Explanations
1. A
The formula for hydrogen sulfide is H2S;
this means that its molar mass is equal to 2(1) + 32 = 34 g/mol.
The closest answer choice, and the correct answer, is A,
which is thus slightly more than 1 mole. Don’t try to use the standard
molar volume for a gas in this problem, even though the problem
asks about a gas, because the volume of the gas is not mentioned at
all.
2. C
This is a simple question: round the atomic masses
that you get from the periodic table to calculate the molar mass
of C2H5OH. The answer
is 2(12) + 6(1) + 16, which is equal to slightly more than 46 g/mol.
3. A
Because there are two starting amounts, you might have
guessed that one of them would act as a limiting reagent. The reaction
is written below, and you’ll need to start by finding all the molar
masses to make your determinations. Round off the numbers. The number of
moles of nitrogen is 14 g (1 mol/28 g/mol) = 0.5 mol, and hydrogen
= 15 g (1 mol/2 g/mol) = 7.5 mol. Now check to see which of the
reactants is the limiting reagent. If you started with 0.5 mole
of N2, in order to consume this amount of
nitrogen completely, you’d need to use 3(0.5) = 1.5 moles of hydrogen.
You have much more hydrogen than that (you have 7.5 moles), so nitrogen
is the limiting reagent. The question asks you how much ammonia
would be produced, so now you know which reactant amount to use
to calculate that, and the answer is 2(0.5 mol) = 1.0 moles, which
is equal to 17 g of NH3, answer choice A.
| Molar masses |
28 |
2 |
17 |
| Reaction |
N2 + |
3H2  |
2NH3 |
| No. of moles |
0.5 mol available
limiting reactant |
1.5 mol used
6.0 moles excess!
7.5 mol available |
1.0 mole produced |
| Amounts |
14 g |
15 g |
Slightly more than 17 g |
4. D
This question gives you the grams of water but asks
for atoms of hydrogen. You’ll need to find the number of water molecules
and then double it since H2O contains 2 atoms
of H for every 1 molecule of water. The molar mass of water is 2(1)
+ 16 = 18 g/mol. The number of moles of water, then, is 12 g H2O
(1 mol/18 g/mol) = .66 mol of 6 (as in 6.02
1023) is 4
1023 molecules
of water. Doubling that you get 8
1023 hydrogen
atoms.
5. B
This question is a percent composition question, so
you’ll be looking for the molecule that has the largest mass of
H compared to its whole molar mass. Examine the analysis below, and
it’s clear that water has the highest percent composition of hydrogen.
-
Hydrogen’s atomic weight = 1, chlorine’s atomic weight
= 35, so hydrogen makes up 1/36, or 2.7%, of the compound.
-
Hydrogen’s atomic weight = 1, oxygen’s atomic weight
= 16; there are two hydrogens here, so it makes up 2/18 of the total
weight, or 11%.
-
Hydrogen’s atomic weight = 1, phosphorus’s weight
= 31, oxygen’s weight = 16, so after you add up the total weight
to get 3(1) + 31 + 4(16) = 98, hydrogen makes up 3/98 = 3%.
-
Hydrogen’s weight = 1, sulfur’s = 32, and oxygen’s
= 16. The total weight is equal to 2(1) + 32 + 4(16) = 98, and hydrogen
makes up 2% of this.
- Hydrogen’s
weight = 1, fluorine’s weight = 19, so hydrogen makes up 1/19, or
5.2%, of the molecule by weight.
As you can see, choice B, in which hydrogen
makes up 11% of the total compound, is the correct answer.
6. F, T
(Do not fill in CE.) The first statement
is false—in most chemical reactions, most times the number of moles
of the compounds involved is not equal. The second statement, however, is
true. Once a limiting reagent has been consumed in the course of
a reaction, the reaction can no longer proceed.
7. A
To solve this problem, first turn the percentages into
grams: C becomes 96 g carbon, so the remaining 4% is hydrogen, so
there are 4 g of hydrogen present. Now convert these masses into
moles: the moles of C = 96 g (1 mol/12 g/mol) = 8 mol. For hydrogen,
H = 4 g (1 mol/1 g/mol) = 4 mol. The C:H ratio is 8:4; remember
that the empirical formula is the formula that shows the relative
numbers of the kinds of atoms in a molecule, so this simplifies
to 2:1, and the empirical formula is C2H.
8. E
Turn the percentages into grams and find the number
of moles of each element. Since the compound is 48% C and 4% H,
it must also be 48% O to make the percentages total 100%. Simplify
the mole:mole ratio to get the empirical formula. Calculate the
empirical molar mass. The molecular mass will be some multiple of
the empirical mass. Mol C = 48 g (1 mol/12 g/mol) = 4 mol. Mol H
= 4 g (1 mol/1 g/mol) = 4 mol. Mol O = 48 g (1 mol/16 g/mol) = 3
mol. The empirical formula is then C4H4O3.
The empirical molar mass = 4(12) + 4(1) + 3(16) = 100, so the molecular
formula is twice the empirical formula, or C8H8O6.
9. B
This is not a limiting reactant problem since only
one amount is given here, 7.0 g of ethene. First find the number
of moles of the substance, in this case ethene: 7 g (1 mol/28 g/mol)
= 0.25 mol, then use mole:mole to determine the moles of the rest
of the compounds involved in the reaction. To find the grams of
CO2, you would do the following: 1 mol
44 g/mol = 44 g CO2.
| Molar mass |
28 |
32 |
44 |
18 |
| Balanced equation |
C2H4 + |
3O2  |
2CO2 + |
2H2O |
| No. of moles |
0.25 |
1.25 |
0.5 |
0.5 |
| Amount |
7 g |
|
22 g |
|
10. D
Here, determine the moles of methane and use the mole:mole
ratio to determine the number of moles, then grams of CF4 that
can be produced.
| Molar mass |
16 |
38 |
88 |
20 |
| Balanced equation |
CH4 + |
4F4  |
CF4 |
4HF |
| No. of moles |
0.5 |
2 |
0.5 |
2 |
| Amount |
8 g |
|
44 g |
|
