Chemistry 1
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Hydrate Worksheet.  Scroll down for Answers:


 


1) Name the following hydrates:


a. Na2S2O3 * 5 H2O                  b. CaSO4 * 2 H2O                     c. MgSO4 * H2O                       d. Mn(NO3)2 * 4 H2O


2) Write the formulas of the following hydrates:


a. magnesium nitrate hexahydrate                       b. iron (II) sulfate heptahydrate c. copper (II) nitrate trihydrate


d. tin (II) chloride dihydrate


3) What is the formula for a hydrate that is 90.7g SrC2O4 and 9.30g H2O?


a.  formula of hydrate = ____________________      b.   name of hydrate = __________________________


4) What is the formula of a hydrate that is 86.7% Mo2S5 and 13.3% H2O?


a.  formula of hydrate = ____________________        b.  name of hydrate = __________________________


5) During lab, 1.62 g of CoCl2 x H2O were heated. After heating, only 0.88 g of CoCl2 remained.


     What was the formula of the original hydrate?


a.  formula of hydrate = ____________________        b.  name of hydrate = __________________________


6) During lab, 1.04 g of NiSO4 x H2O were heated. After heating, only 0.61 g of NiSO4 remained.


     What was the formula of the original hydrate?


a.  formula of hydrate = ____________________        b.  name of hydrate = __________________________


7) a.)  Determine the percent of WATER in K2S · 5 H2O.           b.)  What is the name of this hydrate?    


8.)  Epsom salts, a strong laxative used in veterinary medicine, is a hydrate.  The formula for Epsom salts can be written


      as MgSO4xH2O, where x indicates the moles of water for every mole of magnesium sulfate.  When 5.061 g of this hydrate is


      heated to 250oC, all the water of hydration is lost, leaving 2.472 g of MgSO4. What is the value of x?


 


Answers to the Hydrate Worksheet:


1) Name the following hydrates:


a. Na2S2O3 * 5 H2O   Sodium Thiosulfate Pentahydrate               b. CaSO4 * 2 H2  Calcium Sulfate Dihydrate                c. MgSO4 * H2Magnesium Sulfate Monohydrate                      d. Mn(NO3)2 * 4 H2Manganese (II) Nitrate Tetrahydrate (Manganese is below stairstep in middle of periodic table, need Roman.  Nitrate is a -1 charge (x 2 of them))


 


2) Write the formulas of the following hydrates:


a. magnesium nitrate hexahydrate     Mg(NO3)2                  


b. iron (II) sulfate heptahydrate   FeSO4 * 7 H2O    


c. copper (II) nitrate trihydrate Cu(NO3)2 * 3H2O


d. tin (II) chloride dihydrate   SnCl2*2H2O


3) What is the formula for a hydrate that is 90.7g SrC2O4 and 9.30g H2O?


 SrC2O4 =  Sr = 1 x 87.62    =  87.62


                  C = 2 x 12.0107  =  24.0214


                  O = 4 x 15.9994  =  63.9976


                                              175.639


 


SrC2O4 = 90.7 g x 1 mole   =        0.516404232 moles / 0.516228446 = 1


                            175.6376 g


 


H2O = 9.30 g x  1 mole    =  0.516228446 moles  / 0.516228446 = 1


                        18.01528 g


 


a.  formula of hydrate = __SrC2O4*H2O____      b.   name of hydrate = __Strontium Oxalate Monohydrate________________________


 


4) What is the formula of a hydrate that is 86.7% Mo2S5 and 13.3% H2O?


 


   Mo2S5 = Mo= 2 x 95.96 =  191.92


                  S  = 5 x 32.065 = 160.325


                                           352.245


 


Mo2S5 = 86.7 g x  1 mole    =      0.246135502 / 0.246135502 = 1


                           352.245 g


 


H2O = 13.3 g x 1 mole   =   0.738262186 / 0.246135502 = 3


                       18.01528 g


 


a.  formula of hydrate = ___Mo2S5*3H2O_______        b.  name of hydrate = ___Molybdenum (V) Sulfide Trihydrate   (Molybdenum is under stairstep, but not in first two columns.  Romans required.)


 


5) During lab, 1.62 g of CoCl2 x H2O were heated. After heating, only 0.88 g of CoCl2 remained.


