5 - gravimetric stoich NG_2 by stariya

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									                     Chemistry 20 – Gravimetric Stoichiometry NG

Name: __________________________                              Date: ______________

Read pages 286 – 290 of your textbook to complete the following notes guide.

Review – types of reactions:

         reaction type                              general description
formation                          element + element → compound
                                   2 H2(g) + O2(g) → 2 H2O(g)
decomposition                      compound → element + element
                                     2 H2O(g) → 2 H2(g) + O2(g)
(hydrocarbon) combustion           substance + oxygen → __ CO2(g) + __ H2O(g)
                                      CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
                                   (methane, CH4(g), is a hydrocarbon because it consists
                                   of only hydrogen and carbon)
single replacement                 compound + element → element + compound
                                    2 HCl(aq) + Mg(s) → H2(g) + MgCl2(aq)
double replacement                 compound + compound → compound + compound
                                   H2SO4(aq) + 2 NaOH(aq)→ Na2SO4(aq) + 2 HOH(l)



What does the coefficient in front of a formula in a chemical equation tell you?
e.g.,   N2(g) +                 3 H2(g)                  →           2 NH3(g)
                               it tells you PROPORTIONS!
        1 part                  3 parts                              2 parts
        1 molecule              3 molecules                          2 molecules




        1 mole                 3 moles                               2 moles
                               (also called the mole ratio)
   6.02 × 1023 particles       3 (6.02 × 1023 particles)        2 (6.02 × 1023 particles)
What is a mole? see page 51 of Nelson textbook
1 mole of Na is 6.02 × 1023 Na atoms
1 mole of chlorine is 6.02 × 1023 chlorine molecules
1 mole = 6.02 × 1023 molecules
Avogadro’s number is a constant, just like 1 dozen = 12, 1 mole = 6.02 × 1023
molecules
So, for our example above, the mole ratio is 1: 3: 2.


Stoichiometry is a method for predicting or analyzing the quantities of reactants and
products involved in a chemical reaction.
   ○ it is simply using proportions to calculate the amounts of chemicals reacted and
       produced using chemical equations


Gravimetric Stoichiometry is the method of predicting or analyzing the masses of
reactants and/or products involved in a chemical reaction.
    ○ stoich dealing with mass and mole quantities
    ○ also called “mass to mass” stoich


There are four basic steps involved in solving all gravimetric stoichiometry calculations:
   1.   balanced chemical reaction
   2.   mass → mole conversion
   3.   mole ratio
   4.   mole → mass conversion



To convert between mass → moles or moles → mass, you must use the molar mass.
   ○ molar mass is found on your periodic table for each element and can be calculated
      for any compound if you can correctly provide the chemical formula
                                                        g
   ○ molar mass is in units of grams per mole, or
                                                      1 mol
                                     1 mol
   ○ this value can be inverted to         when   necessary to cancel out appropriate units
                                       g

               mass → moles           or     moles → mass
                         1 mol                 g
                 g   ×                               × mol
                           g                 1 mol
                     = mol                           =g
   ○ underneath the equation you can write the given information (mass of malachite),
     the required information (mass of copper(II) oxide) and any other values you will
     need (molar masses, in this case, which are the conversion factors)




         Convert the mass of malachite to moles of malachite (its chemical amount).
                                                       m        1.00 g
                                                or n =   =
                                                       M 221.13g /mol

        nCu2(OH)2CuCO3(s) = 0.0045222267 mol                = 0.0045222267 mol




         Convert moles of malachite to moles of copper(II) oxide
         using the mole ratio.

                                                  2 mol CuO(s)
0.0045222267 mol Cu 2  OH 2 CuCO3 s  
                                             1 mol Cu 2 (OH)2CuCO3 (s)
= 0.0090444533 mol CuO(s)
Convert the moles of copper(II) oxide into the mass of copper(II) oxide.

                                                  79.55 g
nCu2(OH)2CuCO3(s) = 0.0090444533 mol CuO(s)              = 0.719 g
                                                  1 mol

or use m = nM = (0.0090444533 mol CuO(s) ) ( 79.55 g/ mol )= 0.719 g



Alternatively, this entire question can be done all in one step, as follows:




= 0.719 g CuO(s)


                          Double check the significant digits in your answer. Go back to
                          the original question to find the lowest number of significant
                          digits. In this example, it was 3 digits, given a mass of 1.00 g
                          in the question. Round your answer appropriately. Make sure
                          to include units and the identity of the substance as part of your
                          answer.




There will be times when you will solve problems that do not necessarily involve all four
steps of the stoichiometry process. But, the secret to stoichiometry is to practice,
practice, practice! Show all your work for each and every question to get into good
habits.

								
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