MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_NextPart_01C9062D.2D78DB80" This document is a Single File Web Page, also known as a Web Archive file. If you are seeing this message, your browser or editor doesn't support Web Archive files. Please download a browser that supports Web Archive, such as Microsoft Internet Explorer. ------=_NextPart_01C9062D.2D78DB80 Content-Location: file:///C:/B46592B4/Stoichiometry_Flowchart.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Stoichiometry Flowchart

Homepage:  www.tutor-h= omework.com  (for homework help & tutoring = or find a tutor near you)

 

Flowchart for stoichi= ometry conversions in chemistry:

 

This flowchart can be used by beginning chemistry stud= ents to help them get a feel for how to do conversions in stoichiometry…  Wait a minute!  What the heck is “stoichiometry” anyway and who the heck invented such an awful word?  According to “Wikipedia”, it’s the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.  For example, if I = start with 10 gallons of gasoline, how many liters of oxygen gas would be require= d to burn the 10 gallons of gasoline?  Or maybe a question such as if I have 10 grams of sugar and 10 grams of sulfur= ic acid, which one will be used up first (be the “limiting reactant”).  To answer s= uch questions we will always need the balanced chemical reaction that refers to= the specific situation, so you will need to know what all of the reactants and products are.  Anyway, I don&#= 8217;t know who came up with the word, but that’s what it means.

 

To start lets just look at the flow chart for converti= ng different quantities.  I know = this looks daunting, but we’ll run through a few examples that should help= to clarify things. 

 

Examples are referenced on the flow chart with circled numbers.  Examples are shown d= own below the flowchart.

 

The measurable quantities, are shown in black, the “conversion factors” are shown in red.

 

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =     Volume of a

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =     solution of “A”

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =     (mL, L, etc.)

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =        Ý  Molarity=

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =            ß=   (mol/L = =3D M)

   &nbs= p;            &= nbsp;           &nbs= p;        Density        &= nbsp;           &nbs= p;            &= nbsp;          <= !--[if gte vml 1]> Formula Wt. =             &nb= sp;            =           Avogadro’s #            &n= bsp;           Chemical

            = ;            &n= bsp;           (g/= mL)        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;    (g/mol)        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;   (6.022x1023)                =             f= ormula

Volume of pure &n= bsp;           ¾¾®        &= nbsp;   Mass of pure      =           ¾¾®        &= nbsp;   Moles of       &nbs= p;            &= nbsp;  ¾¾®        &= nbsp;   Molecules         = ¾¾®        &= nbsp;   Atoms of  “X”

substance “A”            =    ¬¾¾        &= nbsp;   substance “A”    &nbs= p;          ¬¾¾        &= nbsp;   substance “A”    &nbs= p;          ¬¾¾        &= nbsp;   of  “A”            =       ¬¾¾        &= nbsp;   in sample of

(mL, L, etc.) &nb= sp;            =                   = ;         (g, kg, etc.)     &nbs= p;             =                 =         (mol)        &= nbsp;           &nbs= p;                       =                 &= nbsp;           &nbs= p;                   =         “compound A”

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =          Ý

XXXXXXXXXXXXX= XXXXXXXXXXXXXXXXXXXXXXXXXXXXX          Ý   *Stoichiome= tric    XXXXXXXXXXXXXXX= XXXXXXXXXXXXXXX

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =          ß   coefficients*

            = ;            &n= bsp;            Density        &= nbsp;           &nbs= p;            &= nbsp;          <= !--[if gte vml 1]> Formula Wt.     ß      &n= bsp;            = ;         Avogadro’s #            &n= bsp;           Chemical

            = ;            &n= bsp;           (g/= mL)        &= nbsp;           &nbs= p;            &= nbsp;           &nbs= p;    (g/mol)        &= nbsp;         ß      &n= bsp;            = ;         (6.022x1023)                =             f= ormula

