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    <title>Recent Posts in 'Chemistry - The Mole Concept' | sgForums.com</title>
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      <title>Chemistry - The Mole Concept replied by UltimaOnline @ Sun, 06 Jul 2008 00:02:44 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by tut4nkh4m3n:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;i agree that it's one of the most fun subjs, but you must put in
quite a lot of effort indeed&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&lt;br /&gt;
Well, every subject must put in a lot of effort at 'A' levels and
beyond. But when you enjoy a subject very much, it becomes
effortless to score distinctions. &lt;img src=
"/images/emoticons/kde-3.5.8/redones/angel.png" alt=
"angel.png" /&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by bonkysleuth:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;Thanks so much!&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Oh, you're welcome, Bonkysleuth. &lt;img src=
"/images/emoticons/kde-3.5.8/redones/angel.png" alt=
"angel.png" /&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</description>
      <pubDate>Sun, 06 Jul 2008 00:02:44 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220959</guid>
      <author>UltimaOnline</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by tut4nkh4m3n @ Sat, 05 Jul 2008 23:57:37 +0800</title>
      <description>&lt;p&gt;i agree that it's one of the most fun subjs, but you must put in
quite a lot of effort indeed&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 23:57:37 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220948</guid>
      <author>tut4nkh4m3n</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by UltimaOnline @ Sat, 05 Jul 2008 23:55:18 +0800</title>
      <description>&lt;p&gt;&lt;a href="http://www.sgforums.com/users/109100" rel=
"nofollow"&gt;FrozernFire&lt;/a&gt;&amp;nbsp;is correct.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;a href="http://www.sgforums.com/users/123442" rel=
"nofollow"&gt;tut4nkh4m3n&lt;/a&gt;, trust me, Chemistry is the most fun of
all subjects to take at 'A' levels.&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 23:55:18 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220940</guid>
      <author>UltimaOnline</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by UltimaOnline @ Sat, 05 Jul 2008 23:19:02 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by bonkysleuth:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;0.3kg of sodium hydroxide is dissolved in 5 litres of water. It
is used as a titrant with 20.0cm3 of sulphuric acid of a
concentration of 0.5mol/dm3. what is the volume of sodium hydroxide
solution that is required to neutralize all the sulphuric acid?&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Solution :&lt;/p&gt;
&lt;p&gt;1) Find the molarity of OH-. (no. of moles / volume of
soln).&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;3) Find the no. of moles of protons H+ in 20cm3 of 0.5M of
H2SO4.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;4) Hence state the no. of moles of OH- required for
titration.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;5) Let the required volume of NaOH be&amp;nbsp;v&amp;nbsp;(algebraic
variable) dm3.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;6) Solve for v in the
equation&amp;nbsp;:&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp; No. of moles of OH-
required = (molarity of OH-) x (v dm3).&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 23:19:02 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220857</guid>
      <author>UltimaOnline</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by tut4nkh4m3n @ Sat, 05 Jul 2008 22:59:46 +0800</title>
      <description>&lt;p&gt;chem needs practice... like a lot a lot a lot... sian.&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 22:59:46 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220787</guid>
      <author>tut4nkh4m3n</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by bonkysleuth @ Sat, 05 Jul 2008 21:20:19 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by FrozernFire:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;find the concentration of the NaOH solution in mol dm-3. find
the no. of moles of NaOH needed to neutralise the H2SO4, then find
the volume of NaOH solution needed&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&lt;br /&gt;
How do i get the no. of moles of NaOH needed to neutralise H2SO4?
my idea is that you write down a balanced chemical equation and
compare the mole ratio of NaOH and H2SO4.(because neutralisation
takes place with the reaction of the 2 products) is my assumption
correct or wrong? please rectify if it's incorrect.&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 21:20:19 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220502</guid>
      <author>bonkysleuth</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by FrozernFire @ Sat, 05 Jul 2008 20:29:24 +0800</title>
      <description>&lt;p&gt;find the concentration of the NaOH solution in mol dm-3. find
the no. of moles of NaOH needed to neutralise the H2SO4, then find
the volume of NaOH solution needed&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 20:29:24 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220412</guid>
      <author>FrozernFire</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by bonkysleuth @ Sat, 05 Jul 2008 19:58:56 +0800</title>
      <description>&lt;blockquote&gt;
&lt;div class="quote_from"&gt;Originally posted by UltimaOnline:&lt;/div&gt;
&lt;div class="quote_body"&gt;
&lt;p&gt;First, write an equation for the combustion of CxHy, and work
out the coefficients of the remaining terms (O2, CO2 and H2O) in x
and y. It'll only take you 10 seconds to do so (do
NOT&amp;nbsp;memorize the coefficients! I always tell my students to
understand and work it out yourself instead of blindly memorizing!