 


CoCl2 = Co = 1 x 58.933195 = 58.933195


             Cl =   2 x 35.4527    = 70.9054  


                                              129.838595


 


Total = 1.62 g


 


CoCl2 = 0.88 g x      1 mole          =  0.006777646 / 0.006777646 = 1


                            129.838595 g


 


H2O = 1.62 g - 0.88 g = 0.74


 


            0.74 g x  1 mole = 0.041076242 / .006777646  = 6 


                       18.01528 g


 


     What was the formula of the original hydrate?


a.  formula of hydrate = ____CoCl2 * 6 H20_______________        b.  name of hydrate = ___Cobalt (II) Chloride Hexahydrate_____


 


6) During lab, 1.04 g of NiSO4 x H2O were heated. After heating, only 0.61 g of NiSO4 remained.


 


NiSO4 = Ni = 1 x 58.6934 = 58.6934


              S =  1 x 32.065 = 32.065


              O = 4 x 15.9994 = 63.9976


                                        154.756 g


 


NiSO4 = 0.61 g x  1 mole   =   0.003941689 / 0.003941689 = 1


                        154.756 g


 


H2O = 1.04 - 0.61 = 0.43 g


 


           0.43 g x  1 mole   = 0.023868627 / 0.003941689 = 6


                       18.01528 g


     What was the formula of the original hydrate?


a.  formula of hydrate = ____NiSO4 * 6 H2O__________        b.  name of hydrate = ____Nickel (II) Sulfate Hexahydrate___________


7) a.)  Determine the percent of WATER in K2S · 5 H2O.  


K2S*5H2O =


    K = 2 x 39.0983 =      78.1966 g


    S = 1 x 32.065 =        32.065


   H2O = 5 x 18.01528 = 90.0764


                                     200.338 g


 


%H2O = 90.0764  x 100 = 44.96% water


              200.338


         b.)  What is the name of this hydrate?   Potassium Sulfide Pentahydrate.  (Potassium in 1st 2 columns, names only, no Romans) 


 


8.)  Epsom salts, a strong laxative used in veterinary medicine, is a hydrate.  The formula for Epsom salts can be written


      as MgSO4xH2O, where x indicates the moles of water for every mole of magnesium sulfate.  When 5.061 g of this hydrate is


      heated to 250oC, all the water of hydration is lost, leaving 2.472 g of MgSO4. What is the value of x?


 


MgSO4 = Mg = 1 x 24.3050 = 24.3050


                S =  1 x 32.065 = 32.065


                O = 4 x 15.9994 = 63.9976


                                          120.3676 g


 


MgSO4 = 2.472 g   x    1 mole   = 0.020537088  / 0.020537088 = 1


                                120.3676 g


 


H2O = 5.061g - 2.472g = 2.589 g


 


        = 2.589 g x  1 mole  = 0.143711338 / 0.020537088 = 7


                      


ANSWER:  MgSO4*7H2O = Magnesium Sulfate Heptahydrate


 


Empirical Formula Worksheet.  Scroll down for answers.


Empirical and Molecular Formula Worksheet


1.       Determine the empirical formula of a compound that contains 36.5% sodium, 25.4% sulfur, and 38.1% oxygen.


2.       An organic compound has an empirical formula of CH and a molecular mass of 78 u.  What is the molecular formula?


      3.   Determine the empirical formula given the following data for each compound:


                  a) Fe = 63.53%, S = 36.47%                              b) Fe = 46.55%, S = 53.45%


      4.  A compound contains 21.6% sodium, 33.0% chlorine, 45.1% oxygen. Determine the empirical formula of the compound.


      5. What is the empirical formula for a compound that is 43.4 % C, 1.2% H, 38.6% O, and 16.9% N?


      6. Calculate the % composition for magnesium hydroxide.


      7. A compound made of nitrogen and oxygen is found to be 74.1 % oxygen. Find the empirical formula. List three


          possible molecular formulas for this compound.


      8. How many nitrogen atoms are in 59.0 g of ammonium phosphate?


      9. For the following compounds, determine three or four possible molecular formulas for each given empirical formula:


           A. CH2                       B. N2O            C. P2O5                            D. CH3O


      10. Can a compound have the same empirical formula and molecular formula? Explain.


      11. The compound methyl butanoate (an ester) smells like apples. It’s percent composition is: 58.8 % C, 9.9 % H, and


            31.3 % O. If it’s molecular mass is 306 g/mole, what is the molecular formula?


      12. A forensic scientist is given a white substance that is thought to be cocaine. The substance is found to have the


            following percent composition: 49.48% C, 5.19 % H, 28.85 % N, and 16.48 % O. What is the empirical formula for


            this substance. Is the substance cocaine? (The formula for cocaine is C17H2NO4) Bonus: What is the substance?


