | 28 |

"Universal Acceptance" of Atomic Theory

In terms of Combining Weights, we can state the Law of Multiple
Proportions as:
"If an element shows more than one combining weight, these
weights will differ among themselves by ratios of small
whole numbers.

Checking back to appropriate spots in the table on the previous page
will easily show this for the oxides of carbon and of nitrogen. In
more modern terms, we state it again:
If two elements form more than one (binary) compound, then
the different weights of one which combines with the same
weight of the other are in the ratio of (small) whole
numbers.

In other words, Dalton showed - by analyzing large amounts of data
from many sources - that the Combining Weights differ by small
whole numbers because the atoms combine in small whole numbers.
The acceptance of atomic theory was rapid and widespread.

Gay-Lussac began to study the volumes of reacting gases. He found the following results:

- Equal volumes of ammonia gas (NH
_{3}) and HCl gas form solid ammonium chloride - Two volumes of hydrogen react with one volume of oxygen to form two volumes of steam (gaseous water).
- Three volumes of hydrogen react with one volume of nitrogen
to form two volumes of gaseous ammonia (NH
_{3}). **One volume of hydrogen reacts with one volume of chlorine to form two volumes of HCl gas.**

Dalton agreed that if the volume of HCl gas (for example) is twice the volume of either hydrogen or of chlorine, there must be half as many molecules per volume in HCl. Hence, the (number) density of HCl per unit volume is n/2 for every n molecules of hydrogen or n molecules of chlorine that react.

Amedeo Avagadro proposed, on the other hand, that **equal volumes
of (gaseous) molecules at the same temperature and pressure contain
equal numbers of molecules.** He published these ideas in 1811.
One consequence of **Avogadro's Hypothesis** is that all
molecules in the HCl reaction - since they are at the same
temperature and pressure would contain the same number of molecules
per unit volume. Since volume was proposed to be proportional to
the number of molecules, stoichiometric coefficents of the chemical
reaction notation first developed by Dalton actually were proportional
to volume. Thus, according to Gay-Lussac's data, we would write the
chemical reaction as:

Thus, by applying Dalton's own ideas of the **conservation of elements
(atoms)**, i.e. that atoms are neither created nor destroyed, one
hydrogen molecule must contain two (2) hydrogen atoms and one chlorine
molecule must contain two (2) chlorine atoms. In other words, both
gaseous hydrogen and gaseous chlorine must be diatomic (i.e. H_{2} and
Cl_{2}). The correct (balanced) reaction would read:

Note that if hydrogen is now presumed to be diatomic and - because
of the implied proportionality of number of molecules and volume -
Gay-Lussac's data would predict the following reaction:

In order to conserve atoms - without changing the coefficients (since
these are already determined by the Gay-Lussac's volume measurements)
the formula for water must be H_{2}) and the formula for oxygen must be
O_{2}, i.e. gaseous oxygen is also predicted to be diatomic. The
predicted reaction, according to Avogadro, would be:

Because of this new predicted proportion of oxygen to hydrogen in
water, Dalton's original assessment of the (relative) atomic
weight of oxygen would be:

8 grams of oxygen combine with 1 gram of hydrogen, but the molecule
contains two atoms hydrogen for each oxygen atom.

Thus: if the (relative) weight of hydrogen is still assigned a value
of "1", then the (relative) weight of hydrogen in a water molecule
is 2 and thus the (relative) weight of an oxygen atom - in order to
maintain the weight ratio of 8:1 - must be 16. So, oxygen has not
only replaced phlogiston but now has a new (relative) atomic weight
of 16!

Unfortunately, Avogadro's hypothesis of equal volumes implying an
equal number of molecules was judged "shaky" due, in part, to the
prediction of diatomic - single element - molecules (
O_{2}, H_{2}, Cl_{2},
etc.). At that time ideas of chemical bonding were based almost
entirely on electrical attraction and repulsion. It was thus
difficult for scientists to understand how two identical atoms could
do anything but repel one another. Also, if they did attract one
another why did not elemental hydrogen (for example) from larger
molecules: H_{3}, H_{4}, etc.? Thus, a half a century of confusion
existed in chemistry, due to the rejection of Avagadro's ideas.

