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42 Cards in this Set
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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?

True.


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

True.


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

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 .


Whenever you see parentheses in a molecule, what must you do?
Do it to the following: (NH^4)^2SO^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 

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

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.


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

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


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

Then we know that it is the limiting reactant.


Why would a chemist add excess reactant?

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!) 

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?

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. 

What did the French chemist named Joseph Louis GayLussac determine?

The stoichiometric coefficients in a chemical equation do not relate just the number of moles of each substance in a chemical equation......
Joseph Louis GayLussac determined that these coefficients also relate the volumes of gases in a chemical equation. 

GayLussac's Law can be used in stoichiometry only if everything you are using in the problem is a gas.
True? Or False? 
True!!!


What GayLussac'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? 
True.


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

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


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?

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


Why are grams used instead of moles?

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.


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


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

convert to moles.


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

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. 

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)

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.


Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.
Al^6O^9 
In order for a chemical formula to be empirical, it must represent a simple, wholenumber 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. 

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.
PbO^2 
In order for a chemical formula to be empirical, it must represent a simple, wholenumber 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. 

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.
MgCO^2 
In order for a chemical formula to be empirical, it must represent a simple, wholenumber 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. 

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.
C^8H^18O^2 
In order for a chemical formula to be empirical, it must represent a simple, wholenumber 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. 

Is the following molecular formulas an empirical formulas? If the formula is not an empirical formula, write the empirical formula.
Si^3H^9O^2 
In order for a chemical formula to be empirical, it must represent a simple, wholenumber 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. 

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

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


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?

True.


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

You have to find the common factor and divide by it


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

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.


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?

The molar mass, therefore, is 16.0 grams.


The molar mass of KCl, for example, is 74.6 grams.
What is its molecular mass? 
That means its molecular mass is 74.6 amu.


Why do chemists use the term “molar mass”?

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. 

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

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.


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

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. 

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

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. 

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? 
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. 

State GayLussac's Law and explain how it is used.

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. 

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 GayLussac's Law to relate to one another? 
The chemist can only use GayLussac's Law to relate the volumes of HCl and CO^2, since those are
the only gases in the equation. 

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?

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.


What is the difference between molecular formulas and empirical formulas?

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

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

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

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


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

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