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Test 2 – Study Guide – Analytical Chemistry


Analytical Chemistry 1 – Test 2

Topic 1:  Determine the empirical formula:
% –>  (assume 100g  sample)  –> grams –> moles –>  (divide by smallest mole) Mole RATIOs
Mole Ratios are the ratio of atoms in a chemical formula! that must be unique for each
                         compound = Law of Definite Proportions – Joseph Proust
Question 1 : Analytical Chemistry I – determining chemical formulas.pdf                                                                                             
Topic 2: Determine the molecular formula (if a molecule) :
*Molecular Formula is the actual ratio of the molecule! Salts (PbCl2) are only Empirical!
Example:   H2O2 is the actual number of atoms in this particle (molecular!) but in the lab if we use % by mass composition techniques  we will get an empirical formula (lowest ratio) of HO.
                                                       Empirical Formula                                                Molecular Formula     
                                                                                                   multiple of 2
                                                                      HO                              —–>                                  H2O2                            
Must be have the empirical formula AND Molecular Mass (Mass per mole of the compound).  I call this the BOX method!
The key here is to FIND the Empirical Mass (mass per 1 mole of the lowest ratio) and divide the molecular mass by the empirical mass TO GET A WHOLE NUMBER (a multiple of the empirical)!!!!!  TAKE THIS WHOLE number and multiply by THAT WHOLE NUMBER (multiple) to get actual ration of the molecule.
Topic 1 example from Analytical Chemistry I – determining chemical formulas.pdf worksheet.

Topic 2 example from Analytical Chemistry I – determining chemical formulas.pdf worksheet.
Analytical Chemistry I – determining chemical formulas.pdf
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Analytical Chemistry I KEY- determining chemical formulas.pdf
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APPLICATION: –  LAB 5 – Determination of the empirical formula of an Unknown Hydrate:
*Determine the chemical formula of a hydrate:   We will find the whole number that goes in the blank!

Because of the conservation of mass by Antoine Levassuer, the mass of  water that is released from the crystal (hydrate) is the same mass that was is in the crystal.  THus the moles of water lost is the water that are in the crystal (in the formula ) .  

Formulas of compounds are fixed! (Thanks Joe Proust – Law of definite proportions)
Once we determine the mass of water lost, we at the same time determine the mass of remaining part of the salt called the annhydrate.  Convert grams of both parts of the hydrate to moles  and then get a ratio of how many (moles) by dividing by the lowest number of moles.  That ratio is what we will use for the fixed formula!
In this Lab like all analytical labs we determine the theoretical value from pencil and paper stoichiometry.  In this case we found the experimental percent by mass of water in hydrate (that was lost when we heated the compound) by using the empirical formula that you found by measuring the mass of the anhydrate and mass of the original hydrate.  

                                        MgSO4  · ___H20 (s)   →    MgSO4 (s)   +    ___H2O (g)

                                                                      Mass of Hydrate                Mass of anhydrate

                                                                      Mass of Hydrate     –     Mass of anhydrate = water
                                                                                mass of water    moles of water
                                                                         mass of anhydrate    moles of anhydrate
                  Find the Mole Ratios (divide by the lowest ratio) —–> Empirical Chemical Formula
                                                                                                                                                                                                                            Topic 3:  Determine the empirical formula from COMBUSTION ANALYSIS:
The goal is still the same as topic 1 but instead of percentages you will need to figure out these percentages or grams of the different elements in the TESTED CHEMICAL that is being burned or combusted. 
                                              TESTED CHEMICAL              +                     O2         —–>        CO2        +       H2O
      (HAS Carbon, Hydrogen, and possibly other elements)           EXCESS
                                grams of original compound given                                    ——->     grams CO2    +    grams of  water
The basic idea is that ALL the C atoms from the TESTED CHEMICAL will now be found in the CO2  AND all 
the H atom from the TESTED CHEMICAL will now be found in the H2O.  
This occurs if we combust or burn in excess oxygen.  
The grams of CO2 and H2will be given and one basic way to solve if calculate the percent by mass of each Element C and H from the products and determine the grams of the C and H in CO2  and H2O.
Once you have these grams of the element you can add them together and subtract this value from original TESTED CHEMICAL’s mass because of Antione Levassuer – Law of Conservation of Mass. 
We would subtract because there is normally a another atom in the TESTED CHEMICAL that we cannot get from the products.  This is usually oxygen that is used to combust or other elements that are in the TESTED CHEMICAL.
Example: Question 4 – Analytical Chemistry I – determining chemical formulas.pdf worksheet.
APPLICATION: –  We did not do a lab on this technique.
Topic 4:  Basic Stoichiometry Problems and Percent Yield 
We have already seen that stoichiometry (mole ratios) are used to determine empirical and molecular formulas AND they can also be used to determine information about one chemical in a chemical reaction FROM information from another chemical in a balanced chemical reaction based on the law of conservation of mass.
Remember that we reacted hydrogen with oxygen to create water:
                                                                 2 H2     +       O2     —->    2 H2O
                                                        2          :         1          :          2
                    For every 2 H2 molecules 1 molecule of O2  is reacted to make 2 H2O molecules.
Because of this we can use information about one of the chemicals and determine something about the other. Remember that these are How Many Ratios, so we must be in moles!!!!!
                        Percent Yield =                           experimental mass of product produced             x     100
                                                                    Theoretical mass of product calculated by stoichiometry
Example: Stiochiometry 1 – balance yield.pdf worksheet

