Honors Chemistry – Mr. Montero – 2005-2006
Problem Set
Playing with moles and empirical formulas.
Clearly write answers on separate sheets. Show all work and units.
1. Calculate the percentage by mass of the indicated element in the following
compounds:
a) Carbon in acetylene, C2H2, a gas used in welding.
Molecular mass of C2H2 = 2(C) + 2(H) = 2(12) + 2(1) = 26 u
%C =
24
× 100 = 92.3% C
26
b) Hydrogen in ammonium sulfate, (NH4)2SO4, a fertilizer
Formula mass of (NH4)2SO4 = 2[1(N) + 4(H)] + 1(S) + 4(O)
= 2[14 + 4(1)] + 32 + 4(16) = 132 u
%H =
8
× 100 = 6.06% H
132
c) Oxygen in ascorbic acid, C6H8O6, also known as vitamin C
Molecular mass of C6H8O6 = 6(C) + 8(H) + 6(O) = 6(12) + 8(1) + 6(16) = 176 u
%O =
6 × 16
× 100 = 54.5% H
176
2. If Avogadro’s number of pennies is divided equally among 250 millions men,
women, and children of the United States, how many dollars would each receive?
Avogadro’s number = 6.022x1023
1 dollar
6.022 × 10 23
= 2.408 × 1013 dollars
×
6
100 cents 250 × 10 people
That means that each person will receive a fortune 600 times greater than Bill Gates’.
Honors Chemistry – Mr. Montero – 2005-2006
3. The molecular formula of aspartame, also known as NutraSweet®, is C14H18N2O5
a) What is the molar mass of aspartame?
Molar mass = 14(C) + 18(H) + 2(N) + 5(O)
= 14(12) + 18(1.01) + 2(N) + 5(16) = 294.2 g/mol
b) How many moles of aspartame are present in 1.00 mg of Aspartame?
1 mol Aspartame
1g
× 3
× 1 mg = 3.4 × 10 −6 moles Aspartame
294.2 g
10 mg
c) How many molecules of aspartame are present in 1.00 mg of Aspartame?
6.02 × 10 23 molecules Aspartame
× 3.4 × 10 −6 mol Aspartame = 2.05 × 1018 molecules Aspartame
1 mol
d) How many Hydrogen atoms are present in 1.00 mg of Aspartame?
18atoms H
× 2.05 × 1018 molecules Aspartame = 3.7 × 1019 H atoms
1 molecule
4. At least 25µg (micrograms) of tetrahydrocannabinol (THC), the active ingredient
in marijuana, is required to produce intoxication. The molecular formula of THC
is C21H30O2. How many moles of THC does this 25µg represent?
C21H30O2 molar mass: 21(C) + 30(H) + 2(O) = 21(12) + 30(1) + 2(16) = 314.53 g/mol
1 mol THC
1g
× 6
× 25µg = 7.95 × 10 −8 mol THC
314.53 g 10 µg
How many molecules?
6.02 × 10 23 molecules THC
× 7.95 × 10 −8 mol THC = 4.8 × 1016 molecules THC
1 mol THC
Honors Chemistry – Mr. Montero – 2005-2006
5. Determine the empirical formula of each of the following compounds if a sample
contains:
a) 0.104 mol K, 0.052 mol C, and 0.156 mol O
K=
0.052
0.156
0.104
= 2; O =
= 3; C =
= 1 K2CO3
0.052
0.052
0.052
b) 5.28 g Sn and 3.37 g F
5.28 g Sn ×
3.37 g F ×
F=
1mol Sn
= 0.0445 mol Sn
118.71 g Sn
1mol F
= 0.1775 mol Sn
18.98 g F
0.1775
0.0445
= 4; Sn =
=1
0.0445
0.0445
SnF4
c) 87.5 % N and 12.5% H by mass.
Assume a basis of 100g so you will have 87.5 g of N and 12.5 g of H
87.5 g N ×
1mol N
= 6.25 mol N
14 g N
12.5 g H ×
1mol H
= 12.376 mol H
1.01 g N
H=
NH2
12.376
6.25
= 2; N =
=1
6.25
6.25
6. Ibuprofen, a headache remedy, contains 75.69% C, 8.80 % H, and 15.52 % O by
mass and its molecular weight is 206 g/mol. Find its Empirical formula and its
molecular formula
Assuming a basis of 100g of ibuprofen so you will have 75.69 g of C, 8.8 g of H and
15.52 g of O.
