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<document xmlns="http://cnx.rice.edu/cnxml" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:m="http://www.w3.org/1998/Math/MathML" xmlns:bib="http://bibtexml.sf.net/" id="id8478196">
  <name>The Atomic Molecular Theory</name>
  <metadata>
  <md:version>1.6</md:version>
  <md:created>2004/10/15 18:14:34 GMT-5</md:created>
  <md:revised>2007/07/17 12:22:12.529 GMT-5</md:revised>
  <md:authorlist>
      <md:author id="jshutch">
      <md:firstname>John</md:firstname>
      <md:othername>Steven</md:othername>
      <md:surname>Hutchinson</md:surname>
      <md:email>jshutch@rice.edu</md:email>
    </md:author>
  </md:authorlist>

  <md:maintainerlist>
    <md:maintainer id="jshutch">
      <md:firstname>John</md:firstname>
      <md:othername>Steven</md:othername>
      <md:surname>Hutchinson</md:surname>
      <md:email>jshutch@rice.edu</md:email>
    </md:maintainer>
    <md:maintainer id="jsilv">
      <md:firstname>Jeffrey</md:firstname>
      <md:othername>M</md:othername>
      <md:surname>Silverman</md:surname>
      <md:email>JSilverman@astro.berkeley.edu</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>atom</md:keyword>
    <md:keyword>Atomic Molecular Theory</md:keyword>
    <md:keyword>compounds</md:keyword>
    <md:keyword>elements</md:keyword>
    <md:keyword>formula</md:keyword>
    <md:keyword>Law of Conservation of Mass</md:keyword>
    <md:keyword>Law of Definite Proportions</md:keyword>
    <md:keyword>Law of Multiple Proportions</md:keyword>
    <md:keyword>particulate</md:keyword>
  </md:keywordlist>