Now from the percent yield and the amount of a product experimentally obtained ,we can we can work backwards and determine the amount of grams of the original reactant that was needed to calculate a certain product with the given percent yield:

Example: Stoichiometry 1 – balance yield.pdf worksheet

In the above problem I solved for the theoretical yield because I WAS GIVEN THE % Yield and thus I need to calculate the mass of the reactant, C7H6Othat was needed to make the aspirin, C9H8OBEFORE there was error.  I need to use the theoretical yield from the first calculation because I need to consider the error built into the problem.  If there is 87% yield then there is some reactant not becoming product and thus we need to start from a value that has no error (theoretical yield) to work back to the original amount reactant, before the error occurred. 

If this is confusing you then think of it this way.  If we used 75 g of aspirin (product) and did stoichiometry to calculate the original reactant (salicylic acid) that was used to make the product (aspirin) then we would get an UNDERVALUED amount of Salicyclic acid because there was more of it initially that did not react!!! This is because there was error in the process as reported by the 87% yield. So in order to achieve all the reactant originally even with error I needed to start with the theoretical value that has no error to calculate the reactant before there was error in the process.
Stiochiometry 1 – balance yield.pdf
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Stiochiometry 1 – balance yield key.pdf
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Topic 5:  Solutions Concentrations and ion concentrations
We begin our work with the skills that deal with solutions AND precipitation analysis. In the Topic 5 we must be aware that there is a difference between the molarity of the compound and the molarity of the ion that came from the dissolved compound. 
Now before we can begin we must make sure we know what molarity is and all of its synonyms. 
M = mol of compound dissolved / Liter of solution
M = Molarity
M = molar 
M = [NaCl]
Remember that molarity is a concentration term that is a RATIO of moles of dissolved particles  TO Liter of total solution and is NOT total moles.  WE can have a 1.0 molar concentration in 100 ml an we can have 1.0 molar concentration in 10 Liters.  The ratio of mole to L is the same but it takes more moles of dissolved particles to make a 10 L of a 1.0 M solution that it does to make 100 ml of 1.0 M solution.
Now what if have a 2.5 M of CaCl2 and we want to determine the concentration of the ions that are actually dissolving.  Remember that CaCl2 is not a molecule but is just a formula of the lowest ratio of ion in that ionic crystal.  Because of this it makes a lot of sense that we should be interested in the actual number of ions dissolves NOT the “fictional molecule” of CaCl2 . 
So 1 one CaClformula there is 1 Ca+2 and 2 Cl ions :  CaCl2    —->    Ca+2   +    2 Cl 
Thus of the molarity of the CaCl is 2.5 M  and the molarity of Ca+2 is 2.5M while the [Cl] = 5.0M
This is another form of stoichiometry because we are using mole ratios!
So just be aware that the molarity of an ion may be different than the molarity of the original compound but they are related by their respective ratios.  Notice that Ca+2 in my example has the same molarity as the compound because there is 1 Ca+2 TO 1 CaCl2 .
Example: Analytical Chemistry II – Mol,Dilution, precipitation analysis.pdf
In the above problem I calculated the Molarity of the K3PO4 first and then I used the ratio of the ions to the salt to calculate the Molarities of the individual ions. Notice that the K+ ions has the 3 times the particles as the formula of the salt thus it has 3 times the molarity of the salt, K3PO4. The phosphate ion , PO4-3  has the same molarity of the salt because the ratio of the salt to the ion is 1 : 1.
Topic 6:  Diluting Solutions:
Now in the second part of the problem b) the molarities of the ions, K+ and PO4-3 (major ionic species), are being asked to be calculated if 42 ml of the solution (of K3PO4)  is added to to volumetric flask and water is added to the 100. ml mark.  All this means is that we are taking 42 ml of the 0.1366 molar solution of K3PO4  and are DILUTING it to 100 ml by adding 58 ml of water TILL the new solution is now 100 ml.
Now there are really 2 ways to find the new molarities when you dilute BUT THEY are really the same:
Lets solve for the moles of  K3PO4  in 42 ml (.042L) of solution first:                               
  Molarity =  moles / Liters of solution     —–>   0.1336 M of  K3PO4    =    moles
                                                                                                                                                        .042 L
                                                          .00561 = moles of K3PO4 
*Remember that Molarity is a ratio of moles to volume of solution so there is less moles of K3PO4 in less amount of solution (42 ml) than 275 ml EVEN THOUGH the molarity is the same!
