Classification of Matter
Hot-air balloonists must understand the relationships between pressure,
temperature, and volume of gases. Experimental studies to derive these
relationships were the beginning of chemistry.
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.
Physical change
A change in the form of a substance, for instance, from solid to liquid or liquid to
gas or solid to gas, without changing the chemical composition of the substance.
As we will see later, chemical bonds are not broken in a physical change.
Examples: Boiling of water and the melting of ice.
Chemical change
The change of a substance into another substance, by reorganization of the atoms,
i.e. by the making and breaking of chemical bonds. In a chemical change a
chemical reaction takes place.
Examples: Rusting of iron and the decomposition of water, induced by an electric
current, to gaseous hydrogen and gaseous oxygen.
One classification scheme for matter can follow the diagram below.
Mixture
Two or more substances, combined in varying proportions - each retaining its
own specific properties. The components of a mixture can be separated by
physical means, i.e. without the making and breaking of chemical bonds.
Examples: Air, table salt thoroughly dissolved in water, milk, wood, and concrete.
Heterogeneous Mixture
Mixture in which the properties and composition are not uniform throughout the
sample.
Examples: Milk, wood, and concrete.
Homogeneous Mixture
Mixture in which the properties and composition are uniform throughout the
sample. Such mixtures are termed solutions.
Examples: Air and table salt thoroughly dissolved in water.
Pure Substance
A substance with constant composition. Can be classified an either an element or
as a compound.
Examples: Table salt (sodium chloride, NaCl), sugar (sucrose, C12H22O11), water
(H2O), iron (Fe), copper (Cu), and oxygen (O2).
Element
A substance that cannot be separated into two or more substances by ordinary
chemical (or physical) means. We use the term ordinary chemical means to
exclude nuclear reactions. Elements are composed of only one kind of atom.
Examples: Iron (Fe), copper (Cu), and oxygen (O2).
Compound
A substance that contains two or more elements, in definite proportion by weight.
The composition of a pure compound will be invariant, regardless of the method
of preparation. Compounds are composed of more than one kind of atom. The
term molecule is often used for the smallest unit of a compound that still retains
all of the properties of the compound.
Examples: Table salt (sodium chloride, NaCl), sugar (sucrose, C12H22O11), and
water (H2O).
Below, we list two of the Laws which helped elucidate the above classification scheme:
Conservation of Mass
Usually attributed to Lavoisier (in 1789). "Matter (mass) is neither created nor
destroyed". In other words, in a closed system (nothing escapes), any process will
not change the total "matter content" (i.e. mass) of the system.
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Law of Definite Proportion or Constant Composition
Usually attributed to Dalton and/or Proust (circa 1808). "Regardless of the
method of separation, a pure compound will always contain the same elements, in
the same proportion by mass." Dalton's specific contribution to this law is with
regard to the inclusion of the consequences of the atomic hypothesis
(Democritus, circa 400 B.C.). Dalton reasoned that since these elements were
indivisible atoms, "each pure compound should contain the same proportion of
these atoms, regardless of the method of preparation."
Example Problem: Three samples of a substance containing sodium (Na) and Oxygen
(O), were prepared in different ways:
Sample
A
B
C
Total mass
1.020 g
1.548 g
1.382 g
Mass of Na in sample
0.757 g
1.149 g
1.025 g
Mass of O in sample
0.263 g
0.399 g
0.357 g
Are samples A, B, and C the same substance?
Solution: A few comments first:


Since the sum of the individual masses of Na and of O in each sample add up to
the total mass of the sample, we can be sure that Na and O are the only
components present.
The proportion by mass of a species in a sample can be expressed as a mass
percent (% ). The definition is:
.
We can use any unit of mass, but the same units must be used in the numerator
and in the denominator.
Sample A:
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.
Gases - A "Simple" Place to Start
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).
Pressure and its Measurement
Pressure has dimensions of force per Area. The SI unit of pressure is the Pascal (Pa). A
Pascal = 1 Newton/m² = kg/(m· s² ). The relationship between atmospheres (atm.) and
Pascals is:
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:
1. If a barometer, consisting of Hg(l) as the "working fluid", is exposed to a pressure
of 1.00 atm. (at 25° C), what will be the height (in cm and in ft.) of the Hg column
that is supported?
2. If a barometer, consisting of H2O(l) as the "working fluid", is exposed to a
pressure of 1.00 atm. (at 25° C), what will be the height (in cm and in ft.) of the
H2O column that is supported?
<>Solution to Example:
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.
For Hg(l), d = 13.6 g/cm³. Thus, solving for h, and using dimensional analysis, we have:
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.