 


ANSWERS to Empirical Formula Worksheet:


 


1.  Na = 36.5 g x    1 mole     =  1.587662756  /  0.79211626 = 2.00 = 2


                           22.989769 g


 


     S =  25.4 g     x       1 mole       =   0.79211626  /  0.79211626 =  1


                                   32.066 g


 


    O =  38.1 g  x    1 mole       =   2.3813393    /  0.79211626 =  3.01 = 3


                            15.9994 g


 


        ANSWER=  Na2SO3


 


2.     CH=   C:  (1 x 12.0107) = 12.0107


                   H:  (1 x 1.00794) =  1.00794


                                                  13.01864 u


       


           78 u          =  6            Answer = C6H6


     13.01864 u


 


3.  a.) Fe:  63.53 g  x  1 mole  = 1.137613036  / 1.137341733 = 1.00 = 1


                                55.845 g


 


          S:   36.47 g  x     1 mole  =   1.137341733 /  1.137341733 = 1


                                   32.066 g


 


        ANSWER=  FeS


 


     b.)  Fe:  46.55 g  x     1 mole  =  0.833557167  / 0.833557167  =  1


                                    55.845 g


 


           S:   53.45 g  x  1 mole  = 1.666874571 /  0.833557167 = 2.00 = 2


                                 32.066 g


 


        ANSWER=  FeS2


 


4.)  Na = 21.6 g    x  1 mole    = 0.93954837  / 0.930817681 =  1.01  = 1


                            22.989769 g


 


      Cl = 33.0 g    x    1 mole        =    0.930817681  / 0.930817681  = 1


                               35.4527 g


 


  O =  45.1 g  x   1 mole   =   2.818855707 /  0.930817681    =  3.03 = 3


                        15.9994 g


 


      ANSWER = NaClO3


5.  C:    43.4 g  x  1 mole     =   3.613444679    /  1.190547056     =  3.04


                           12.0107 g


 


    H:  1.2 g     x       1 mole     =   1.190547056    /  1.190547056     =  1


                             1.00794


 


    O =   38.6 g  x   1 mole     =   2.412590472  /  1.190547056    =  2.03


                            15.9994 g


 


    N =   16.9 g  x    1 mole       =   1.206561984  /  1.190547056  = 1.01


                            14.00674 g


 


   ANSWER:  C3HO2N


 


6.  Magnesium Hydroxide = Mg(OH)2


 


     Mg =    1 x  24.3050  = 24.3050


     O   =     2 x  15.9994 = 31.9988


     H   =     2 x 1.00794  =   2.01588


                                          58.31968 g


 


     Mg =    24.3050     x    100   =   41.68 % Mg


                 58.31968


 


     O  =       31.9988     x    100   =   54.87 % O


                 58.31968


 


     H  =       2.01588   x    100   =   3.46 % H


                 58.31968


 


ANSWER:  41.68% Mg, 54.87% O, and 3.46% H


 


7.      N = 100%  minus   74.1%   = 25.9% N


 


   N = 25.9 g x    1 mole    = 1.849109786   /  1.849109786  = 1  x 2  = 2


                  14.00674 g


 


  O = 74.1 g x   1 mole   =  4.631423678  /  1.849109786 = 2.50  x 2 = 5


                15.9994 g


 


ANSWER:  Empirical Formula:  N2O5


                     Molecular Formulas:  N4O10 ;  N6O15 ;  N8O20


 


 


8.  Ammonium Phosphate =  (NH4)3PO4 


          N = 3 x 14.00674 =         42.02022 g


          H = 12 x 1.00794 =         12.09528 g


          P =  1  x 30.973761 =      30.973761 g


          O = 4 x 15.9994 =           63.9976   g  


                                                 149.086861 g


 


? N atoms =59.0 g (NH4)3PO4     x           1 mole (NH4)3PO4          x


                                                              149.086861 g (NH4)3PO4     


         

   3 mole N atoms    x  6.022 x 1023 atoms     =     7.15 x 1023  N atoms

 1 mole (NH4)3PO4      1 mole N atoms


 


  


  ANSWER:  7.15 x 1023  N atoms


 


9.  a.)  CH2  =       (x 2) C2H4   =          ( x 3) C3H6   


                         =  (x 4) C4H8   =          (x 5)  C5H10


     b.)  N2O =        (x 2)  N4O2  =         (x 3) N6O3


                           = (x4) N8O4  =            (x5) N10O5


 


10.  Yes.  The mole ratio of the actual molecule can already be at the lowest ratio


       possible.


