Problem Set 2 Solutions
(30 points total)
1. Compound X has the following composition: 32.40% C, 8.16% H, 21.57% O, and 37.87% Si by mass.
Its molar mass is approximately 300 g/mol. Determine the empirical and molecular formulas of
Compound X.
Assume 100 grams of Compound X.
Mass:
Moles:
32.40 g C
2.698
8.16 g H
8.095
21.57 g O
1.348
37.88 g Si
1.349
Mole ratio:
2.00
6.00
1.00
1.00
Whole-number ratio:
2
6
1
1
so the empirical formula is C2H6OSi
This would have a molar mass of 74.16 grams, so we need to multiply by 4 to get the molecular
formula of C8H24O4Si4 , which has a molar mass of 297 g/mol. (3 Points)
2. Compound Y contains only carbon, hydrogen, and oxygen. When a 5.00-gram sample of Compound Y
is subjected to combustion analysis, 10.60 grams of CO2 and 1.63 grams of H2O are produced.
Determine the empirical formula of Compound Y.
1 mol CO 2
1 mol C
12.01 g C
×
×
= 2.89 g C
44.01 g CO 2
1 mol CO 2
1 mol CO 2
1 mol H 2 O
2 mol H
1.008 g H
1.63 g H 2 O ×
×
×
= 0.182 g H
18.016 g H 2 O
1 mol H 2 O
1 mol H
10.60 g CO 2 ×
We can subtract from 5.00 gram total to find: 1.93 g O.
Then set up the table:
Mass:
Moles:
2.89 g C
0.241
0.182 g H
0.181
1.93 g O
0.121
so the empirical formula is
C 4H 3O 2
Mole ratio:
1.99
1.50
1
(3 Points)
Whole-number ratio:
4
3
2
3. Compound Z is known to contain only the elements carbon, hydrogen, nitrogen, and chlorine. When
1.00 gram of Compound Z is dissolved in water and allowed to react with excess AgNO3, all of the
chlorine in Compound Z precipitates as solid silver chloride (AgCl) and 1.95 g AgCl is collected.
When 1.00 gram of Compound Z is completely combusted, 0.900 g CO2 and 0.735 g H2O are collected.
Using all this information, determine the complete empirical formula of Compound Z.
Let's find mass of Cl:
35.45 g Cl
1.95 g AgCl × 143.32 g AgCl = 0.4823 g Cl
Then C and H from the combustion analysis:
12.01 g C
0.900 g CO2 × 44.01 g CO
= 0.2456 g C
2
2.016 g H
0.735 g H2O ×
= 0.0822 g H
18.016 g H2O
Then N by subtraction from:
Total mass = 1.000 g. Subtract (0.4823+0.2456+0.0822) = 0.1899 g N
€
Then set up the usual table:
Mass:
Moles:
Mole ratio:
Whole-number ratio:
0.2456 g C
0.02045
1.51
3
0.0822 g H
0.08155
6.02
12
0.1899 g N
0.01355
1.00
2
0.4823 g Cl
0.01361
1.00
2
so the empirical formula is: C3H12N2Cl2
(3 points)
4. Aluminum will react with bromine (Br2) to form AlBr3. What mass of AlBr3 will be produced if 1.50
gram of Al completely reacts with excess Br2?
1.50 g Al ×
1 mol Al
1 mol AlBr3
266.68 g AlBr3
×
×
=
26.98 g Al
1 mol Al
1 mol AlBr3
14.83 g AlBr3
(3 Points)
5. a) Balance the following equation: Mg (s) + B2O3 (s) ! MgO (s) + B (s)
3 Mg (s) + B2O3 (s) ! 3 MgO (s) + 2 B (s)
(3 points)
b) How many grams of B2O3 (s) are required to produce 20.0 g of B (s)?
1 mol B
1 mol B2O3
69.62 g B2O3
20.0 g B ×
×
×
= 64.4 g B2O3
10.81 g B
2 mol B
1 mol B2O3
(3 points)
c) A mixture of 20.0 g Mg and 20.0 g B2O3 are allowed to react to completion. Identify the mass of
each product, and the mass of any left-over reactants.
Let’s look at the mass of B produced in each case:
•Assuming that all of the Mg is consumed:
1 mol Mg
2 mol B
10.81 g B
20.0 g Mg ×
×
×
= 5.93 g B
24.31 g Mg
3 mol Mg
1 mol B
•Assuming that all of the B2O3 is consumed:
1 mol B2 O3
2 mol B
10.81 g B
20.0 g B2 O3 ×
×
×
= 6.21 g B
69.62 g B2 O3
1 mol B2 O3
1 mol B
So the actual amount of B produced is the lower one, and Mg is the limiting reagent and is
entirely consumed. So we know that 5.93 g of B are produced. How much MgO is produced?