So, by 1860, there was such widespread confusion about atomic weights, that a conference was organized in Karlsruhe, Germany to attempt to reach some agreement. Stanislao Cannizzaro suggested a rigorous method for finding atomic weights based on the rejected work of his countryman Avogadro. Cannizzaro's method is listed below:

- Assume that the atomic weight of hydrogen is 1.0 and that that hydrogen is made of diatomic molecules - Gay-Lussac's data suggested.
- Assume that Avogadro was correct in deducing that oxygen
gas is diatomic (O
_{2}) and hence that the correct molecular formula for water is H_{2}O. As mentioned previously, this would give rise to a (relative) atomic weight of atomic oxygen of 16 and the (relative) molecular weight of O_{2}as 32. - If equal volumes of all gases contain equal numbers of
molecules then the molecular weight (M) of a gas is
proportional to the density (d) of the gas: M= kd.
Use H
_{2}and O_{2}to evaluate the proportionality constant "k". Before we determine the value of "k" we must make some comments about the units of molecular and atomic weights. One outcome of all this work was the realization that the number of molecules present in 2.0 grams of H_{2}(g) or 32.0 grams of O_{2}(g) would be the same and be fixed. This number of molecules - a very important number for chemists - is termed the**mole**. The number was not known at the time of these measurements - but whatever it was it was thus hypothesized to be fixed. Thus, the actual value was not needed - since proportions were used.So, the relative atomic and molecular weights discussed thus far are the masses of 1 mole of such atoms and molecules in grams. Therefore, the units of atomic or molecular weights can be considered as grams per mole (g/mole). Now, list the determination of "k".

Gas Density (in g/l) Molecular Weight (M) (in g/mole) Constant (k) in L/mole H _{2}0.0894 2.0 22.37 O _{2}1.427 32.0 22.42 Average value: 22.4 L/mole - Back to Cannizzaro's procedure. Evaluate the molecular weights
of a series of compounds containing the elements whose atomic
weights are to be determined. Starting with percent composition
(by weight) calculated from gas densities, calculate the weight
of each element
*per molecular unit*. Look over these weights for a given element to see if the numbers can be interpreted as integral multiples of some common factor. This factor may be interpreted as the atomic weight of the element.

Compound | d | M=kd | Elemental mass comp. | Weight per molecular unit | Probable formula | ||||
---|---|---|---|---|---|---|---|---|---|

(g/l) | (g/mole) | C | H | Cl | C | H | Cl | ||

Methane | 0.0714 | 16.0 | 74.8 | 25.0 | --- | 12.0 | 4.03 | --- | CH_{4} |

Ethane | 1.335 | 29.9 | 79.8 | 20.2 | --- | 23.9 | 6.04 | --- | C_{2}H_{6} |

Benzene | 3.47 | 77.8 | 92.3 | 7.7 | --- | 71.8 | 6.00 | --- | C_{6}H_{6} |

Chloroform | 5.32 | 119.1 | 10.05 | 0.844 | 89.10 | 12.0 | 1.01 | 106.2 | CHCl_{3} |

Ethyl Chloride | 2.87 | 64.3 | 37.2 | 7.8 | 55.0 | 23.9 | 5.02 | 35.4 | C_{2}H_{5}Cl |

Carbon tetrachloride | 6.81 | 152.6 g/L | 7.8 | --- | 92.2 | 11.9 | --- | 141.0 | CCl_{4} |

Hence, the Cannizzaro method allows the determination of atomic
weights for any element that appeared in compounds having measurable
vapor densities. With these atomic weights, the percent composition
of a new compound would lead unambiguously to the chemical formula.
It was realized that a mole of any compound would have the same
number of molecules - although the value of that number was not
then known. This number was named Avogadro's number in honor of
Amedeo's pioneering work. This number of particles (whether it
be atoms, molecules, or apples) is also termed the **mole**.
This number is now known to be 6.022 x 10² ³ . Thus,
In hindsight it is not surprising that a mole of any compound
contains the same number of molecules. This comes from the
knowledge of the **ideal gas law**. It can be shown that
the "k" (Cannizzaro's) constant is simply RT/P. Let's see:

Remember that density (PV = nRT Ideal Gas Equation of StatewhereP= pressure (usually in atmospheres)V= volume (usually in Liters)T= temperature (in Kelvin) remember T (in K) = t(in ° C) + 273.15 mass of molecules in gramsgn= number of molecules (in moles) = ----------------------------- = --- molecular weight in g/moleMwhereg= mass of molecules in grams andM= molecular weight of molecules in grams per moleR= ideal gas constant = 0.0821 Liter atmosphere/(mole ° K)

where the "k" from Cannizzaro's procedure is seen to be P/RT. We can now calculate the value of "k" at STP. The value isg MP P d = --- = --- = M --- = M (k);V RT RT

d R T g R T M = ------ = ------ P P V

Let's look at a sample program that makes use of the Cannizzaro method in Maple. As usual, the Maple output is centered and not all Maple output is shown.

