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A chemist will usually classify all "stuff" that has "substance", i.e. matter, in the following manner. Definitions and examples are given below. This classification scheme, although seemingly straightforward, is the result of centuries of investigation and thought! In fact, after we list the results, we will trace how some of the ideas developed - paying attention to the comments expressed above. Classification is usually part of the early stages of understanding science. It allows us to sort our data into unifying categories so that we can wrestle with "what it means" in a more efficient manner.
First, we need to give a working definition of chemical change and of physical change.
Below, we list two of the Laws which helped elucidate the above classification scheme:
Example Problem: Three samples of a substance containing sodium (Na) and Oxygen (O), were prepared in different ways:
Sample Total mass Mass of Na in sample Mass of O in sample A 1.020 g 0.757 g 0.263 g B 1.548 g 1.149 g 0.399 g C 1.382 g 1.025 g 0.357 gAre samples A, B, and C the same substance?
Solution: A few comments first:
.We can use any unit of mass, but the same units must be used in the numerator and in the denominator.
mass % of Na = (0.757 g/1.020 g) x 100% = 74.2% mass % of O = (0.263 g/1.020 g) x 100% = 25.8%Or, since Na and O are the only elements in the compound, we know (based on conservation of mass):
mass % Na + mass % O = 100%So,
mass % of O = 100% - 74.2% = 25.8%which is the same as the result above. Using the same procedure as given above, prove for yourself that:
Sample B: mass % Na = 74.2% mass % O = 25.8% Sample C: mass % Na = 74.2% mass % O = 25.8%According to the above computations, and the Law of Definite Proportions, each of the samples is most likely the same compound. It is unlikely that these are mixtures, since they were prepared in different ways. If they were mixtures, the likelihood of the percentages being identical would be very small.
We are getting a bit ahead of ourselves. How did scientists during this time period know the difference between elements and compounds (atoms and molecules)? How could they quantify these species - they could not perform measurements on individual species (atoms or molecules). They necessarily had to measure bulk properties. How can we relate bulk properties of matter to these elementary building blocks? If these building blocks are pervasive (i.e. make up all macroscopic forms of matter - solid, liquid, gas) where do we begin? At first glance, these three forms of matter seem to be vastly different - yet it was presumed that the building blocks were fundamental to solids, liquids, and gases. It turns out - historically and conceptually - that gases are the easiest place to begin. One reason is that changes of macroscopic (bulk) properties are more "noticeable", i.e. give rise to measurements that are easier to detect - under the prevailing conditions - with gases.
It was realized early on that gases required the fewest macroscopic parameters to quantify their physical state (to a good approximation). Specifically, we will investigate the parameters: pressure, volume, temperature, and amount to see how they quantify the physical state of a gas. Of course, we need to define these parameters and investigate how they can be measured. After we discuss pressure - and the devices used to measure it - we will analyze some original data made on gas (air - actually a mixture of gases) by Robert Boyle in the 1600's. You will have the opportunity to use your newly-acquired MAPLE skills to investigate this data. First we must talk about pressure and some of the devices that are used to measure it (barometers and manometers).
1.00 atm. = 1.013 x 105 Pa
At 25° C, the density of Hg(l) is 13.6 g/cm³ and the
density of H2O(l) is 1.00 g/cm³. Also, the acceleration
due to gravity, g, is 9.8 m/s². From the above information,
answer the following:The relationship between the applied pressure (P) and the height of a fluid is:
P = dgh
where "g" is the acceleration due to gravity and "d" is the
temperature-dependent density of the fluid. The above
relationship provides the "working principle" of the barometer.
Thus, we define 1.00 atm. = 760 mm Hg = 760 torr, where
1 mm Hg = 1 torr. In feet,
The height of mercury is determined by the need
to balance the weight of mercury lifted against the weight
of the air above it. The weight of air presses down
uniformly on everything on the surface of the earth,
including the surface of the pool of mercury in the beaker.
This pool transmits the pressure uniformly through its
volume and therefore maintains the height of mercury in
the glass tube. The 76.0 mm of mercury has a weight equal
to the weight of a column of air with the same cross-sectional
area and a height of 150 km (roughly the height of our atmosphere).
Using a J-tube, we can accurately measure the height of
mercury that balances the atmospheric pressure and hence
create a barometer.
Exercises 1-36 and 1-37 use Maple to explore relationships between temperature, pressure, and volume for gases