 


11.  C = 58.8 g / 12.0107 g =  4.895634726    =  2.50     x 2  = 5


                                                 1.956323363


 


       H = 9.9 g / 1.00794 g = 9.822013215    =  5.02       x 2 = 10


                                             1.956323363


 


       O = 31.3 g / 15.9994 g = 1.956323363   = 1            x 2 = 2


                                                1.956323363


 


      ANSWER= C5H10O2


 


12.   C = 49.48 g / 12.0107 g =  4.11965997    =  4.00  = 4


                                                   1.030038626   


 


       H = 5.19 g / 1.00794 g = 5.149116019      =  5.00 = 5      


                                                1.030038626   


 


       N = 28.85 g / 14.0067 g = 2.059728558   =    2.00 = 2


                                                 1.030038626   


 


       O = 16.48 g / 15.9994 g = 1.030038626   = 1           


                                                1.030038626   


     


     Answer:  C4H5N2O


 


EMPIRICAL FORMULA, PERCENT COMP, AND MOLECULAR FORMULA warm-ups with answers (scroll for answers)


 


1.             A substance contains 35.0 g nitrogen, 5.05 g hydrogen, and 60.0 g of oxygen.  What is the percent composition of this compound?


2.             A sample with a mass of 0.432 grams is analyzed.  The sample contains only Fluorine and Oxygen.  If there is 0.128 grams of Oxygen in this sample, what is it’s percent composition?


3.             A compound of Sulfur and Chlorine contains 9.63 grams of Sulfur and 21.3 grams of Chlorine.  What is the percent composition of this compound?


4.             What is the percent composition of Barium Nitrate (Nitrate = (NO3)-1 ?


5.             Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu.


6.             The compound methyl butanoate (an ester) smells like apples. It’s percent composition is: 58.8 % C, 9.9 % H, and 31.3 % O. If it’s molecular mass is 306 g/mole, what is the molecular formula?


7.             A compound is 26.28% carbon, 3.68% hydrogen, and the rest is oxygen. The molecular weight is 548.32 g/mole. What are the empirical and molecular formulas?



 


ANSWERS:



1.  Find the total mass of the compound:  35.0 g 5.05 g 60.0 g = 100.05 g


       N = 35 / 100.05  x 100 = 34.98 %


      


       H = 5.05 / 100.05 x 100 = 5.05 %


 


       O = 60.0 / 100.05 x 100 = 59.97 %


 


2.                   Fluorine = 0.432 g – 0.128 g = 0.304 g Fluorine


 


        F = 0.304 / 0.432  x 100 =  70.37 %


 


        O = 0.128 / 0.432 x 100 =  29.63 %


 


3.  Total mass = 9.63 g 21.3 g = 30.93 g


 


     S = 9.63 / 30.93 x 100 = 31.13 %


 


     Cl =  21.3 / 30.93 x 100 = 68.87 %


 


4.)  Ba 2 (NO3) -1  =  Ba(NO3)2


 


       Ba = 1 x 137.327 = 137.327


       N =  2 x 14.00674= 28.01348


       O = 6 x 15.9994 =  95.9964


                                      261.33688 g


 


      Ba = 137.327 / 261.33688    x 100 =   52.55 %


       N  =  28.01348 / 261.33688 x 100 = 10.72 %


       O = 95.9964 / 261.33688  x 100 = 36.73 %


 


 


5.  CH =  C = 1 x 12.0107 = 12.0107


                H = 1 x 1.00794 = 1.00794


                                              13.01864 g


 


    78 / 13.01864 = 5.99 = 6


 


    CH x 6 = C6H6


 


6.  C = 58.8  / 12.0107 =   4.895634726  /  1.956323362  =  2.50       x 2 = 5


 


     H = 9.9 / 1.00794 =      9.822013215  /  1.956323362  =  5.02       x 2 = 10


 


     O = 31.3 / 15.9994 =    1.956323362  /  1.956323362    = 1          x 2 = 2


 


    Empirical Formula =   C5H10O2


 


    Molecular Formula =   C =   5 x 12.0107   =   60.0535 g


                                                       H =  10 x 1.00794  =   10.0794 g


                                                       O =   2 x 15.9994 =     31.9988 g


                                                                                         102.1317 g


                          306 / 102.1317 =  3


                             


                         C5H10O2  x 3 =  C15H30O


 


7.  Empirical Formula = C3H5O6


        Empirical Forumla Mass = 137.05g/mol ;    548.32 / 137.05 = 4,


         so Molecular Formula = C12H20O24


 


 


Stoichiometry General Equation


 


Change to:                              moles X                        moles Z                     grams Z


Get from:                          Periodic Table         Balanced Equation          Periodic Table


 


? g Z = given g X x     1 mole X     x       moles Z          x molar mass Z


                               molar mass X           moles X               1 mole Z


 


______ C8H18 (l) __ O2 (g) ® ___ CO2 (g) ___H2O (g)


 


1.           If 196 g of octane burns in oxygen, how many moles of


     CO2 are produced?