You already got so much to memorize, why torture yourself
further?)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;You will then, of course, obtain (see my stickied post on
balancing equations if you are unable to&amp;nbsp;obtain this balanced
equation)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;CxHy + (x + y/4) O2 --&amp;gt; x CO2 + y/2 H2O&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Since you're given&amp;nbsp;the&amp;nbsp;sample mass AND molar
mass&amp;nbsp;(aka relative molecular mass) of the hydrocarbon, work
out the no. of moles of the hydrocarbon.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Hence, you can obtain :&lt;/p&gt;
&lt;p&gt;(i) no. of moles of CO2 produced in combustion, based on
stoichiometry or the&amp;nbsp;balanced equation.&lt;/p&gt;
&lt;p&gt;(ii) no. of moles of CO2 produced in combustion, based on
experimental data (you're given sample mass).&lt;/p&gt;
&lt;p&gt;(iii) no. of moles of H2O produced in combustion, based on
stoichiometry&amp;nbsp;or the&amp;nbsp;balanced equation.&lt;/p&gt;
&lt;p&gt;(iv) no. of moles of H2O produced in combustion, based on
experimental data (you're given sample mass).&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Simply equate (i) and (ii), and you will obtain the value of
x.&lt;/p&gt;
&lt;p&gt;Simply equate (iii) and (iv), and you will obtain the value of
y.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;------------------------------------------------------------&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;(Edited : &lt;a href="http://www.sgforums.com/users/242366" rel=
"nofollow"&gt;jiaxing2&lt;/a&gt;&amp;nbsp;has worked it out correctly.)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;.&lt;/p&gt;
&lt;/div&gt;
&lt;/blockquote&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;Thanks so much! I have one question. You just have to tell
me what the basis of comparison is. 0.3kg of sodium hydroxide is
dissolved in 5 litres of water. It is used as a titrant with
20.0cm3 of sulphuric acid of a concentration of 0.5mol/dm3. what is
the volume of sodium hydroxide solution that is required to
neutralize all the sulphuric acid?&lt;/p&gt;
&lt;p&gt;Right now, I have worked out the concentration of NaOH, which is
mass of solute/vol of solution&lt;/p&gt;
&lt;p&gt;you get 60g/dm3. so to find the volume of added NaOH, do you do
so in terms of the concentration or any other stuff? if it is
concentration, it doesn't make sense because you cant change a
solution's concentration. However, it seems like there isn't any
other way to compare other than concentration. I find the
concentration, changed them to get the no. of moles involved in the
reaction for each reactant and compare using mole ratio. can it be
done this way?&lt;/p&gt;
&lt;p&gt;Please reply ASAP. urgent. thanks.&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 19:58:56 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220384</guid>
      <author>bonkysleuth</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by UltimaOnline @ Sat, 05 Jul 2008 18:44:48 +0800</title>
      <description>&lt;p&gt;&lt;span style="color: #0000ff;"&gt;&lt;span style="color: #0000ff;"&gt;Here
are some 'O' level&amp;nbsp;questions (but 'A' level students
should&amp;nbsp;also try them for your own revision and practice too!),
related questions (variants&amp;nbsp;of &lt;a href="/users/210425" rel=
"nofollow"&gt;&lt;span style=
"color: #0000ff;"&gt;bonkysleuth&lt;/span&gt;&lt;/a&gt;&lt;span style=
"color: #0000ff;"&gt;'s question)&amp;nbsp;for students' practice and
enjoyment&amp;nbsp;:&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style="color: #0000ff;"&gt;&amp;nbsp;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&lt;span style="color: #0000ff;"&gt;1)&amp;nbsp;&lt;/span&gt;&lt;/p&gt;
&lt;p class="" style=""&gt;&lt;span style=""&gt;&lt;span style="color: #ff0000;"&gt;A
sample of 0.43g of an organic compound containing only carbon,
hydrogen and oxygen, was burnt in excess oxygen. The combustion
produced 1.10g of carbon dioxide and 0.45g of
water.&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p class="" style=""&gt;&lt;span style=""&gt;&lt;span style=
"color: #ff0000;"&gt;&lt;span style=
