UNIT IV – STOICHIOMETRY
REVIEW
I. Good things to know
 stoichiometry, chemical formula
 ionic compound  formula unit
 hydrate, anhydrous
 molecular compound  molecule
 mole, Avagadro’s Number
 formula mass, molar mass, molecular mass
 Standard Temperature and Pressure (STP)
 percent yield
 limiting reactant, excess reactant
 Law of Definite Proportions, Law of Multiple Proportions
 empirical formula, molecular formula
 mass spectrometer
 titration, standard solution, unknown solution
II. Problems
1) Calculate the formula mass for:
a) HF
b) SiO2
1 + 19 = 20.0
28.1 + 32 = 60.1
c) SiF4
28.1 + 4(19) = 104
2) Calculate the number of moles of HF molecules in:
a) 0.385 g HF
b) 3.02 x 1024 molecules of HF
0.385 g HF 
1mol HF
 0.0193mol HF
20 g HF
3.02 x10 24 mcs HF 
c) 50.0 mL of a 0.600M HF solution
1mol HF
 5.02mol HF
6.02 x10 23 mcs HF
d) 1.50L of HF at 0.989 atm of pressure
and 23.1oC
(0.600M )(0.050 L)  0.0300mol HF n 
PV (0.989atm)(1.50 L)

 0.0610mol HF
RT (0.0821)( 296.25K )
3) Answer the following questions using the following equation:
4
HF(g) + 1
SiO2(s) 
1
SiF4(s) + 2
H2O(l)
a) balance the equation
b) How many grams of silicon tetrafluoride will you get if you react 5.00g of silicon dioxide with
excess HF?
5.00 g SiO 2 
1mol SiO 2 1mol SiF4 104 g SiF4


 8.65g SiF4
60.1g SiO 2 1mol SiO 2 1mol SiF4
c) How many grams of silicon tetrafluoride will you get if you react 5.00g of silicon dioxide with
5.00g of HF? How many grams of the excess reactant will be left over?
1mol HF 1mol SiF 4

 0.0625mol SiF 4  LR
20.0 g HF 1mol HF
1mol SiO 2 1mol SiF 4
5.00 g SiO 2 

 0.0832mol SiF 4
60.1g SiO 2 1mol SiO 2
5.00 g HF 
0.0625mol SiF 4 
104 g SiF 4
 6.50 g SiF 4
1mol SiF 4
0.0625mol SiF 4 
1mol SiO 2 60.1g SiO 2

 3.76 g SiO 2  5.00  3.76  1.24 g SiO 2 left
1mol SiF 4 1mol SiO 2
UNIT IV – STOICHIOMETRY
REVIEW
d) If you actually produce 5.92g of silicon tetrafluoride in problem (c), what is the percent yield?
5.92 g
x100%  91.1%
6.50 g
4) Find the percent composition of:
a) SiO2  28.1 + 2(16) = 60.1
28.1
x100  46.8%
60.1
32
%O 
x100  53.2%
60.1
% Si 
b) Ba(NO3)2  137 + 2(14) + 6(16) = 261
137
28
x100  52.5% % N 
x100  10.7%
261
261
96
%O 
x100  36.8%
261
% Ba 
5) How would you prepare 1.00 L of a 4.00 M H2SO4 solution from concentrated H2SO4 (18 M)?
(18M )V  (4.00M )(1.00L)  V  222mL H 2SO 4
1000mL  222  778mL water
6) A sample of 0.6760g of an unknown compound containing barium ions (Ba +2) is dissolved in water
and treated with an excess of Na2SO4. If the mass of the barium sulfate precipitate formed is
0.4105g, what is the percent by mass of barium in the original unknown compound?
BaSO 4  FM  233.4  % Ba 
(0.588)(0.4105 g ) 
137
x100%  58.5%
233.4
0.2415 g Ba +2
x100%  35.7%
0.6760 g cmpd
7) An ion containing only oxygen and chlorine is 31% oxygen by mass. What is the empirical formula?
1O

 1.9375 O

16 g O
1 O
  1.9375
 OCl
1 Cl
1 Cl


69 g Cl 
 1.946 Cl
35.45 g Cl

31g O 
8) A hydrocarbon was found to be 20% hydrogen by weight. If one mole of the hydrocarbon has a
mass of 30 grams, what is its molecular formula?
1C

 6.66 C
12 g C
1 C
30
  6.66 
 CH 3 (FM = 15) 
 2  C2H6
1H
3H
15


12 g H 
 20.0 H 
1g H

80 g C 
UNIT IV – STOICHIOMETRY
REVIEW
9) If you add 20.0 mL of 0.100 M iron(III) nitrate to 20.0 mL of 0.100M sodium hydroxide, how many
grams of precipitate will be formed? Write the net ionic equation for the reaction.
LR
(0.100M )(0.0200 L)  0.00200mol Fe +3 
+3
-1
Fe
+
3OH
 Fe(OH) 3

(0.100M )(0.0200 L)  0.00200mol OH-1  0.002mol 0.002mol
0.00200mol OH-1 
1mol Fe(OH) 3 106.8 g Fe(OH) 3

 0.0712 g Fe(OH) 3
3mol OH-1
1mol Fe(OH) 3
10) If it requires 25.0 L of a 0.500M KI solution to precipitate all of the lead(II) ions out of a 100.0
mL sample, what is the concentration of the lead ions?
1mol Pb 2
(0.500M )( 25 x10 L)  1.25 x10 mol I 
 6.25 x10 6 mol Pb  2
-1
2mol I
6.25 x10 6 mol Pb  2
 6.25 x10 5 M
0.100 L
6
5
-1
11) What is the concentration of each ion in a 0.300 M solution of Na 3PO4? Na3PO4  3 Na+1 + PO4-3
[Na+1] = 0.900M [PO4-3] = 0.300M
12) If you titrate 10.0 mL of an unknown monoprotic acid with a solution of 0.600 M NaOH, and it
requires 25.4 mL of the base to reach the equivalence point, what is the concentration of the acid?
1 H 1
n  MV  (0.600 mol )(0.0254 L)  0.01524mol OH -1 
L
1 OH -1
 0.01524mol H
1
0.0100L
 1.52M
13) What is the concentration of an unknown H2SO4 solution if it requires 156.3 mL of 1.50M NaOH
standard to titrate a 100.0 mL sample of the unknown?
1 H 1 1 H 2 SO 4
n  MV  (1.50 mol )(0.1563L)  0.23445mol OH -1 

L
1 OH -1
2 H
 0.117225mo l H 2 SO 4
 1.17 M
0.100L
14) In your lab, you titrated hydrogen peroxide with potassium permanganate:
6 H+1 + 2 MnO4-1 + 5 H2O2  5 O2 + 2 Mn+2 + 8 H2O
If 36.44 mL of a 0.01652 M KMnO4 solution is required to completely oxidize 25.00 mL of
a H2O2 solution, calculate the molarity of the peroxide solution.
5 H 2O2
1
n  MV  (0.01652 mol )(0.03644 L)  6.02 x10 4 mol MnO 4 
1
L
2 MnO 4

0.001505 H 2 O 2
 0.06020M
0.02500L