2.           If 855 moles of oxygen are consumed, how many grams of


     water are created?


3.           If 456 g of Carbon Dioxide are emitted, how many grams of


     water were also produced?


 


 


 


 


ANSWERS to ABOVE


 


__2__ C8H18 (l) _25_ O2 (g) ® __16_ CO2 (g) _18__H2O (g)


 



  1. If 196 g of octane burns in oxygen, how many moles of

     CO2 are produced?


? mol CO2 =196 g C8H181 mol C8H18  x  16 mol CO2  = 13.7 mol


                                    14.22852 g C8H18   2 mol C8H18



  1. If 855 moles of oxygen are consumed, how many grams of

     water are created?


? g H2O =855 mol O2 x  18 mol H2 x18.01528 g H2O = 11100 g H2O


                                    25 mol O2           1 mol H2O



  1. If 456 g of Carbon Dioxide are emitted, how many grams of

     water were also produced?


? g H2O =456 g CO2  x   1 mol CO2 18 mol H2O  x 18.01528 g H2O =


                                  44.0095 g CO2  16 mol CO2     1 mol H2O


 


      = 210 g H2O


 


   


                  __Fe(s) __O2(g)à  __Fe2O3(s)


 


1.         How many grams of Oxygen are burned to create 4.6 moles of Iron (III) oxide?


2.         How many moles of iron in the above reaction would produce 26.8 grams of Iron (III) Oxide?



 


 


ANSWER:


 


Answer to warm up:


                  _4_Fe(s) _3_O2(g)à  _2_Fe2O3(s)


 


? g O2 = 4.6 mol Fe2O3 x  3 mol O2        x        31.9988 g O2 =    220 g O2


                                           2 mol Fe2O3             1 mol O2


 


? mol Fe = 26.8 g Fe2O3 x           1 mol Fe2O3   x    4 mol Fe    =   0.336 mol Fe


                                              159.6882 g Fe2O3        2 mol Fe2O3


 


 


1.         Calculate the number of grams of oxygen that could be produced by heating 9.70 g of potassium  chlorate? 


          __KClO3(s)  ®  __KCl(s)    __O2(g)



2.         How many grams of chlorine gas must be reacted with excess sodium iodide if 10.0 g of sodium  chloride is produced?  (The reaction is balanced already!!)


2NaI(aq)    Cl2(aq)  ®  2NaCl(aq)    I2(s)


ANSWERS:


1.      Calculate the number of grams of oxygen that could be produced by heating 9.70 g of potassium  chlorate? 


          _2_KClO3(s)  ®   _2_KCl(s)    _3_O2(g)


=  3.80 g O2


 


? g O2 = 9.70 g KClO3   x   1 mol KClO3     x    3 mol O2         x  31.9988 g O2


                                           122.5492 g KClO3     2 mol KClO3         1 mol O2


 


2. How many grams of chlorine gas must be reacted with excess sodium iodide if 10.0 g of sodium  chloride is produced?  (The reaction is balanced already!!)


2NaI(aq)    Cl2(aq)  ®  2NaCl(aq)    I2(s)


? g Cl2 = 10.0 g NaCl  x 1 mol NaCl    x   1 mol Cl2   x  70.9054 g Cl2 = 6.07 g Cl2


                                           58.44 g NaCl     2 mol NaCl        1 mol Cl2


 


__ AlBr3 __ K2SO4 à __ KBr __ Al2(SO4)3


 


If 4.64 moles of AlBr3 are used, how many grams of potassium bromide would be formed?


Answers:


_2_ AlBr3 _3_ K2SO4 à _6_ KBr _1_ Al2(SO4)3


 


If 4.64 moles of AlBr3 are used, how many grams of potassium bromide would be formed?


Steps:


1.      Change to moles


2.      Use balanced equation to change “thing”


3.      Change new “thing” to grams


 


2 moles AlBr3 = 6 moles KBr


                                                                 


? g KBr = 4.64 mol AlBr3 x  6 mol KBr  x   119.0023 g KBr


                                              2 mol AlBr3         1 mol KBr


            


             =  1,657 g KBr = 1660 g KBr



 

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