Honors Chemistry – Mr. Montero – 2005-2006
75.69 g C
= 6.302 mol C
12.011 g / mol
Converting to moles:
8.8 g H
= 8.71 mol H
1.01 g / mol
15.52 g O
= 0.97 mol O
16 g / mol
C=
6.302
8.71
0.97
= 6.5; H =
= 9; O =
=1
0.97
0.97
0.97
Because the value of Carbon is 6.5 we must double all of them to find the
empirical formula: C13H18O2 This is also the molecular formula.
7. Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which
means that a certain number of water molecules are included in the solid structure.
The formula for Epsom salts can be written as MgSO4 y x H2O, where x indicates
the number of moles of water per mole of MgSO4. 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?
MgSO4yxH2O = 5.061 g
MgSO4 = 2.472 g H2O = 2.589 g
Formula mass MgSO4 = 4(16) + 32.066 + 24.3 = 120.36 g/mol
Molar mass H2O = 18 g/mol
2.472 g MgSO4
= 0.0205 mol MgSO4
120.36 g / mol
Converting to moles:
2.589 g H 2 O
= 0.1438 mol H 2 O
18 g / mol
Finding the mole ratio:
moles H 2 O
0.1438
=
=7=x
moles MgSO4 0.0205
Honors Chemistry – Mr. Montero – 2005-2006
8. The dirt in a certain iron mine contains 32.7 % m/m of the iron ore hematite (ferric
oxide). Hematite can be processed to obtain pure iron which is the main component of
steel. If a steel production plant requires 1 million pounds of pure iron, how much of this
specific dirt will it need to buy?
1 kg dirt 100 g dirt 159.69 g Ore 1 mol Fe2 O3
1 mol Fe
453.6 g Fe
×
×
×
×
×
× 10 6 lb Fe
3
2 mol Fe
55.845 g Fe
1lb Fe
10 g dirt 32.7 g Ore 1 mol Fe2 O3
=1.98x106 kg of dirt or 4.37x106 lb of dirt
9. In the early 20th century, carbon tetrachloride was widely used as a dry cleaning
solvent, as a refrigerant, and in fire extinguishers. However, once it became apparent that
carbon tetrachloride exposure had severe adverse health effects, safer alternatives were
found for these applications, and its use in these roles declined from about 1940 onward.
Carbon tetrachloride persisted as a pesticide to kill insects in stored grain, but in 1970, it
was banned in consumer products in the United States. If you found a 1 gallon bottle of
carbon tetrachloride how many chlorine atoms will you have? The density of carbon
tetrachloride is 1.5842 g/ml?
1 mol CCl 4
1.5842 g CCl 4 10 3 ml 3.785 l
4 Cl atoms 6.02 × 10 molecule CCl 4
×
×
×
×
×
× 1 gal
1 molecule CCl 4
1 mol CCl 4
153.8 g CCl 4
1 ml CCl 4
1l
1 gal
23
= 9.388x1025 atoms of Cl
10. Direct reaction of iodine (I2) and chlorine (Cl2) produces an iodine chloride, IxCly a
bright yellow solid. If you completely used up 0.678 g of iodine and produced 1.246 g of
IxCly what is the empirical formula of the compound? A later experiment showed that the
molar mass of IxCly was 467 g/mol. What is the molecular formula of the compound?
I 2 + Cl 2 ⎯
⎯→ I x Cl y
0.678 g of I2 react with 0.568 g of Cl2 to produce
1.246 g of IxCly
1mol I 2
= 0.002676 mol I 2
253.4 g I 2
1mol Cl 2
0.568 g Cl 2 ×
= 0.0080 mol Cl 2
70.9 g Cl 2
Converting to moles: 0.678 g I 2 ×
moles Cl 2
0.0080
=
= 3 = y So the empirical formula is ICl3
moles I 2
0.002676
The molecular formula is I2Cl6