Name: _________________________________________
Date:____________________________
Optional Homework 3 – do not turn in!!!
1.) Use the standard enthalpies of formation (online or in appendix of text) to calculate the standard enthalpy
change for each of the following chemical reactions:
a.
NH3 (g)
+
HCl (g)
-45.90 kJ/mole
→
-92.31 kJ/mole
NH4Cl (s)
Hrxn =???
-314.55 kJ/mole
H = [1moleNH4Cl( - 314.55 kJ/mole)] – [1 mole NH3( -45.90 kJ/mole) + 1 mole HCl( -92.31 kJ/mole)]
H = -176.34 kJ
b.
MgCl2 (s)
+
-641.62 kJ/mole
Ca (s)
0 kJ/mole
→
Mg (s)
0 kJ/mole
+ CaCl2 (s) Hrxn = ???
-795.8 kJ/mole
H = [1moleCaCl2( - 795.8 kJ/mole) + 1mol Mg(0)] – [1 mole MgCl2( -641.62 kJ/mole) + 1 mole Ca( 0)]
H = -154.2 kJ
2.) Write balanced heat of formation reactions for each of the following. Do not forget to include the Hf values from
the table for the phases listed. Use your appendix (H values) to write balanced formation equations determine
whether these reactions are exo or endothermic.
a.
CaCl2 (s)
Hf = -795.8 kJ/mole EXOTHERMIC
Ca (s) + Cl2 (g) → CaCl2 (s)
a. NaHCO3 (s)
Hf = -947.7 kJ/mole EXOTHERMIC
Na (s) + ½ H2 (g) + C (graphite) + 3/2 O2 (g) → NaHCO3 (s)
b.
HNO3 (l)
Hf = -173.2 kJ/mole EXOTHERMIC
½ H2 (g) + ½ N2 (g) + 3/2 O2 (g) → HNO3 (l)
3.)
Calculate the mass, in grams, of 8.622 x 1023 chlorine atoms
8.622 x 1023 Cl atoms x
34.45grams Cl
1 mole Cl atoms
x
=
23
1 mole Cl
6.022x10 Cl atoms
50.76 grams Cl
4.) Calculate the % mass of each element in the following compounds:
a. H2C2O4
2.02
x 100 =
90.04
24.02
%C=
x 100 =
90.04
64.00
%O=
x 100 =
90.04
H: 2 x 1.01 = 2.02
%H=
C: 2 x 12.01 = 24.02
O: 4 x 16.00 = 64.00
90.04
2.24 % H
26.67 % C
70.08 % O
Check: 2.24 + 26.67 + 71.08 = 99.99 
b. SnF4
Sn: 1 x 118.7 = 118.7
% Sn =
F: 4 x 19.00 = 76.00
%F=
194.70
Check: 60.97 + 39.03 = 100.00 
5.) Give formulas for the following molecular compounds:
N2O4
a.
dinitrogen tetraoxide:
b.
sulfur hexachloride:
SCl6
c.
diiodine heptoxide:
I2O7
118.7
x 100 =
194.70
76.00
x 100 =
194.70
60.97 % Sn
39.03 % F
6.) Complete the following table:
IONS
nitrate
sulfite
hydrogen carbonate
phosphate
sodium
NaNO3
Na2SO3
NaHCO3
Na3PO4
ammonium
NH4NO3
(NH4)2SO3
NH4HCO3
(NH4)3PO4
calcium
Ca(NO3)2
CaSO3
Ca(HCO3)2
Ca3(PO4)2
aluminum
Al(NO3)3
Al2(SO3)3
Al(HCO3)3
AlPO4
7.) Name the following compounds
Species
SrCl2
Name
Strontium chloride
Species
NO
Name
Nitrogen monoxide
Ga2S3
Gallium sulfide
NO2
Nitrogen dioxide
Zn(ClO4)2
Zinc perchlorate
PBr5
Phosphorus pentabromide
CrO3
Chromium (VI) oxide
PBr3
Phosphorus tribromide
HINT: prove/rationalize your answer for CrO3 by writing the equation of the ions coming together to make the
product CrO3. Put your suspected charges on the line to MAKE sure you have the correct species!