Now take the  .00561 = moles of K3PO4 and solve for the NEW molarity at new volume of 100. ml. Remember that in the problem we diluted the 42 ml of original solution to 100. ml (by adding 58 ml of water).
                                                            M =  .00561 = moles of K3PO4    =    .056 M
                                                                          .100 L (new volume)
The new Molarity makes sense since we added water we lowered the ratio of moles to volume of total solution.  Now all we have to do is adjust the molarity of the K3PO4 to the molarity of the ions using the ratios we found in the first part of the problem (a).
                                                                    .056 M x 3 = 0.168 M K+
                                                                    .056 M x  1 = .056 M  PO4-3
Now we could also Use the DILUTION FORMULA which would of accomplished the same calculation.
                                                                 M    V     =  M    V
                                                                 (0.1336) (.042 L)   =     M    (0.100 L)       
                                                     Solve for M =       (0.1336)(.042)      =  .056 M   
This is exactly what we did in the steps above but if this better for you go for it.
Topic 7:  Making a Molar Solution:
In question 3 of the Analytical Chemistry II – Mol,Dilution, precipitation analysis.pdf you asked to make a molar solution. Here is the exact question.
3. How many grams of silver nitrate must be added to enough water to make 1.0 L of a 0.1 M solution?
We want to make s 1.o M solution and make 1.0 Liter of it. We need to know how much silver nitrate is needed to be dissolved in enough water to make a liter of solution.  Using your Molarity formula you can see we need to solve for moles first:
                                      M =      moles of AgNO3    =    0.1 M  =  moles of AgNO     =       0.1 moles of AgNO3
                                                   Liter of solution                                  1.0 L
Convert to grams:     0.1 moles of AgNO3  x  170 grams (formula mass of AgNO3)  = 17 grams of AgNO3
                                                                                    1 mol
*Remember that we need 17 grams of the salt to be dissolved in enough water to make 100. ml of solution. WE DO NOT ADD 17 grams to 100. of water!!!!
Topic 8:  Predicting the mass of a precipitate – Intro to Precipitation Analysis:
In question 4 of the Analytical Chemistry II – Mol,Dilution, precipitation analysis.pdf you are asked to calculate the mass of a precipitate given the volume and Molarity of a solution. Here is the exact question.
4. What mass of silver chloride will be recovered if excess sodium chloride solution is added to 595 ml of an aqueous solution containing 1.75 x 10-2 M Ag+?
Hmmm..We are given the volume and Molarity of the silver ion, Ag+ .  Like most analytical questions you need to convert to moles of the ion.   When you have Molarity and Volume of a solution you should now know that you need to convert to moles by multiplying moles and volume (in Liters) like we have done in TOPIC 6 above,
So             Molarity moles of Ag+   —–>   moles of Ag+   =  Molarity    x  Liter of solution  
                                  Liter of solution
                                                                                 moles of Ag+   =  1.75 x 10-2  x  .595 L 
                                                                                  moles of Ag+   = .0104 
Now comes the precipitation analysis part of the question. Now that we know the moles of the Agin solution we can calculate the mass of the precipitate that WILL be made from the Cl ions that will be delivered by the sodium chloride, NaCl.  
                                                             NaCl    —–>    Na+   +    Cl   
From the Agalready in solution the Cl–  will form a precipitate:
                                                                Ag+ (aq)   +      Cl  (aq) —->   AgCl (s)
                                                                      1           :         1           :          1
**This is the precipitation analysis part!!!  There is 1 Ag+ for every 1 AgCl so the 
    moles of Ag is equal to the moles of AgCl!!!
                                                       So moles of Ag+   = .01041 = moles of AgCl
So now all we have to do convert the moles of AgCl to grams.
                                                      .0104 mol AgCl  x  143 grams (formula mass)  =  1.49 grams of precipitate
                                                                                              1 mol
Topic 9:  Combustion analysis with Precipitation Analysis to determine Empirical Formula:
In question 5 of Analytical Chemistry II – Mol,Dilution, precipitation analysis.pdf, you are asked to to determine the empirical formula of a compound that has C, Cl and H atoms only.  We have covered combustion analysis and we will get the C atom information from the CO2 formed in the combustion test but now we are adding precipitation analysis AS A SECOND  separate test to gather data for the Cl (ions). Here is the exact question:

