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<h1><!--mstheme--><font color="#000099"><img src="../../media/chemcomp/chemcomptitle.gif" alt="chemcomptitle.gif" width="370" height="103"><!--mstheme--></font></h1>
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<!--mstheme--></font><!--msthemelist--><table border="0" cellpadding="0" cellspacing="0" width="100%">
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#weight">Molecular Weight, Percent Composition, Empirical
Formula</a>
<!--msthemelist--><tr><td valign="baseline" width="42"><img src="../../../../_themes/education/blbull1.gif" width="15" height="15" hspace="13"></td><td valign="top" width="100%"><!--mstheme--><a href="#form">Molecular formula from Empirical Formula</a>
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<a name="weight"></a>
<center>
<h2><!--mstheme--><font color="#0033CC">Molecular Weight, Percent Composition, Empirical Formula<!--mstheme--></font></h2>
</center>
<br>
<p>The molecular weight of a substance is the sum of the atomic weight of
the atoms in a molecule. The atomic weight is the average atomic mass for
a naturally occurring element. This means that molecular weight is the
average mass of a
molecule of a substance. Molecular weight is expressed in atomic mass
units. For example, we might want to find the molecular weight of a
molecule of water. We have 2 atoms of H, with each hydrogen atom weighing 1
amu. We multiply 2 H atoms by 1 amu a piece to get 2 amu. We add the 16
amu from one O atom to the 2 amu from the oxygen to get a total of 18 amu
for one molecule of water.</p>
<center>H<sub>2</sub>O = 2 atoms of hydrogen and 1 molecule of oxygen</center>
<br>
<center>2 H * 1 amu = 2 amu</center>
<center>1 O * 16 amu = 16 amu</center>
<br>
<center>2 amu + 16 amu = 18 amu = molecular weight of water</center>
<br>
<br>
<p>The formula weight of a substance is the sum of the atomic weights of
all atoms in a formula unit of the compound. Formula weight doesn't depend on
whether
or not the substance is a molecule.For example, sodium
chloride, which is NaCl, has a formula weight of 58.44 amu. This results
from having 22.99 amu from Na and 35.45 amu from Cl. You would use the
formula weight for substances which are not molecules, such as ionic
compounds.</p>
<p><h3><!--mstheme--><font color="#6666CC">Gram Formula Weight Examples:<!--mstheme--></font></h3>
<p>Find the gram formula weight of H<sub>2</sub>SO<sub>4</sub>
<ol>
<li><!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->2H = 2 x 1 = 2<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->1S = 1 x 32 = 32<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->4O = 4 x 16 = 64<!--mstheme--></font></td>
</tr>
<tr>
<td align=right><!--mstheme-->98g/mole<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>Find the gram formula weight of Na<sub>2</sub>CO<sub>3</sub>
<sup>.</sup> 10 H<sub>2</sub>O
<li><!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->2Na = 2 x 23 = 46<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->1C = 1 x 12 =12<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->3O = 3 x 16 = 48<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->10H<sub>2</sub> = 10 x 18 = 180<!--mstheme--></font></td>
</tr>
<tr>
<td align=right><!--mstheme-->286g/mole<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
</ol>
<br>
<br>
<p>Sometimes we want to find out what the formula of a compound would be.
To figure this out, we analyze the compound into amounts of the elements
for a given amount of the compound. This is expressed as the percent
composition which is the mass percentages of each different element in a
compound. We must know the molecular weight of the compound in order to
determine the molecular formula.</p>
<p>Say we have an element X in a compound. This element X is just part of
the whole compound. We define the mass percentage of X as the parts of X
per hundred parts of the total, by mass. That is:</p>
<br>
<center>
Mass % X = (mass of X in the whole)/(mass of the whole) * 100%
</center>
<br>
<p><h3><!--mstheme--><font color="#6666CC">Percent Composition Example:<!--mstheme--></font></h3>
<p>Calculate the percent composition of Mg(NO<sub>3</sub>)<sub>2</sub>
<!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->1Mg = 1 x 24 = 24<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->2N = 2 x 14 = 28<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->6O = 6 x 16 = 96<!--mstheme--></font></td>
</tr>
<tr>
<td align=right><!--mstheme-->148g/mole<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->%Mg = 24/148 x 100 = 16.2%<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->%N = 28/148 x 100 = 18.9%<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->%O = 96/148 x 100 = 64.0%<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>The percentage composition of a compound leads directly to its empirical
formula. An empirical formula for a compound is the formula of a substance
written with the lowest integer subscripts. For example, hydrogen
peroxide has the molecular formula H<sub>2</sub>O<sub>2</sub>. The
molecular formula tells us the precise number of atoms of different
elements in the substance. The empirical number tells us ratio of numbers
of atoms in the compound. The empirical formula of hydrogen peroxide is
HO, while the molecular formula is H<sub>2</sub>O<sub>2</sub>. Compounds with different molecular formulas can have the same empirical
formulas and such substances will have the same percentage composition.