#Construct an array in which each row refers to the information #for one of the (six) compounds. Each row will consist of the #data in the following order: # # [density, mass of compound, mass of element #1 in comp'd,...] # #Since we are given elemental composition in mass percents, we can #assume that we have 100.0 g of compound. # #Also, we will set Digits:=4 so that we can sell all of the #entries aligned properly in the array. # restart; Digits:=4;

a:= array([[0.714,100,74.8,25.0,0], [1.335,100,79.8,20.2,0], [3.47,100,92.3,7.70,0], [5.32,100,10.05,0.844,89.10], [2.87,100,37.2,7.8,55.0],[6.81,100,7.8,0,92.2]]);

#Now we will determine our average value of our proportionality #constant (k) by use of the relationship that molecular weight #(M) equals k times the density (d), i.e. M = kd. Thus, #k = M/d. So if we have some gaseous substances of known #molecular weight and measured density under the prevailing #conditions, we can determine k. According to Cannizzaro, #we presume that hydrogen (H) and oxygen (O) exist as #diatomic gases, H2 and O2. Given the atomic weight of #H as 1.01 g/mole and O as 16.00 g/mole, we can determine #k from the measured gas density of H2 and O2 and hence use #the averaged k for our analysis. The measured gas density #of H2 and of O2 - at the prevailing conditions - are given #as 0.0894 g/L and 1.427 g/L, respectively. Remember, the #molecular weight of diatomic hydrogen and diatomic oxygen # are 2.02 g/mole and 32.00 g/mole, respectively. # k1:=2.0/0.0894; k2:=32.00/1.427; # #The averaged value of k is (k1+k2)/2 - as expected. # k:=(k1+k2)/2; # #Now, what we want to do is to multiply each density - in #column 1 of array "a" by k (i.e. M = k x d). We can #accomplish with a "for" statement. We tell Maple to print #the results. Convince yourself that the units of M are #g/mole. # for i from 1 to 6 do print(k*a[i,1]) od;

#Now we will define another array, "b", that will contain the #results of performing the Cannizzaro procedure on "a". First #we will set up the array - to get a look - and then we will #"fill" it. The array will have the same dimensions as "a". # b:=array(1..6,1..5);

#To show that the array has been set up, we use the "evalm(...)" #command to evaluate the array (or matrix). # evalm(b);

#Recalling the Cannizzaro procedure, we want to find the mass of #each element per mole of the compound. Since, according to #Dalton, there must be a whole number of atoms in each compound, #i.e. a whole number of moles of atoms for each mole of compound, #then these amounts should be a whole number of some smallest #mass (Greatest Common Factor, GCF). This smallest mass - for #each element - should be the atomic weight of that element. # #First, to determine the grams of each element per mole of each #compound, a simple dimensional analysis argument reveals that #if we compute the number of grams of each element per gram of #compound (i.e. entry a[i,3] or a[i,4] or a[i,5] divided by #a[i,2]) and multiply this result by the molecular weight (M) #of the compound (i.e. k*a[i,1] - as we did above) we will #have the number of grams of the particular element per mole #of this compound. BE SURE TO CONVINCE YOURSELF OF THIS! # #So, the Maple command is: # for i from 1 to 6 do for j from 1 to 5 do b[i,j]:=a[i,j]* (k*a[i,1]/a[i,2]) od od; evalm(b);

#By inspection of each of the last three columns above, we #look for an entry that is a multiple of the others, i.e. #an entry that when multiplied by appropriate whole numbers #will yield all of the other entries. This is the Greatest #Common Factor (GCF). This GCF should be the smallest #combining mass of the element in g/mole. Convince yourself #of this. Hence, for C it is about 12.0 (11.90), for H it #is about 1.0 (1.006), and for Cl it is about 35.4 (35.36). #So, we divide each of the last three columns by the appropriate #GCF and this will allow us to determine how atoms of each #element appears in the compound, i.e. the probable molecular #formula of the compound. # #So, we tell Maple to divide all of the entries in column 3 of #"b" (i.e. b[i,3]) by the GCF of 11.90 (i.e. the "atomic weight" #of C). Similarly, we tell Maple to divide all of the entries #in column 4 of "b" (i.e. b[i,4]) by the GCF of 1.006 (i.e. the #"atomic weight" of H). Finally, we tell Maple to divide all #of the entries in column 5 of "b" (i.e. b[i,5]) by the GCF #of 35.36 (i.e. the "atomic weight" of Cl). We instruct #Maple to print out only these last three columns - after #they have been processed as stated above. Remember, by the #use of the Cannizzaro method, the GCF for a given element - for #a range of compounds containing the element - is a good estimate #of the probable atomic weight of the element. # for i from 1 to 6 do print((b[i,3]/11.90),(b[i,4]/1.006), (b[i,5]/35.36)) od;

#Each of the entries above can be easily rounded to the nearest #whole number. These rounded whole numbers tell us - for each #of the six compounds (represented by the six rows above) - the #(whole) number of C atoms, of H atoms, and of Cl atoms in a #molecule of each of the compounds. In other words, we have #determined the probable molecular formula of each of these #compounds! The rounded-off values agree with the results #shown in the text. They are listed below. #Hopefully, you can easily apply this Maple-aided Cannizzaro #method to your homework.

larryg@truth.hep.upenn.edu

Sat Feb 25 18:05:36 EST 1994

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