5. A pure substance was known to contain only C, H, and Cl. When a 4.00-g sample was burned, 4.34g

of CO2 was produced. In a separate experiment, the chlorine contained in 0.125 g of the compound was converted
to 0.334 g of AgCl. Determine the empirical formula of the compound.
Now I always say in questions that have multiple steps, ” What are we doing last? ”  Once we know what we are doing last we know how to proceed to get to the last step.  
So in this question they are asking for the empirical formula so we have calculate the mole ratios of the 3 atoms in the unknown compound last.  THUS we need to gather moles of each atom, C, Cl, and H.  Because we have only have 3 atoms in the unknown compound, we only have to find 2 out of the three since we have the initial mass of the compound and can subtract to find the 3rd atom.  
****Now there are 2 separate tests, combustion analysis to test for C and precipitation analysis to test for Cl. In these separate tests they used 2 different grams of the unknown.  Because of this it would be wise to find the percentage of the mass of the C and Cl in both tests AND then convert these masses to grams by assuming a 100 gram sample.
Combustion Analysis – (for determining C in unknown)
All of the C that was in the original unknown compound is locked up in the CO2 product of combustion:
Assuming that an excess of O2 was reacted with unknown.
                      Unknown (with C, Cl, H atoms)     +      O2          —–>       CO2     +     H2O
                         4.34 g of CO2 was produced.  Lets find the percent by mass of the C in CO2 .
                           % by mass  =      12   (atomic mass of C)      x 100 =  27.2 % of the mass of CO2 is C atoms!
                                                     44   (molecular mass of CO2)
Now lets find the percentage of the COthat was collected that was C atoms that came from the unknown:
                                       4.34 g of CO2 X  .272   = 1.18 g of C      from the 4.00 g sample   
Now lets find the percent by mass of C in the original 4.00 gram sample:
                                                                     1.18 g C     X  100 = 29.5% of 1st tested sample  is C
                                                               4.00 g sample
We will keep this value for the end but now go after the Cl from precipitation analysis:
In a separate 2nd tested.125 sample, precipitation analysis, generated a precipitate with a mass of .344 grams of AgCl.
Although we do not know the chemical formula of the unknown, which is the purpose of this problem we do know that ALL of the Cl atoms that are in the precipitate CAME originally from the unknown. 
SO all the moles of the precipitate (AgCl) will equal the moles of the Cl atoms because there is 1 AgCl for every 1 Cl.
                                                                      AgCl     —–>   Ag+    +      Cl–     
                                                                        1           :         1        :        1
*Careful this may not always be the case, like in Topic 5, there was 1 K3PO4 for 3 K 
So lets go find the moles of Cl from the precipitate:
                                             .334 grams AgCl  x    1 mol       x     1 Cl       =    0.00234 moles of Cl   
                                                                                                      143 grams         1 AgCl
Lets convert the moles to grams of Cl:
                        0.00234 moles of Cl  x  35.4 g (atomic mass, Cl) = 0.0827 g of Cl in the precipitate
                                                                                       1 mole
Lets find the percent of Cl in the original 0.124 gram tested sample: 
                                           0.0827 g Cl      x     100    =    66.2 % of Cl from 2nd tested Sample 
                                       .125 grams of precipitate
You must understand that we have 2 separate tested samples for each of the two analytical tests.
                            1st sample: 4.00 grams from the combustion analysis —–>  29.5%   C
                               2nd sample: .125 grams from the precipitation analysis —–>   66.2 % of Cl
        The missing percentage is the H atoms.  So 100% – (66.2% + 29.5%)  = 4.3 % of H
NOW we assume a 100 gram sample and our percentages become grams and we convert to moles.