Volume of pure &n= bsp;           ¾¾®        &= nbsp;   Mass of pure      =           ¾¾®        &= nbsp;   Moles of       &nbs= p;            &= nbsp;  ¾¾®        &= nbsp;   Molecules         = ¾¾®        &= nbsp;   Atoms of  “Y”

substance “B”            =    ¬¾¾        &= nbsp;   substance “B”    &nbs= p;          ¬¾¾        &= nbsp;   substance “B”    &nbs= p;          ¬¾¾        &= nbsp;   of  “B”            =        ¬¾¾        &= nbsp;   in sample of

(mL, L, etc.) &nb= sp;            =             &nb= sp;            =   (g, kg, etc.)     &nbs= p;             =                = ;         (mol)        &= nbsp;           &nbs= p;                      = ;                 &= nbsp;           &nbs= p;                  =          “compound B”

 

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =         Ý  Molarity=

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             ß=   (mol/L = =3D M)

   &nbs= p;            &= nbsp;       The “XXXXX” boundary cann= ot be crossed     &n= bsp;            = ;      Volume of a

   &nbs= p;            &= nbsp;       without going through  !!!!           &nbs= p;            &= nbsp;           &nbs= p;            solution of “B”    &nbs= p;         * The key to converting from a quantitiy of one chemical

   &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =             &nb= sp;            =     (mL, L, etc.)      = ;           to a quanti= ty of another.  (i.e., grams A to gr= ams B)

 

Download the = flowchart as a graphic: Stoichiometry_Flowchart.jpg

 

        Converting from mass to volume and vo= lume to mass:  Density is our conve= rsion factor.

   &nbs= p;        This one you’ve probably done before, let’s convert, say 45.9 mL of liquid mercury (Hg) to grams of mercury.&n= bsp; We need to look up the density of mercury, 13.5 g/mL.

   &nbs= p;       

 

   &nbs= p;        Note: make sure that you have the right units where they belong.  You want mL to “cancel out&#= 8221;.

 

   &nbs= p;        Let’s try that going the other way, we’ll convert 1.50 kg of gasoline into = kg of gasoline.  Well, we need to= know that 1.50 kg =3D 1500 g, but that’s easy enough.

 

   &nbs= p;       

   &nbs= p;        And of course, 2030 mL =3D 2.03 L

 

   &nbs= p;        Now then, what I recommend is setting up your conversions in all one step, more like this…

 

   &nbs= p;       

 

        Converting from mass to moles and from moles to mass:  Molar mass is = our conversion factor.

   &nbs= p;        Molar mass is often called molecular weight or formula weight, depending on the t= ype of compound (molecular or ionic).

 

   &nbs= p;        Lets just convert 100 g of water to moles of water.

 

   &nbs= p;       

 

   &nbs= p;        This is a simple enough calculation and I could have left out “H2O”, but I find it helpful when you’re converting from the mass of one compound to the mass of another, so I often include the chemical formula.

   &nbs= p;        Likewise you can convert, say 10.5 mole of methane, CH4, to grams of meth= ane (Try this yourself).  The answ= er is 168 g CH4.

   &nbs= p;        A good tool to calculate molar masses is: http://www.tutor-homework.com/Chemistry_Help/molar_mass_calculator.ht= ml

 

        Converting from moles of one compound= to moles of another:  We use the = stoichiometric coefficients.

            <= /span>It should be noted that this is pr= obably the most important step in stoichiometric calculations, but one that studen= ts often forget or miscalculate.

 

   &nbs= p;        First, we need a balanced reaction.  Let’s keep it simple…

 

   &nbs= p;        4 Al(s) + 3 O2(g) à 2 Al2O3(s)

 

   &nbs= p;        How many moles of aluminum oxide can be made from 18.7 moles of Aluminum metal, assuming we have more than enough oxygen gas to react?

 

   &nbs= p;       

 

   &nbs= p;        And how many moles of oxygen gas are needed (“consumed”)?