Cr+ ____6________
+
O- ________2_________ → CrO3
8.) One of the most widespread environmental carcinogens is benzo[]pyrene (mm = 252.30 g/mole). It is found
in coal dust, cigarette smoke, and in charcoal grilled meat. Analysis of this hydrocarbon shows 95.21 mass %
carbon and 4.79 mass % hydrogen. What is the empirical formula of benzo[]pyrene? What is the molecular
formula of benzo[]pyrene?
95.21 % = 95.21 grams and 4.79% = 4.79 grams since the total percentage = 100
95.21 grams C x
1 mole C
=
12.01g C
7.92756 moles C
4.79 grams H x
1 mole H
= 4.74257 moles H
1.01g H
Remember now we need to divide through by the lowest number of moles. This sets someone at a “1”
which will give us the lowest whole number ratio, thus we are solving for the empirical formula!
7.92756mole C
= 1.67 moles C
4.74257mole
4.74257mole H
= 1 mole H
4.74257mole
These need to be WHOLE numbers, so we need to multiply both by 3
1.67 x 3 = 5 moles C
1 x 3 = 3 moles H
Empirical Formula = C5H3
but is that the molecular formula? Let’s check. Does this molar mass match the mass given??
C5H3= (5 x 12.01) + (3 x 1.01) = 63.08 grams/mole NO
So the molecular formula must be some whole number multiple of the formula we have now such that
the molar mass = 252.30 grams/mole
252.30/63.08 = 4 The whole number multiple is 4!
C20H12 is the molecular formula
(check it by determining the molar mass of this compound!)
C20H12 = (20 x 12.01) + (12 x 1.01) = 252.3 grams/mole 
9.) A dry cleaning solvent (mm = 146.99 g/mole) contains C, H, and Cl. When a 0.250 gram sample was studied
by combustion analysis, 0.451 grams of CO2 and 0.0617 grams of H2O were formed. Calculate the molecular
formula
CwHxCly + O2 → CO2 +
0.250 g
0.451 g
H2O
0.0617 g
The CO2 will give us the carbon, the H2O will give us the hydrogen but since the chlorine does
not appear with the information given, we will need to wait on the chlorine just a bit!
0.451 grams CO2 x
1 mole CO 2
1 mole C
= 0.1231 grams C
x
44.01g CO 2 1 mole CO 2
0.0617 grams H2O x
1 mole H 2 O
2 mole H
= 0.00692 grams H
x
18.01g H 2 O 1 mole H 2 O
Since the total mass of the compound must equal the grams H + grams C + grams Cl we can
now find out the mass of Cl in the compound
0.250 grams = grams C + grams H + grams Cl
0.250 grams = 0.1231 g + 0.00692 g + grams Cl
Grams Cl = 0.120 grams
Now that we have our grams of species, we can calculate the number of moles, divide through
by the lowest number of moles, and get that empirical formula:
0.1231 grams C x
1 mole C
= 0.012049 moles C
12.01g C
0.00692 grams H x
1 mole H
= 0.00685 moles H
1.01g H
0.120 grams Cl x
1 mole Cl
= 0.00339 moles Cl
35.45g Cl
0.012049 moles C
=3
0.00339 moles
0.00685 moles H
=2
0.00339 moles
0.00339 moles Cl
=1
0.00339 moles
The empirical formula = C3H2Cl
But is that the molecular formula? We need to find the mass of the empirical formula and see!
Molar Mass C3H2Cl = (3 x 12.01) + (2 x 1.01) + (1 x 35.45) = 73.5 grams/mole
Nope, it’s not the molecular formula: but we can figure out the multiplier to turn the empirical formula
into the molecular formula!
146.99/73.5 = 2
Therefore the molecular formula = C6H4Cl2
Check just to be sure!
Molar Mass C6H2Cl2 = (6 x 12.01) + (4 x 1.01) + (2 x 35.45) = 147 grams/mole 