                                                        29.5 g of C / 12 g = 2.46 moles C
                                                        66.2 g of Cl / 35.4 g = 1.87 moles of Cl
                                                          4.3 g of H / 1 g    = 4.3 moles H
Divide by the smallest mole value to get the atom ratio.
                                                        2.46 mol C/1.87 = 1.3 C       x    3  =  about 4
                                                        1.87 mol Cl/1.87 = 1.0 Cl     x    3  =    3 
                                                        4.3 mol H/ 1.87  = 2.30 H   x    3  =  about 7
 We multiplied out values to get the lowest possible whole number Ratios.
Empirical Formula:      C4H7Cl3         
*There were other ways to solve this but I just showed you one.  
Topic 10:  Determining an Empirical Formula with Precipitation Analysis – Hydrate 
Here is another example of a empirical formula determined from the precipitation analysis and heating a hydrate. Both of these chemical analytical tests are examples of gravimetric titrations.
In question 6 of the Analytical Chemistry II – Mol,Dilution, precipitation analysis.pdf  we are looking for the formula of a salt that has water inside it crystal structure.  We will use the concepts from Lab 5 to find the mass of the annhydrate (dried salt) to find the mass of the water (that is the missing mass from the original hydrate mass).
6. A student discovered an old bottle labeled “copper sulfate, hydrated” on a laboratory shelf. When the student heated 0.500 g of this compound in an oven for 1 hour, it lost its waters of hydration. The resulting residue had a mass of 0.320 g. When the student dissolved 0.300 g of the compound in water and added excess BaCl2 solution, 0.280 g of solid BaSO4 was produced. Find the empirical formula of the hydrated copper sulfate.
Lets start with the Hydrate Test: . 500 gram sample
                                                            Hydrate    +   Heat —–>   Anhydrate   +  water
                                                             .500 g                              =        .320 g          +   .180 g
So by subtracting the .320 from .500 I get the mass of the water that was driven out 
of the crystal = .180 g = H2O.  
Percent by mass of water in original compound:  (.180 /.500 gram sample) x 100 = 36%
We will use this a bit later to get the empirical formula.
Now lets deal with the second test which was precipitation analysis.300 gram sample
From this sample, a precipitate of 0.280 g of solid BaSO4 was formed.
Now we must assume like in ALL precipitation analysis that all of the SO4-2 (sulfate ion) became locked into the insoluble precipitate.  Since there is Sulfate ions in the original compound lets find the percent by mass of SO4-2 in the precipitate.
Percent by mass of SO4-2 in BaSO4 :
             Part – SO4-2    =  Lets use the fraction in 1 mole  =     (32 + 64) = 96g   =   .412
        Whole – BaSO4                                                                               233 g
Lets find the grams of SO4-2 in the precipitate.
                                                         .412    x   .280 g    =   .1154 grams of SO4-2  in precipitate.
Now we have to relate this back to the the original compound so lets find the percent by mass of SO4-2   from the tested compound:
 Percent by mass of SO4-2 ( .1154 grams of SO4-2  in precipitate /  .300 gram sample)  x 100  =  38.45 % 
                                1st sample:    .500 g sample of unknown ——–>  36 %  H2O
                                2nd sample:  .300 g sample of unknown ——–> 38.47 %  SO4-2   
    The missing percentage is copper atoms:  100 – (38.45 + 36) = 25.55 % Cu
NOW we assume a 100 gram sample and our percentages become grams and we convert to moles.
                                                        36 g of H2O/ 18g  = 2 moles H2O
                                                 38.47 of SO4-2 / 96  = .40 moles SO4-2
                                                 25.55 of Cu  /  64  =  .399 mole Cu
Divide by the smallest mole value to get the atom ratio.
                                                               2 mole H2O /.399 = 5
                                                       .40 moles SO4-2 / .399  = 1
                                                        399 mol Cu / .399 = 1
                                            Empirical Formula:      CuSO4 · 5H2O      😀