 

   &nbs= p;       

 

        Converting between moles and number of molecules: Use Avogadro’s number, 6.022x1023.

 

   &nbs= p;        This is very much like converting between “dozen eggs” and just eggs.  If you have 10 doz. egg= s, then you have 120 eggs, if you have 3.5 doz. eggs, you have 42 eggs.

 

   &nbs= p;        If you have 10.3 moles of hydrogen gas…=   Let’s convert it to moles…

 

   &nbs= p;       

 

        Converting between molecules and atoms.  Use the molecular form= ula

        &= nbsp;   We’ll pick up from where we left off above…  How many atoms of hydrogen are the= re from the step above?

 

   &nbs= p;       

 

        Converting between a volume of a solu= tion and the moles present: We use molarity (the moles/L) as the conversion fact= or.

   &nbs= p;        If you have 50.3 mL of 0.206 M solution of sodium chloride, how many moles of sodium chloride are there?

 

   &nbs= p;       

 

 

Combining it all…

 

Suppose we have 103.2 lbs of a compound… say [Au(CH2)2P(C= 6H5)2]2Cl2   How many atoms of gold are i= n that compound?

&n= bsp;

To simp= lify things, the condensed formula is Au2C28H28= P2Cl2 and its molar mass is 891.33 g/mol.

&n= bsp;

 

Notice that we first converted pounds to grams, then g= rams to moles, moles to molecules, and finally molecules to atoms.  There are no shortcuts to go from = the first step to the last, that’s why I made that flowchart above to show how to go from one quantity to another.

 

Grams of comp= ound “A” to grams of compound “B”.  This is one of the most common typ= es of problems!

 

If we have 31.8 grams of hydrogen gas, and more than e= nough oxygen gas, how many grams of water can be made?

 

The balanced reaction:  2 H2 + O2 à 2 H2O

 

 

Limiting reac= tants a.k.a. limiting reagents (pronounced “re-agents”).

 

If you have 100.0 g of Al and 100.0 g of O2= , how much Al2O3 can be made.

 

4 Al(s) + 3 O2(g) à 2 Al2<= /sub>O3(s)

 

These are the steps to follow:

(a)  Conv= ert to moles…

 

   &nbs= p;        100.0 g Al (1 mol / 26.98 g) =3D 3.706 mol Al

 

   &nbs= p;        100.0 g O2 (1 mol / 32.00 g) =3D 3.125 mol O2

 

(b)  Divi= de the moles by their stoichiometric coefficients in the balanced reaction.  (Test for the limiting reactant).<= /p>

 

   &nbs= p;        Al:  3.706 / 4 =3D 0.9265

 

   &nbs= p;        O2:  3.125 / 3 =3D 1.042

 

   &nbs= p;        You can see that the aluminum has the smallest mole/stoichiometric coefficient ratio; aluminum is the limiting reactant.

 

(c)    =     The amount of products made and reactants remaining all depends on the limiting reactant.

 

   &nbs= p;       

 

Solution Stoichiometry (Titrations)

 

Titrations are fairly straightforward when the reactan= ts are in a one-to-one mole ratio in the balanced chemical reaction, but sometimes they are not.

 

Let’s take a look at this…  H2SO4(aq) + 2 NaOH(aq) à 2 H2O(l) + Na2SO4(aq)

 

If it takes 38.56 mL of 0.05004 M NaOH to titrate 50.0= 0 mL of H2SO4, what is the concentration of the acid?

 

Titration is just another way to say “neutralization”, but done in such a way that we can analyze the concentration of a solution, sulfuric acid in this case.

 

The easiest way to answer this is with a formula, the = “titration formula”:

 

   &nbs= p;       

 

C: concentration (molarity)        V: volume (mL, L)     = ;   subscript a: acid      =       subscript b: base      =      a: the acid stoichiometric coefficient &n= bsp;         b: the base stoichiometric coefficient.

 

Solve for Ca…

   &nbs= p;       

   &nbs= p;            &= nbsp;      

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