Lesson: Chapter - 7

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.

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.

1. Hydrogen’s atomic weight = 1, chlorine’s atomic weight = 35, so hydrogen makes up 1/36, or 2.7%, of the compound.
2. 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%.
3. 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%.
4. 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.
5. 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.

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.

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