Apologia Chemistry Module 06 Review

Front 1

Using the mole concept, you can actually calculate the amount of product formed in a chemical reaction if you are given the number of moles of reactant material. True or False?

Back 1

True.

Front 2

chemical equations are really the same as mathematical equations. Anything you can do to one, you can do to the other. True or False?

Back 2

True.

Front 3

stoichiometry must always relate the amount of reactant to the amount of product. True or False?

Back 3

False. This simply isn't true. Like mathematical equations, chemical equations can be used to relate the left side of the equation to the right side of the equation, and vice versa .

Front 4

Whenever you see parentheses in a molecule, what must you do?

Do it to the following:

(NH^4)^2SO^4

Back 4

You must distribute the subscript after the parentheses to every atom within the parentheses. That's the only way you will know how many of each atom is in the molecule.

N^2H^8SO^4

Front 5

In more complex chemical equations, the relationships between quantities of substances depends only on.... what?

Back 5

the quantity of the limiting reactant . As a result, you can essentially ignore the quantities of the other reactants, because the limiting reactant determines how much product can be formed in the reaction.

Front 6

What does it mean when we say that the reactants “reacted completely?”

Back 6

This means that all reactants were added in just the right quantities necessary to be completely used up.

Front 7

If we know that the substance reacts completely, then what else do we know about the substance?

Back 7

Then we know that it is the limiting reactant.

Front 8

Why would a chemist add excess reactant?

Back 8

To make sure that he doesn't waste any money, the chemist might want to use a lot extra of a certain reactant to be confident that all of the limiting factor reacted fully and none was wasted.

(Thus, it would make sense if the limiting factor were the more expensive one!)

Front 9

In stoichiometry, when using the relationship given by the chemical equation to convert the number of moles of one substance into the number of moles of another substance, what three broad things are we trying to learn?

Back 9

1. Sometimes we are trying to determine how much reactant is needed to react fully with another reactant.

2. Sometimes we are trying to see how much product would be formed.

3. Sometimes we are trying to determine how much starting material we would need in order to end up with a certain amount of product.

In each case, however, we used the same mathematical tools to determine the answer.

Front 10

What did the French chemist named Joseph Louis Gay-Lussac determine?

Back 10

The stoichiometric coefficients in a chemical equation do not relate just the number of moles of each substance in a chemical equation......

Joseph Louis Gay-Lussac determined that these coefficients also relate the volumes of gases in a chemical equation.

Front 11

Gay-Lussac's Law can be used in stoichiometry only if everything you are using in the problem is a gas.

True? Or False?

Back 11

True!!!

Front 12

What Gay-Lussac's Law allows us to do is also answer the problem if liters (or any other volume unit) is used instead of moles, so long as the substances we are interested in are all gases.

True or False?

Back 12

True.

Front 13

Gay-Lussac's Law is a useful tool, but there are many things it cannot do. Give the one example listed in the book.

Back 13

One thing it cannot do is convert from moles to liters.

Front 14

Usually, when asked to determine “how much” of a certain substance you have, the easiest thing to do is to measure the substance's mass. So what is used, usually, instead of moles?

Back 14

We usually report the quantity of a substance in grams, not moles.

Front 15

Why are grams used instead of moles?

Back 15

It turns out that you cannot purchase chemicals by the mole. Instead, when you purchase chemicals, you do it by the gram. Thus, if you wanted to order the necessary quantities of reactants, you would have to determine how many grams of each you required.

Front 16

Stoichiometry allows us to relate the number of moles of one substance in a chemical equation to the number of moles of another substance in the chemical equation.

Thus, in order to use stoichiometry, we must have our quantities in moles. Therefore, if a problem gives us the amount of a substance in grams, what do you do?

Back 16

We must first convert to moles before we can use stoichiometry.

Front 17

In order to determine the mass relationship between one chemical and another in a chemical equation, we must do what?

Back 17

convert to moles.

Front 18

In order to determine the mass relationship between one chemical and another in a chemical equation what are the steps you take?

Back 18

1. Since the chemical equation gives me the conversion relationship in moles, I must first convert from grams to moles.

2. Then I can use the chemical equation to convert from one chemical into another.

3. At that point, I have the moles of the new chemical, not the mass. Therefore, I must then convert from moles back into grams.

Front 19

How can we determine a molecule's chemical formula without ever seeing the atoms that make it up? In other words, how can you determine a chemical formula? (Such as H^2O)

Back 19

First of all, a decomposition reaction is carried out which separates a compound into its constituent elements. This tells us what atoms go in the chemical formula and allows us to write an unbalanced chemical equation for the decomposition. The masses of each individual element then allow us to calculate the moles of each element, and that tells us the stoichiometric coefficients which go next to each element on the products side of the equation. Once we have that, then we can figure out the chemical formula necessary to make the equation balance.

Front 20

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.

Al^6O^9

Back 20

In order for a chemical formula to be empirical, it must represent a simple, whole-number ratio of atoms. Therefore, if a chemical formula is an empirical formula, the subscripts cannot have a common factor.

This is NOT an empirical formula, because 6 and 9 have a common factor of 3. In order to get the empirical formula, you would divide by that common factor and get Al^2O^3.

Front 21

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.

PbO^2

Back 21

In order for a chemical formula to be empirical, it must represent a simple, whole-number ratio of atoms. Therefore, if a chemical formula is an empirical formula, the subscripts cannot have a common factor.

This IS an empirical formula, because 1 and 2 have no common factors.

Front 22

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.

MgCO^2

Back 22

In order for a chemical formula to be empirical, it must represent a simple, whole-number ratio of atoms. Therefore, if a chemical formula is an empirical formula, the subscripts cannot have a common factor.