  <md:abstract>A development of the atomic molecular theory from the law of multiple proportions and law of definite proportions.</md:abstract>
</metadata>
  <content><section id="id8487474"><name>Foundation</name><para id="id8487485">There are over 18 million known substances in
	our world. We will begin by assuming that all materials are
	made from <term>elements</term>, materials which cannot be
	decomposed into simpler substances. We will assume that we
	have identified all of these elements, and that there a very
	small number of them. All other pure substances, which we call
	<term>compounds</term>, are made up from these elements and
	can be decomposed into these elements. For example, metallic
	iron and gaseous oxygen are both elements and cannot be
	reduced into simpler substances, but iron rust, or ferrous
	oxide, is a compound which can be reduced to elemental iron
	and oxygen. The elements are not transmutable: one element
	cannot be converted into another. Finally, we will assume that
	we have demonstrated the <term>Law of Conservation of
	  Mass</term>.
	<rule type="law" id="cosvmass">
	  <name>Law of Conservation of Mass</name>
	  <statement><para id="cnsmassp">The total mass of all products of a chemical reaction is equal to the total mass of all reactants of that reaction.</para></statement></rule> These statements are summaries of many observations, which required a tremendous amount of experimentation to achieve and even more creative thinking to systematize as we have written them here. By making these assumptions, we can proceed directly with the experiments which led to the development of the atomic-molecular theory.</para></section>
<section id="id8569858"><name>Goals</name><para id="id8569873">The statements above, though correct, are actually more vague than they might first appear. For example, exactly what do we mean when we say that all materials are made from elements? Why is it that the elements cannot be decomposed? What does it mean to combine elements into a compound?  We want to understand more about the nature of elements and compounds so we can describe the processes by which elements combine to form compounds, by which compounds are decomposed into elements, and by which compounds are converted from one to another during chemical reactions.</para><para id="id8522734">One possibility for answering these questions is to assume that a compound is formed when indestructible elements are simply mixed together, as for example, if we imagine stirring together a mixture of sugar and sand. Neither the sand nor the sugar is decomposed in the process. And the mixture can be decomposed back into the original components. In this case, though, the resultant mixture exhibits the properties of <emphasis>both</emphasis> components: for example, the mixture would taste sweet, owing to the sugar component, but gritty, characteristic of the sand component.</para><para id="id8522767">In contrast, the compound we call iron rust bears little resemblance to elemental iron: iron rust does not exhibit elemental iron's color, density, hardness, magnetism, etc. Since the properties of the elements are not maintained by the compound, then the compound must not be a simple mixture of the elements. </para><para id="id8522786">We could, of course, jump directly to the answers to these questions by stating that the elements themselves are comprised of atoms: indivisible, identical particles distinctive of that element. Then a compound is formed by combining the atoms of the composite elements. Certainly, the Law of Conservation of Mass would be easily explained by the existence of immutable atoms of fixed mass.</para><para id="id8472995">However, if we do decide to jump to conclusions and assume the existence of atoms without further evidence (as did the leading chemists of the seventeenth and eighteenth centuries), it does not lead us anywhere. What happens to iron when, after prolonged heating in air, it converts to iron rust? Why is it that the resultant combination of iron and air does not maintain the properties of either, as we would expect if the atoms of each are mixed together? An atomic view of nature would not yet provide any understanding of how the air and the iron have interacted or combined to form the new compound, and we can't make any predictions about how much iron will produce how much iron rust. There is no basis for making any statements about the properties of these atoms. We need further observations.</para></section>
    <section id="id8473035"><name>Observation 1: Mass relationships
	during chemical reactions</name><para id="id8473051">The
	Law of Conservation of Mass, by itself alone,
	does not require an atomic view of the elements. Mass
	could be conserved even if matter were not atomic. The
	importance of the Law of Conservation of Mass is
	that it reveals that we can usefully measure the masses of the
	elements which are contained in a fixed mass of a
	compound. As an example, we can decompose copper
	carbonate into its constituent elements, copper, oxygen, and
	carbon, weighing each and taking the ratios of these
	masses. The result is that every sample of copper
	carbonate is 51.5% copper, 38.8% oxygen, and 9.7%
	carbon. Stated differently, the masses of copper,
	oxygen, and carbon are in the ratio of 5.3 : 4 : 1, for every
	measurement of every sample of copper
	carbonate. Similarly, lead sulfide is 86.7% lead and
	13.3% sulfur, so that the mass ratio for lead to sulfur in
	lead sulfide is always 6.5 : 1. Every sample of copper
	carbonate and every sample of lead sulfide will produce these
	elemental proportions, regardless of how much material we
	decompose or where the material came from. These results
	are examples of a general principle known as the <term>Law of
	  Definite Proportions</term>.
	<rule type="law" id="defprop"><name>Law of Definite
	  Proportions</name>
	  <statement><para id="defpopp">When two or more elements
	combine to form a compound, their masses in that compound are
	in a fixed and definite ratio.</para></statement></rule></para><para id="id8556953">These data help justify an atomic view of
	matter. We can simply argue that, for example, lead
	sulfide is formed by taking one lead atom and combining it
	with one sulfur atom. If this were true, then we also
	must conclude that the ratio of the mass of a lead
	<emphasis>atom</emphasis> to that of a sulfur
	<emphasis>atom</emphasis> is the same as the 6.5 : 1 lead to
	sulfur mass ratio we found for the bulk lead
	sulfide. This atomic explanation looks like the
	definitive answer to the question of what it means to combine
	two elements to make a compound, and it should even permit
	prediction of what quantity of lead sulfide will be produced
	by a given amount of lead. For example, 6.5g of lead
	will produce exactly 7.5g of lead sulfide, 50g of lead will
	produce 57.7g of lead sulfide, etc.</para><para id="id8467682">There is a problem, however. We can
	illustrate with three compounds formed from hydrogen, oxygen,
	and nitrogen. The three mass proportion measurements are
	given in the following <cnxn target="table1">table</cnxn>. First we examine nitric oxide, to find
	that the mass proportion is 8 : 7 oxygen to nitrogen. If
	this is one nitrogen atom combined with one oxygen atom, we
	would expect that the mass of an oxygen atom is 8/7=1.14 times
	that of a nitrogen atom. Second we examine ammonia,
	which is a combination of nitrogen and hydrogen with the mass
	proportion of 7 : 1.5 nitrogen to hydrogen. If this is
	one nitrogen combined with one hydrogen, we would expect that
	a nitrogen atom mass is 4.67 times that of a hydrogen atom
	mass. These two expectations predict a relationship
	between the mass of an oxygen atom and the mass of a hydrogen
	atom. If the mass of an oxygen atom is 1.14 times the mass of
	a nitrogen atom and if the mass of a nitrogen atom is 4.67
	times the mass of a hydrogen atom, then we must conclude that
	an oxygen atom has a mass which is 1.14 × 4.67 = 5.34
	times that of a hydrogen atom. </para> <para id="id8467734">But there is a problem with this
	calculation. The third line of the following <cnxn target="table1">table</cnxn> shows that the compound
	formed from hydrogen and oxygen is water, which is found to
	have mass proportion 8:1 oxygen to hydrogen. Our
	expectation should then be that an oxygen atom mass is 8.0
	times a hydrogen atom mass. Thus the three measurements
	in 
	 the following <cnxn target="table1">table</cnxn> appear to lead to contradictory
	expectations of atomic mass ratios. How are we to
	reconcile these results?</para>