1 mol Mg
3 mol MgO
40.31 g MgO
20.0 g Mg ×
×
×
= 33.16 g MgO
24.31 g Mg
3 mol Mg
1 mol MgO
In order to find how much B2O3 is left-over unreacted, we need to find how much reacted and
then subtract that from how much we had to begin with:
1 mol Mg
1 mol B2O3
69.62 g B2O3
20.0 g Mg ×
×
×
= 19.09 g B2O3 reacted
24.31 g Mg
3 mol Mg
1 mol B2O3
mass of B2O3 left over = 20.0 g initially – 19.09 g reacted = 0.91 g B2O3 left over
So, at the end of the reaction, we have:
0 g Mg
0.9 g B2O3
5.93 g B
33.2 g MgO
(3 points)
6. Mercury (II) nitrate will react with potassium iodide to form mercury (II) iodide and potassium nitrate.
If 2.00 grams of mercury (II) nitrate react completely with 2.00 grams of potassium iodide, determine the
mass of each product, and the mass of any left-over reactants.
First we need a balanced equation:
Hg(NO3)2 + 2 KI ! HgI2 + 2 KNO3
Now let’s calculate the moles of HgI2 which will be produced in each case:
•Assuming all the Hg(NO3)2 is consumed:
1 mol Hg(NO3 )2
1 mol HgI2
2.00 g Hg(NO3 )2 ×
×
= 0.006161 mol HgI2
324.61 g Hg(NO3 )2
1 mol Hg(NO3 )2
•Assuming all the KI is consumed:
1 mol KI
1 mol HgI2
2.00 g KI ×
×
= 0.006024 mol HgI2
166.01 g KI
2 mol KI
€
So the actual amount of HgI2 produced is the lower one, so KI is the limiting reagent, it is
completely consumed, and 0.006024 mol of HgI2 are formed.
€In addition, 0.012048 mol of KNO3 would be formed, and 0.006024 moles of Hg(NO3)2 would be
consumed. Convert all of these to masses, and we have:
For the products:
2.74 g HgI2 and 1.22 g KNO3
We would have consumed 1.96 g Hg(NO3)2, leaving
0.04 g Hg(NO3)2 remaining
(3 points)
7. You are given 10.00 grams of a mixture of FeO and Fe2O3. Complete reduction of this mixture yields
7.28 grams of pure Fe. Calculate the mass of FeO and Fe2O3 in the original sample.
Let x = mass FeO and y = mass Fe2O3.
Then one equation is simply: x + y = 10 g
For the other we need to do stoichiometry:
1 mol FeO
1 mol Fe
55.85 g Fe
x g FeO × 71.85 g FeO × 1 mol FeO × 1 mol Fe = 0.7773(x) g Fe (from FeO)
1 mol Fe2O3
2 mol Fe
55.85 g Fe
y g Fe2O3 × 159.70 g Fe O × 1 mol Fe O × 1 mol Fe = 0.6994(y) g Fe (from Fe2O3)
2 3
2 3
So the other equation is total mass of Fe:
0.7773x + 0.6994y = 7.28g
If we solve these simultaneous equations for x and y, we get:
x = 3.67
y = 6.33
That means the original mixture had:
3.67 g FeO and 6.33 g Fe2O3
(3 points)
8. A sample of calcium metal is contaminated by a certain amount of highly toxic, radioactive thorium
metal (Th). A 10.00-gram sample of this mixture is dissolved in acid, and reacted with excess sodium
oxalate to form a precipitate which is a mixture of calcium oxalate, CaC2O4, and thorium oxalate,
Th(C2O4)2. The total mass of this precipitate is 31.03 grams. Determine the mass of each metal in the
original 10.00-gram mixture.
Let
then
x = mass of Ca
y = mass of Th
x + y = 10.00 g
(Eqn 1)
How much calcium oxalate do we get?
1 mol Ca 1 mol CaC2O4 128.10 g CaC2O4
x g Ca × 40.08 g × 1 mol Ca
× 1 mol CaC O
2 4
= 3.196(x) g CaC2O4
How much thorium oxalate do we get?
1 mol Th 1 mol Th(C2O4)2 408.08 g Th(C2O4)2
y g Th × 232.04 g ×
× 1 mol Th(C O )
1 mol Th
2 4 2
= 1.759(y) g Th(C2O4)2
Therefore, the total mass of product precipitate is 3.196x + 1.759y = 31.03 g
Rearrange Eqn 1 to y = 10 – x
and substitute it in Eqn 2:
3.196x + 1.759(10 – x)
= 31.03
3.196x + 17.59 – 1.759x
= 31.03
1.437x
= 13.44
Solve for
x = 9.35 g Ca
and
y = 0.65 g Th
(3 points)
(Eqn 2)