An example is acetylene, C<sub>2</sub>H<sub>2</sub> and benzene,
C<sub>6</sub>H<sub>6</sub>. In order to obtain the molecular formula of a
substance, you need to know the percent composition and the molecular
weight. The molecular weight allows us to choose the correct multiple of
the empirical formula for the molecular formula.</p>
<p><h3><!--mstheme--><font color="#6666CC">Empirical Formula Example:<!--mstheme--></font></h3>
<p>Determine the empirical formula for a compound which is 54.09% Ca,
43.18% O, and 2.73% H
<p>Divide each percent by that element's atomic weight. To get the
answers to whole numbers, divide through by the smallest one.
<!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->Ca = 54.09/40 = 1.352<!--mstheme--></font></td>
<td><!--mstheme-->1.352/1.352 = 1<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->O = 43.18/16 = 2.699<!--mstheme--></font></td>
<td><!--mstheme-->2.699/1.352 = 2<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->H = 2.73/1 = 2.73<!--mstheme--></font></td>
<td><!--mstheme-->2.73/1.352 = 2<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p>CaO<sub>2</sub>H<sub>2</sub> =
<p>Ca(OH)<sub>2</sub>
<p>Calcium Hydroxide
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<center>
<h2><!--mstheme--><font color="#0033CC">Molecular formula from Empirical Formula<!--mstheme--></font></h2>
</center>
<p>The molecular formula of a compound is a multiple of its empirical
formula. The empirical formula is the simplest formula of a substance,
written with the smallest integers. For example, the molecular formula of benzene
C<sub>6</sub>H<sub>6</sub>is equivalent to the empirical formula, (CH)<sub>6</sub>. This means
that the molecular weight is some multiple of the empirical formula
weight. The empirical formula weight is obtained by summing the atomic
weights from the empirical formula. For any kind of molecular compound, we can
write:</p>
<br>
<center>Molecular weight = <i>n</i> * empirical formula weight</center>
<br>
where <i>n</i> is the number of empirical formula units in a compound. We
can get the molecular formula by multiplying the subscripts of the
empirical formula by whatever <i>n</i> is. We can calculate this from the equation:</p>
<br>
<center>n = (molecular weight)/(empirical formula weight)</center>
<br>
<p>Once we determine the empirical formula of a compound, we can calculate
is empirical formula weight. If we have an experimental determination of
its molecular weight, we can calculate <i>n</i> and then its molecular
formula.</p>
<p><h3><!--mstheme--><font color="#6666CC">Molecular Formula Example:<!--mstheme--></font></h3>
<p>A hydrocarbon is 84.25% carbon and 15.75% hydrogen and has a molecular
weight of 114. What is its molecular formula?
<!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->C 84.25/12 = 7.021<!--mstheme--></font></td>
<td><!--mstheme-->7.021/7.021 = 1 x 4 = 4<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->H 15.75/1 = 15.75<!--mstheme--></font></td>
<td><!--mstheme-->15.75/7.021 = 2.25 x 4 = 9<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<!--mstheme--></font><table border=0>
<tr>
<td><!--mstheme-->4 C = 48<!--mstheme--></font></td>
</tr>
<tr>
<td><!--mstheme-->9 H = 9<!--mstheme--></font></td>
</tr>
<tr>
<td align=right><!--mstheme-->57<!--mstheme--></font></td>
</tr>
</table><!--mstheme-->
<p> 114 / 57 = 2
<p>2(C<sub>4</sub>H<sub>9</sub>) = C<sub>8</sub>H<sub>18</sub>
<p>Octane
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