""&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;/span&gt;
a) Calculate the empirical formula of the
compound.&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p class="" style=""&gt;&lt;span style=""&gt;&lt;span style=
"color: #ff0000;"&gt;&lt;span style=
""&gt;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&lt;/span&gt;
b) Given that this compound has a relative molecular mass of 250g,
deduce its molecular formula.&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style=""&gt;&lt;span style="color: #000000;"&gt;&lt;span style=
"color: #ff0000;"&gt;&lt;span style="color: #0000ff;"&gt;Ans : C15 H30
O3&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style="color: #0000ff;"&gt;&amp;nbsp;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style="color: #0000ff;"&gt;&lt;span style=""&gt;&lt;span style=
"font-size: x-small; color: #000000;"&gt;&lt;span style=
""&gt;2)&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&amp;nbsp; &lt;span style=""&gt;&lt;span style=
"font-size: x-small; color: #000000;"&gt;&lt;span style=""&gt;&lt;span style=
""&gt;&lt;span style="color: #ff0000;"&gt;20cm&lt;sup&gt;3&lt;/sup&gt; of a gaseous
hydrocarbon was mixed with 100cm&lt;sup&gt;3&lt;/sup&gt; of oxygen so that the
hydrocarbon was completely burnt. The volume of gas remaining at
the end of the combustion was 70cm&lt;sup&gt;3&lt;/sup&gt;. After passing
through soda lime, the volume was reduced to 10cm&lt;sup&gt;3&lt;/sup&gt;. All
gases were measured at r.t.p. Determine the formula of the
hydrocarbon.&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style="color: #0000ff;"&gt;Ans : C3 H6&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style="color: #0000ff;"&gt;3)&lt;/span&gt; &lt;span style=
"color: #ff0000;"&gt;&lt;span style=""&gt;A mixture of
MgSO&lt;sub&gt;4&lt;/sub&gt;.7H&lt;sub&gt;2&lt;/sub&gt;O and
CuSO&lt;sub&gt;4&lt;/sub&gt;.5H&lt;sub&gt;2&lt;/sub&gt;O is heated until a mixture of the
anhydrous salts, is obtained. If 5.0g of the hydrated mixture when
heated gives 3.0g of the anhydrous salts, calculate the % by mass
of CuSO&lt;sub&gt;4&lt;/sub&gt;.5H&lt;sub&gt;2&lt;/sub&gt;O in the initial hydrated
mixture.&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style=""&gt;&lt;span style="color: #0000ff;"&gt;Ans :
73.9%&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style=""&gt;&lt;span style="color: #0000ff;"&gt;4) (Ok, this isn't
as closely related to BonkySleuth's qn as the above 3, but is
nonetheless a fun classic that my students enjoy very much, so here
goes)&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style=""&gt;&lt;span style=""&gt;&lt;span style="color: #ff0000;"&gt;When
Fe and Fe&lt;sup&gt;3+&lt;/sup&gt; are mixed together, a reaction occurs in
which Fe&lt;sup&gt;2+&lt;/sup&gt; is produced. What is the ratio of Fe to
Fe&lt;sup&gt;3+&lt;/sup&gt; required to produce equal moles of Fe&lt;sup&gt;2+&lt;/sup&gt;
and Fe&lt;sup&gt;3+&lt;/sup&gt; when the reaction is
complete?&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style=""&gt;&lt;span style=""&gt;&lt;span style="color: #0000ff;"&gt;Ans
: (For fun's sake, I'll leave it to you to work out the answer. You
can post your answer to check with me if you
like).&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&lt;span style=""&gt;&lt;span style=""&gt;&lt;span style=
"color: #0000ff;"&gt;&lt;span style="color: #ff00ff;"&gt;Everyone Loves
Chemistry!&lt;/span&gt; &lt;img src=
"/images/emoticons/kde-3.5.8/redones/angel.png" alt=
"angel.png" /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 18:44:48 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220339</guid>
      <author>UltimaOnline</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
    </item>
    <item>
      <title>Chemistry - The Mole Concept replied by UltimaOnline @ Sat, 05 Jul 2008 18:29:13 +0800</title>
      <description>&lt;p&gt;First, write an equation for the combustion of CxHy, and work
out the coefficients of the remaining terms (O2, CO2 and H2O) in x
and y. It'll only take you 10 seconds to do so (do
NOT&amp;nbsp;memorize the coefficients! I always tell my students to
understand and work it out yourself instead of blindly memorizing!