This IS an empirical formula, because 1, 1, and 2 have no common factors.

Front 23

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.

C^8H^18O^2

Back 23

In order for a chemical formula to be empirical, it must represent a simple, whole-number ratio of atoms. Therefore, if a chemical formula is an empirical formula, the subscripts cannot have a common factor.

This is NOT an empirical formula, because 8, 18 and 2 have a common factor of 2. In order to get the empirical formula, you would divide by that common factor and get C^4H^9O.

Front 24

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.

Si^3H^9O^2

Back 24

In order for a chemical formula to be empirical, it must represent a simple, whole-number ratio of atoms. Therefore, if a chemical formula is an empirical formula, the subscripts cannot have a common factor.

This IS an empirical formula, because 3, 9, and 2 have no common factors.

Front 25

Which gives you more detailed information: empirical formulas or molecular formulas?

Back 25

Even though empirical formulas do not contain as much detailed information about a molecule as do molecular formulas, they are better than nothing.

Front 26

It is possible to convert from a molecule's empirical formula to its molecular formula, if you happen to know the molecule's mass. True or False?

Back 26

True.

Front 27

To go from a molecular formula to an empirical formula, what do you have to do?

Back 27

You have to find the common factor and divide by it

Front 28

If we want to turn an empirical formula into a molecular formula, what will we have to do?

Back 28

We will have to multiply by the common factor. How do we know what that common factor is? We can determine it from the molecule's mass.

Front 29

the molecular mass of CH^4 is 16.0 amu. That means it takes 16.0 grams to make one mole. So what is its molar mass?

Back 29

The molar mass, therefore, is 16.0 grams.

Front 30

The molar mass of KCl, for example, is 74.6 grams.

What is its molecular mass?

Back 30

That means its molecular mass is 74.6 amu.

Front 31

Why do chemists use the term “molar mass”?

Back 31

We use it because molar mass is typically what we measure in the lab.

It is hard to measure the mass of a single molecule, because a single molecule has very little mass (on the order of 10 -24 grams). Thus, the molecular mass is hard to measure. It can be measured; it just requires some pretty sophisticated equipment.

However, it is pretty easy to measure the mass of a mole of molecules, because a mole of molecules has a mass that is somewhere between several grams and a few hundred grams. That's something you can measure on a typical lab scale.

Thus, we typically measure the molar mass of a compound and then infer from that its molecular mass.

Front 32

When solving problems, if the stoichiometric coefficients turn out not to be integers, what do you do?

Back 32

You can simply divide by the smallest of them, and that will give you a reasonable chemical equation. Balancing the equation then gives you the chemical formula. If the chemical formula is not an empirical formula, you can then convert it to one.

Front 33

Define stoichiometry and explain why it is such a useful tool for chemists.

Back 33

Stoichiometry is the method used to relate the quantities of substances in chemical equations. It is
useful because with the quantity of just one substance in the equation, you can learn something about
the quantities of all other substances in that equation.

Front 34

Explain what a limiting reactant is and why it is important in stoichiometry.

Back 34

A limiting reactant is the reactant in a chemical equation that runs out first. It is important because
its quantity determines the amount of products produced.

Front 35

A chemist experiments with the following reaction:
14HCl + K^2Cr^2O^7 -----> 2KCl + 2CrCl^3 + 3Cl^2 + 7H^2O
If the chemist adds 15 moles of HCl to 1 mole of K^2Cr^2O^7, what is the limiting reactant?

Back 35

The limiting reactant is K^2Cr^2O^7 because 14 moles of HCl react with 1 mole of K^2Cr^2O^7. Since 15
moles of HCl are added, the 1 mole of K^2Cr^2O^7 will run out and there will still be 1 extra mole of HCl.

Front 36

State Gay-Lussac's Law and explain how it is used.

Back 36

Gay Lussac's Law says that the stoichiometric coefficients in a chemical equation relate the volume of gases in that equation as well as the number of moles. It can be used to relate the quantities of
gaseous substances in a chemical equation if those quantities are given in volume units.

Front 37

A chemist wants to perform the following reaction:
CaCO^3 (s) + 2HCl (g) ------> CaCl^2 (aq) + CO^2 (g) + H^2O (l)
Which substances can she use Gay-Lussac's Law to relate to one another?

Back 37

The chemist can only use Gay-Lussac's Law to relate the volumes of HCl and CO^2, since those are
the only gases in the equation.

Front 38

If a stoichiometry problem gives you the quantity of a substance in grams, what is the first thing you must do to solve the problem?

Back 38

Since stoichiometry can only be done in moles (or volume in the case of gases), the first thing you must do is convert to moles.

Front 39

What is the difference between molecular formulas and empirical formulas?

Back 39

Molecular formulas provide the exact number of each type of atom in the molecule. Empirical
formulas provide only a simple, whole-number ratio of atoms in the molecule.

Front 40

. Which of the following are empirical formulas?
a. C^12H^24O^10
b. K^2S^2O^3
c. CaN^2O^6

Back 40

The formulas given in (b) and (c) are empirical formulas, because the subscripts have no common factor. The formula given in (a) is not an empirical formula because the subscripts have a common
factor of 2

Front 41

A compound has a molecular formula of C^14H^21O^7. What is its empirical formula?

Back 41

The common factor between the subscripts is 7; thus we must divide each subscript by 7 to get C^2H^3O.

Front 42

A compound has a molecular formula of H^3PO^4. What is its empirical formula?

Back 42

The molecular formula is also the empirical formula, since the subscripts have no common factor. Thus, the empirical formula is also H^3PO^4.