2045C Exam #1 (Chapters 1-3)
 
1. State number of significant digits:  (5)
   (a) 4.010x10-3   (b) 2.70x106   (c) 0.0012200   (d) 19.040   (e) 1005

2. Provide result with correct significant digits and uncertainty: (5)

    (a) 25.666 - 1.3               (b) 11.11 + 22.22 + 4           (c) (125 ± 3) (3.0 ± 0.3)
    (d) (45.0) ÷ (9.0 ± 0.5)    (e) (8 + 6 + 4) (4.0 ± 0.6)


3. Perform calculations:    (12)
    (a) 24.0 in/s to ft/min
    (b) 2.00 ft3 to in3
    (c) Volume uncertainty for 8.00 cm x 8.0 cm x 8 cm
 

4.  For the following measurements, calculate to proper significant digits and uncertainty:   (16)
    (a) Density (g/mL) for 14.50g and 8.0 ± 0.5 mL.
    (b) Velocity (cm/s) for 125 ± 5cm in 15 ± 2s.
    (c) Volume (cm3) for cube with each side measured as 5.0 ± 0.5 cm.
    (d) Volume if water level in graduate cylinder 20.0 ± 0.2 mL before and 27.5 ± 0.2 mL
          after object immersed.
 
5.   (a) Identify cation/anion for Ba(ClO3)2 and  (NH4)2CO3                 (4)
      (b) Provide numberof  atoms of each element present in Ca3(PO4)2  and  Al(ClO4)3 (4)
      (c) Write formulas for the four ionic compounds possible using   Ag+1    Pb+2    PO4-3    SO4-2   (4)
      (d) Name: KNO3    SnBr4    CaCl2.2H2O   N3S5


6.  (a) Calculate atoms present in 75.0g of iron (Fe). (4)
     (b) Calculate total number of atoms of C, H, and O in 0.600 mol C5H10O2.  (4)
     (c) Calculate grams of O combined with 4.50x1023 atoms of S to from the compound S2O3.  (4)
 

7.  (a) Calculate percent composition (by mass) for C10N4H16     (6)
     (b)  Determine molecular formula for compound consisting of 29.1% Na, 40.5% S, 30.4% O (molar mass = 316). (8)
     (c)  When 15.00g of compound containing CHO subjected to combustion, 33.00g CO2 and   (8)
            18.00g H2O obtained.   Determine empirical formula.


8.     4Al   +  3O2   2Al2O3      (16)
     (a) Calculate mol Al  needed to react with 20.0 mol O2.
     (b) Calculate grams Al  needed to react with 50.0g O2.
     (c) Calculate grams Al2O3 obtained from the reaction of 10.8g Al  with 11.2g O2.
     (d) Calculate % yield if 18.5g Al2O3 actual yield from part (c).
 

Answers
1.  (a) 4     (b) 3     (c) 5     (d) 5     (e) 4

2.  (a) 24.4 ± 0.1    (b) 37 ± 1     (c) 380 ± 40    (d) 5.0 ± 0.3     (e) 72 ± 16

3.  (a) (24.0in/1s)(1ft)/12in)(60s/1min)= 120. ft/min
      (b) (2.00ft3)(1728in3/1ft3) = 3460 in3
      (c) (1/8)(83) = 500±64
 
 

4.  (a) 1.8 ± 0.1 g/mL
     (b) 8.3 ± 1.2 cm/s
    (c) 130 ± 20 cm3
    (d) 7.5 ± 0.4 mL


5.  (a)  cation: Ba+2   anion: ClO3-1 cation: NH4+1      anion: CO3-2
      (b) Ca3(PO4)2:    3 Ca    2P    8 O
           Al(ClO4)3:    1 Al    3 Cl    12 O
      (c)  Ag3PO4    Ag2SO4   Pb3(PO4)2  PbSO4
     (d) potassium nitrate     tin (IV) bromide    calcium chloride dihydate    trinitrogen pentasulfide
 

6.  (a) (75.0g Fe) (6.02x1023 atom/55.8g Fe) = 8.09x1023 atom

     (b) (0.600 mol) (6.02x1023 molecule/1 mol) (17 atom/1 molecule) = 6.14x1024 atom
     (c) (4.50x1023 atom S)(1mol S/6.02x1023 atom S ) = 0.748 mol S
           (0.748 mol S) (3 mol O/2 mol S) (16.0 g O/1 mol O) = 18.0g O
 

7.  (a)  mass 10 C = 120             % C = (120/192)x100 = 62.5%
           mass   4 N =  56.0          % N = (56.0/192)x100 = 29.2%
           mass 16 H =  16.0          % H = (16.0/192)x100 =   8.3%
                                192

    (b)  mol Na  = 29.1/23 = 1.27     mol S = 40.5/32 = 1.27    mol O = 30.4/16 = 1.90
          Na1.27S1.27O1.90    or   Na1S1O1.50   or   Na2S2O3 (molar mass = 158)
           Since molar mass 316, the molecular formula is Na4S4O6

    (c)   CHO   CO2    +   H2O
           15.00g           33.00g         18.00g

             mass C = (33.00g)(12/44) = 9.00g       mass H = (18.00g)(2/18) = 2.00g        mass O = 15.00-9.00-2.00 = 4.00g
             mol C = 9.00/12 = 0.750                         mol H  = (2.00/1) = 2.00                         mol O = 4.00/16 =  0.250
        C0.750H2.00O0.250   or  C3H8O1

8.    (a)  (20.0 mol O2) (4 mol Al/3 mol O2) = 26.7 mol Al
       (b) (50.0g O2) (1 mol O2/32g O2) (4 mol Al/3 mol O2) (27g Al/1 mol Al) = 56.3g Al
       (c) If Al LR:  (10.8g Al) (1 mol Al/27g Al) (2 mol Al2O3/4 mol Al) (102g Al2O3/1 mol Al2O3) = 20.4g  Al2O3
            If O2 LR: (11.2g O2) (1 mol O2/32g O2) (2 mol Al2O3/3 mol O2) (102g Al2O3/1 mol Al2O3) = 23.8g Al2O3
           Al Limiting Reactant and 20.4g Al2O3 obtained
      (d)  % yield = (18.5/20.4) x 100 = 90.7%