   
	    
      <table id="table1">
<name>Mass Relationships for Hydrogen, Nitrogen, Oxygen	  Compounds</name>
<tgroup cols="8" align="center"><colspec colname="col1" colnum="1" colwidth="50pt"/>
	    <colspec colname="col2" colnum="2" colwidth="27pt"/>
	    <colspec colname="col3" colnum="3" colwidth="50pt"/>
	    <colspec colname="col4" colnum="4" colwidth="45pt"/>
	    <colspec colname="col5" colnum="5" colwidth="42pt"/>
	    <colspec colname="col6" colnum="6" colwidth="55pt"/>
	    <colspec colname="col7" colnum="7" colwidth="55pt"/>
	    <colspec colname="col8" colnum="8" colwidth="55pt"/>

	    <thead>
	      <row>
		<entry>Compound</entry>
		<entry>Total Mass</entry>
		<entry>Mass of Hydrogen</entry>
		<entry>Mass of Nitrogen</entry>
		<entry>Mass of Oxygen</entry>
		<entry>"Expected" Relative Atomic Mass of Hydrogen</entry>
		<entry>"Expected" Relative Atomic Mass of Nitrogen</entry>
		<entry>"Expected" Relative Atomic Mass of Oxygen</entry>
	      </row>
	    </thead>
	    <tbody>
	      <row><entry>Nitric
		  Oxide</entry><entry>15.0
		  g</entry><entry>-</entry><entry>7.0
		  g</entry><entry>8.0 g</entry><entry>-
		</entry><entry>7.0 </entry><entry>8.0
		</entry></row><row><entry>Ammonia</entry><entry>8.5
		  g</entry><entry>1.5 g </entry><entry>7.0
		  g</entry><entry>- </entry><entry>1.5</entry><entry>7.0
		</entry><entry>-</entry></row><row><entry>Water</entry><entry>9.0
		  g</entry><entry>1.0 g</entry><entry>- </entry><entry>8.0
		  g</entry><entry>1.0</entry><entry>-</entry><entry>8.0</entry></row>
	    </tbody>
	  