You already got so much to memorize, why torture yourself
further?)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;You will then, of course, obtain (see my stickied post on
balancing equations if you are unable to&amp;nbsp;obtain this balanced
equation)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;CxHy + (x + y/4) O2 --&amp;gt; x CO2 + y/2 H2O&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Since you're given&amp;nbsp;the&amp;nbsp;sample mass AND molar
mass&amp;nbsp;(aka relative molecular mass) of the hydrocarbon, work
out the no. of moles of the hydrocarbon.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Hence, you can obtain :&lt;/p&gt;
&lt;p&gt;(i) no. of moles of CO2 produced in combustion, based on
stoichiometry or the&amp;nbsp;balanced equation.&lt;/p&gt;
&lt;p&gt;(ii) no. of moles of CO2 produced in combustion, based on
experimental data (you're given sample mass).&lt;/p&gt;
&lt;p&gt;(iii) no. of moles of H2O produced in combustion, based on
stoichiometry&amp;nbsp;or the&amp;nbsp;balanced equation.&lt;/p&gt;
&lt;p&gt;(iv) no. of moles of H2O produced in combustion, based on
experimental data (you're given sample mass).&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Simply equate (i) and (ii), and you will obtain the value of
x.&lt;/p&gt;
&lt;p&gt;Simply equate (iii) and (iv), and you will obtain the value of
y.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;------------------------------------------------------------&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;(Edited : &lt;a href="http://www.sgforums.com/users/242366" rel=
"nofollow"&gt;jiaxing2&lt;/a&gt;&amp;nbsp;has worked it out correctly.)&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 18:29:13 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220326</guid>
      <author>UltimaOnline</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
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      <title>Chemistry - The Mole Concept replied by jiaxing2 @ Sat, 05 Jul 2008 18:04:17 +0800</title>
      <description>&lt;p&gt;hydrocarbon undergo complete combustion to form CO2 and H20&lt;/p&gt;
&lt;p&gt;general formula for the combustion goes like this:&lt;/p&gt;
&lt;p&gt;CxHy + (x+y/4) O2 -----------&amp;gt;&amp;nbsp;&amp;nbsp;&amp;nbsp; XCO2 + (y/2)
H20&lt;/p&gt;
&lt;p&gt;notice Rhs balance wif Lhs&lt;/p&gt;
&lt;p&gt;we know that the products consist of 4.4g CO2....Mr (CO2) =
44&lt;/p&gt;
&lt;p&gt;thus x = 0.10&lt;/p&gt;
&lt;p&gt;Mr (H2O) = 18&lt;/p&gt;
&lt;p&gt;y/2 = 3.6/18&lt;/p&gt;
&lt;p&gt;y = 0.4&lt;/p&gt;
&lt;p&gt;ratio of x:y = 1 : 4........empirical formula and molecular
coincide...CH4 ( Mr = 16 )&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 18:04:17 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220271</guid>
      <author>jiaxing2</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
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      <title>Chemistry - The Mole Concept replied by bonkysleuth @ Sat, 05 Jul 2008 17:51:33 +0800</title>
      <description>&lt;p&gt;Need a little help with this. I don't really understand the
question and what it is asking for.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;A 1.50g sample of hydrocarbon undergoes complete combustion to
produce 4.40g of CO2 ad 3.60g of H2O. What is the empirical formula
of this &lt;strong&gt;compound&lt;/strong&gt;? In addition, its molercular
weight has been determined to be about 16. What isthe molecular
formula?&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Hint : hydrocarbon contains only H&amp;nbsp; &amp;amp; C. Find mass of
carbon and hydrogen first.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;It seems a little strange to me. What's the compound they are
talking about? Is it the combination of CO2 and H2O? But how do you
get the mass of carbon and hydrogen? through simultaneous equation?
That is just a quaint suggestion of mine.&lt;/p&gt;
&lt;p&gt;&amp;nbsp;&lt;/p&gt;
&lt;p&gt;Please help as I haven't a single idea how to get the mass of
carbon and hydrogen out... or perhaps the empirical formula. I
should be able to tackle the rest on my own. THANKS!&lt;/p&gt;</description>
      <pubDate>Sat, 05 Jul 2008 17:51:33 +0800</pubDate>
      <guid isPermaLink="false">www.sgforums.com:2297:323219:8220237</guid>
      <author>bonkysleuth</author>
      <link>http://www.sgforums.com/forums/2297/topics/323219</link>
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