</tgroup>
</table>
<para id="id8569043">One possibility is that we were mistaken in assuming that there are atoms of the elements which combine to form the different compounds. If so, then we would not be surprised to see variations in relative masses of materials which combine.</para>
<para id="id8464001">Another possibility is that we have erred in our reasoning. Looking back, we see that we have to assume how many atoms of each type are contained in each compound to find the relative masses of the atoms. In each of the above examples, we assumed the ratio of atoms to be 1:1 in each compound. If there are atoms of the elements, then this assumption must be wrong, since it gives relative atomic masses which differ from compound to compound. How could we find the correct atomic ratios? It would help if we knew the ratio of the atomic masses: for example, if we knew that the oxygen to hydrogen mass ratio were 8:1, then we could conclude that the atomic ratio in water would be 1 oxygen and 1 hydrogen. Our reasoning seems to circular: to know the atomic masses, we must know the <term>formula</term> of the compound (the numbers of atoms of each type), but to know the formula we must know the masses.</para><para id="id8464046">Which of these possibilities is correct? Without further observations, we cannot say for certain whether matter is composed of atoms or not.</para></section>
<section id="id8464060"><name>Observation 2: Multiple Mass
		Ratios</name><para id="id8464071">Significant insight
		into the above problem is found by studying different
		compounds formed from the same elements. For example,
		there are actually three oxides of nitrogen, that is,
		compounds composed only of nitrogen and oxygen. For
		now, we will call them oxide A, oxide B, and oxide
		C. Oxide A has oxygen to nitrogen mass ratio 2.28 :
		1. Oxide B has oxygen to nitrogen mass ratio 1.14 : 1,
		and oxide C has oxygen to nitrogen mass ratio 0.57 :
		1. </para><para id="id7988391">The fact that there are
		three mass ratios might seem to contradict the
		Law of Definite Proportions, which on the
		surface seems to say that there should be just one
		ratio. However, each mass combination gives rise to a
		completely unique chemical compound with very
		different chemical properties. For example, oxide A is
		very toxic, whereas oxide C is used as an
		anesthesia. It is also true that the mass ratio is not
		arbitrary or continuously variable: we cannot pick
		just any combination of masses in combining oxygen and
		nitrogen, rather we must obey one of only three. So
		there is no contradiction: we simply need to be
		careful with the Law of Definite
		Proportions to say <emphasis>that each unique
		compound</emphasis> has a definite mass ratio of
		combining elements.</para><para id="id7988437">These
		new mass ratio numbers are highly suggestive in the
		following way. Notice that, in each case, we took the
		ratio of oxygen mass to a nitrogen mass of 1, and that
		the resultant ratios have a very simple relationship: 
	<equation id="eqn1">
	  <m:math>
	    <m:apply>
	      <m:eq/>
	      <m:mrow>
		<m:mn>2.28</m:mn>
		<m:mo>:</m:mo>
		<m:mn>1.14</m:mn>
		<m:mo>:</m:mo>
		<m:mn>0.57</m:mn>
	      </m:mrow>
	      <m:mrow>
		<m:mn>2</m:mn>
		<m:mo>:</m:mo>
		<m:mn>1</m:mn>
		<m:mo>:</m:mo>
		<m:mn>0.5</m:mn>
	      </m:mrow>
	      <m:mrow>
		<m:mn>4</m:mn>
		<m:mo>:</m:mo>
		<m:mn>2</m:mn>
		<m:mo>:</m:mo>
		<m:mn>1</m:mn>
	      </m:mrow>
	    </m:apply>
	  </m:math>
	</equation>
The masses of oxygen appearing in these compounds are in simple whole
	number ratios when we take a fixed amount of nitrogen. The
	appearance of these simple whole numbers is very
	significant. These integers imply that the compounds contain a
	multiple of a fixed unit of mass of oxygen. The simplest
	explanation for this fixed unit of mass is that oxygen is
	<term>particulate</term>. We call the fixed unit of mass an
	<term>atom</term>.  We now assume that the compounds have been
	formed from combinations of atoms with fixed masses, and that
	different compounds have differing numbers of atoms. The mass
	ratios make it clear that oxide B contains twice as many
	oxygen atoms (per nitrogen atom) as does oxide C and half as
	many oxygen atoms (per nitrogen atom) as does oxide A. The
	simple mass ratios must be the result of the simple ratios in
	which atoms combine into molecules. If, for example, oxide C
	has the molecular formula 
	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math>, then oxide B has the formula
	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:msub>
	      <m:mi>O</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:mrow>
	</m:math>, and oxide A has the formula 
	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:msub>
	      <m:mi>O</m:mi>
	      <m:mn>4</m:mn>
	    </m:msub>
	  </m:mrow>
	</m:math>. There are other
	possibilities: if oxide B has molecular formula 
	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math>, then oxide A has formula 
	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:msub>
	      <m:mi>O</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:mrow>
	</m:math>, and oxide C has formula
	<m:math>
	  <m:mrow>
	    <m:msub>
	      <m:mi>N</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math>. Or if oxide A has formula
	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math>, then oxide B has formula
	<m:math>
	  <m:mrow>
	    <m:msub>
	      <m:mi>N</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math>
	and oxide C has formula
	<m:math>
	  <m:mrow>
	    <m:msub>
	      <m:mi>N</m:mi>
	      <m:mn>4</m:mn>
	    </m:msub>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math>. These three possibilities are listed in the following 
	<cnxn target="table2">table</cnxn>.</para>

      <table id="table2" frame="all">
<name>Possible Molecular Formulae for Nitrogen	Oxides</name>
<tgroup cols="4" align="center"><colspec colname="col1" colnum="1" colwidth="1*"/>
	    <colspec colname="col2" colnum="2" colwidth="1*"/>
	    <colspec colname="col3" colnum="3" colwidth="1*"/>
	    <colspec colname="col4" colnum="4" colwidth="1*"/>
           <thead>
	    <row>
	      <entry>Assuming that:</entry>
	      <entry>Oxide C is 	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry>
	      <entry>Oxide B is 	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry>
	      <entry>Oxide A is 	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry>
	    </row>
	  </thead>
	  <tbody>
	    <row>
	      <entry>Oxide A
		is</entry><entry>	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:msub>
	      <m:mi>O</m:mi>
	      <m:mn>4</m:mn>
	    </m:msub>
	  </m:mrow>
	</m:math></entry><entry>	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:msub>
	      <m:mi>O</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:mrow>
	</m:math></entry><entry>	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry></row><row>
	      <entry>Oxide B is</entry><entry>	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:msub>
	      <m:mi>O</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	  </m:mrow>
	</m:math></entry><entry>	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry><entry>	<m:math>
	  <m:mrow>
	    <m:msub>
	      <m:mi>N</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry></row><row>
	      <entry>Oxide C is</entry><entry>	<m:math>
	  <m:mrow>
	    <m:mi>N</m:mi>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry><entry>	<m:math>
	  <m:mrow>
	    <m:msub>
	      <m:mi>N</m:mi>
	      <m:mn>2</m:mn>
	    </m:msub>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry><entry>	<m:math>
	  <m:mrow>
	    <m:msub>
	      <m:mi>N</m:mi>
	      <m:mn>4</m:mn>
	    </m:msub>
	    <m:mi>O</m:mi>
	  </m:mrow>
	</m:math></entry></row></tbody>

</tgroup>
</table>
      <para id="id8335445">We don't have a way (from these data) to
      know which of these sets of molecular formulae are right. But we
      can assert that either one of them or one analogous to them is
      right. </para><para id="id8335462">Similar data are found for
      any set of compounds formed from common elements. For example,
      there are two oxides of carbon, one with oxygen to carbon mass
      ratio 1.33:1 and the other with mass ratio 2.66:1. The second
      oxide must have twice as many oxygen atoms, per carbon atom, as
      does the first. The general statement of this observation is the
      <term>Law of Multiple Proportions</term>.
	<rule type="law" id="multprop"><name>Law of Multiple
      Proportions</name>
	  <statement><para id="mpropp">When two elements
      combine to form more than one compound, the mass of element A
      which combines in the first compound with a given amount of
      element B has a simple whole number ratio with the mass of
      element A which combines in the second compound with the same
      given mass of element B.</para></statement></rule>
	  This sounds confusing, but an example
      clarifies this statement. Consider the carbon oxides, and let
      carbon be element B and oxygen be element A. Take a fixed given
      mass of carbon (element B), say 1 gram. The mass of oxygen which
      combines with 1 gram of carbon to form the first oxide is 1.33
      grams. The mass of oxygen which combines with 1 gram of carbon
      to form the second oxide is 2.66. These masses are in ratio 
	<m:math type="inline">
	  <m:apply>
	    <m:eq/>
	    <m:mrow>
	      <m:mn>2.66</m:mn>
	      <m:mo>:</m:mo>
	      <m:mn>1.33</m:mn>
	    </m:mrow>
	    <m:mrow>
	      <m:mn>2</m:mn>
	      <m:mo>:</m:mo>
	      <m:mn>1</m:mn>
	    </m:mrow>
	  </m:apply>
	</m:math>, a simple whole number ratio.</para><para id="id8471532">In explaining our observations of the Law
      of Multiple Proportions for the carbon oxides and the
      nitrogen oxides, we have concluded that the simple mass ratio
      arises from the simple ratio of atoms contained in the
      individual molecules. Thus, we have established the following
      postulates of the <term>Atomic Molecular Theory</term>.
	<rule type="theory" id="amt"><name>Atomic Molecular
      Theory</name>
	  <statement><para id="amtp"><list id="amtlist"><item>the elements are comprised of
      identical atoms</item> <item>all atoms of a single element have
      the same characteristic mass</item> <item>these number and
      masses of these atoms do not change during a chemical
      transformation</item> <item>compounds consist of identical molecules formed of atoms combined in simple whole number ratios</item></list></para></statement></rule> </para></section>
<section id="id8471572"><name>Review and Discussion Questions</name>
      <exercise id="id8471583"><problem><para id="ex1">Assume that matter does not
      consist of atoms. Show by example how this assumption leads to
      hypothetical predictions which contradict the Law of
      Multiple Proportions. Do these hypothetical examples
      contradict the Law of Definite
      Proportions? Are
      both observations required for confirmation of the atomic
      theory?</para></problem></exercise>

<exercise id="id8342841"><problem><para id="ex2">Two compounds, A and
	    B, are formed entirely from hydrogen and carbon. Compound
	    A is 80.0% carbon by mass, and 20.0% hydrogen, whereas
	    Compound B is 83.3% carbon by mass and 16.7%
	    hydrogen. Demonstrate that these two compounds obey the
	    Law of Multiple Proportions. Explain why
	    these results strongly indicate that the elements carbon
	    and hydrogen are composed of
	    atoms.</para></problem></exercise>

<exercise id="id8342872"><problem><para id="ex3">In many chemical
	    reactions, mass does not appear to be a conserved
	    quantity. For example, when a tin can rusts, the resultant
	    rusty tin can has a greater mass than before rusting. When
	    a candle burns, the remaining candle has invariably less
	    mass than before it was burned. Provide an explanation of
	    these observations, and describe an experiment which would
	    demonstrate that mass is actually conserved in these
	    chemical reactions. </para></problem></exercise>

<exercise id="id8342899"><problem><para id="ex4">The following
	    question was posed on an exam:<quote type="block">An
	      unknown non-metal element (Q) forms two gaseous
	      fluorides of unknown molecular formula. A 3.2 g sample
	      of Q reacts with fluorine to form 10.8 g of the unknown
	      fluoride A. A 6.4 g sample of Q reacts with fluorine to
	      form 29.2 g of unknown fluoride B. Using these data
	      only, demonstrate by calculation and explanation that
	      these unknown compounds obey the Law of Multiple
		Proportions.</quote>A student responded with
	    the following answer:<quote type="block">The Law of
		Multiple Proportions states that when two
	      elements form two or more compounds, the ratios of the
	      masses of the elements between the two compounds are in
	      a simple whole number ratio. So, looking at the data
	      above, we see that the ratio of the mass of element Q in
	      compound A to the mass of element Q in compound B is 
	      <m:math type="inline">
		<m:apply>
		  <m:eq/>
		  <m:mrow>
		    <m:mn>3.2</m:mn>
		    <m:mo>:</m:mo>
		    <m:mn>6.4</m:mn>
		  </m:mrow>
		  <m:mrow>
		    <m:mn>1</m:mn>
		    <m:mo>:</m:mo>
		    <m:mn>2</m:mn>
		  </m:mrow>
		</m:apply>
	      </m:math>, which is a simple whole number ratio. This demonstrates that these compounds obey the Law of Multiple Proportions.</quote>Assess the accuracy of the students answer. In your assessment, you must determine what information is correct or incorrect, provide the correct information where needed, explain whether the reasoning is logical or not, and provide logical reasoning where needed.</para></problem></exercise></section>
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