## Lecture Notes

The mole is used so commonly in chemistry to measure the amount of an element or compound that it is sometimes elevated to the level of a concept. In fact, it is just a definition, in the same way that the term ream is defined as 500 sheets of paper or a dozen is defined as 12. Because the mole is part of the everyday language of chemistry we must become entirely familiar with its use. We begin with its origin in atomic weights.

**Atomic Weights**

In the 18th and 19th centuries, chemists began to explore weight relationships between the elements. During that period, combining weights were determined. (The term mass is preferred, but we will frequently use weight and mass interchangeably.) Let's pretend that we are working in a 19th-centruy lab and have just carried out a reaction between zinc and sulfur. Specifically, we mixed 65 grams of elemental zinc with 40 grams of elemental sulfur and heated them in a crucible for several hours. We then dissolved the excess sulfur in methylene chloride. The material that remained after the excess sulfur has been removed showed no sign of elemental zinc and weighed 97 grams.

Thus, in this reaction the combining weights of zinc and sulfur are 65 g and 32 g. The significance of these combining weights became clear in the early 1800's when Dalton proposed that all matter consists of atoms and that reactions involve the combinations of these atoms in whole number ratios. Dalton also assumed that all atoms of the same element have the same mass (an assumption we now know is incorrect).

If we know that one atom of sulfur combines with one atom of zinc in forming the compound zinc sulfide, we can assume that if 32 g of sulfur react with 65 g of zinc, then an atom of zinc is 65/32 times heavier than an atom of sulfur. The following chemical equation represents this reaction:

Zn + S → ZnS

In this equation the symbols for the elements are used to represent the elemental forms, and the compound is written as ZnS to indicate that there is a one to one ratio of zinc and sulfur atoms. We can read the equation as "One atom of elemental zinc reacts or combines with one atom of elemental sulfur to form the compound zinc sulfide."

Problem Two

Suppose that it were true that one atom of zinc combines with two atoms of sulfur. Which of the following represents this reaction?

Problem Three

Assuming that the same weights of each element--65 g zinc and 32 g sulfur--were consumed in this reaction of one atom of zinc with two atoms of sulfur, what would be the combining weight of sulfur?

Atoms of different elements combine with sulfur (and other elements) in different ratios. For example, silicon combines with two sulfur atoms (the formula of silicon sulfide is SiS_{2}); two atoms of sodium combine with one atom of sulfur (the formula of sodium sulfide is Na_{2}S). Thus, it is important to determine the ratio of atoms in each compound. We will not pursue this here, but methods such as the use of the combining volumes of gases can provide this kind of information.

So, let us assume that eventually we put together a list of relative weights of atoms, such as the following:

Hydrogen | 5 |

Carbon | 60 |

Boron | 50 |

Silver | 535 |

Because hydrogen is the lightest element, let us assign it a weight of 1. Calculate the combining weights of the other elements relative to hydrogen. Your table should now look like this:

Hydrogen | 1 |

Carbon | 12 |

Boron | 10 |

Silver | 107 |

Problem Four

According to your new table, how much heavier is an atom of carbon than an atom of boron?

a) 12 times heavier b) 12/10 times heavier c) 10/12 times heavier

Problem Five

What mass of silver contains the same number of atoms as 12 g of carbon?

Problem Six

What mass of silver contains the same number of atoms as 6 g of carbon?

Near the beginning of the 20th century, mass spectrometers, which allowed chemists to determine the mass of individual atoms, were produced. Several important observations followed the use of this instrument. First, it is was found that not all atoms of the same element have the same mass. (This is why we have used the phrase "average atom" above.) Atoms of the same element that have different masses are called **isotopes**. For example, there are two important isotopes of boron; both have 5 protons in the nucleusRemember that the nucleus of the atom contains protons, which are positively charged, and neutrons, which have no charge. These two nuclear particles are sometimes called **nucleons** and they have almost identical masses. The number of protons in the nucleus is called the **atomic number** of the element. This number distinquishes one element from another. For example, two elements, such as boron and nitrogen, could have the same number of neutrons or they could have the same number of electrons, but boron will always have 5 protons in the nucleus and nitrogen will always have 7 protons in the nucleus., but one has 5 neutrons in the nucleus while the other has 6 neutrons. The sum of the number of protons and neutrons in the nucleus is called the mass number of the isotope. This number is designated as a superscript before the symbolChemists use a type of shorthand to write and talk about the different elements. The element is represented by its chemical symbol, usually the first or first two letters in the element's name. For example, the symbol for fluorine is F, that for carbon is C. Sometimes the first two letters of the element's Latin name are used; such as Au for gold (from aurus) and Pb for lead (from plumbus). of the element. For example, the isotope of boron with 6 protons is designated as ^{11}B.

The electron is the other particle in the atom that is of great interest to the chemist. The electrons are outside of the nucleus, they are negatively charged, and they have a much smaller mass than either the proton or the neutron. The charge on an electron is equal to, but opposite in sign, to the charge on a proton. Consequently, in a neutral atom the number of electrons is equal to the number of protons.

1 | 18 | ||||||||||||||||

1 H 1.0079 |
2 | 13 | 14 | 15 | 16 | 17 | 2 He 4.0026 |
||||||||||

3 Li 6.941 |
4 Be 9.0122 |
5 B 10.811 |
6 C 12.011 |
7 N 14.007 |
8 O 15.999 |
9 F 18.998 |
10 Ne 20.180 |
||||||||||

11 Na 22.990 |
12 Mg 24.305 |
3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 Al 26.982 |
14 Si 28.086 |
15 P 30.974 |
16 S 32.065 |
17 Cl 35.453 |
18 Ar 39.948 |

19 K 39.098 |
20 Ca 40.078 |
21 Sc 44.956 |
22 Ti 47.867 |
23 V 50.942 |
24 Cr 51.996 |
25 Mn 54.938 |
26 Fe 55.845 |
27 Co 58.933 |
28 Ni 58.693 |
29 Cu 63.546 |
30 Zn 65.409 |
31 Ga 69.723 |
32 Ge 72.64 |
33 As 74.922 |
34 Se 78.96 |
35 Br 79.904 |
36 Kr 83.798 |

37 Rb 85.468 |
38 Sr 87.62 |
39 Y 88.906 |
40 Zr 91.224 |
41 Nb 92.906 |
42 Mo 95.94 |
43 Tc (98) |
44 Ru 101.07 |
45 Rh 102.91 |
46 Pd 106.42 |
47 Ag 107.87 |
48 Cd 112.41 |
49 In 114.82 |
50 Sn 118.71 |
51 Sb 121.76 |
52 Te 127.60 |
53 I 126.90 |
54 Xe 131.29 |

55 Cs 132.91 |
56 Ba 137.33 |
57 - 71 | 72 Hf 178.49 |
73 Ta 180.95 |
74 W 183.84 |
75 Re 186.21 |
76 Os 190.23 |
77 Ir 192.22 |
78 Pt 195.08 |
79 Au 196.97 |
80 Hg 200.59 |
81 Tl 204.38 |
82 Pb 207.2 |
83 Bi 208.98 |
84 Po (209) |
85 At (210) |
86 Rn (222) |

87 Fr (223) |
88 Ra (226) |
89 - 103 | 104 Rf (261) |
105 Db (262) |
106 Sg (266) |
107 Bh (264) |
108 Hs (277) |
109 Mt (268) |
110 Ds (271) |
111 Rg (272) |
|||||||

57 La 138.91 |
58 Ce 140.12 |
59 Pr 140.91 |
60 Nd 144.24 |
61 Pm (145) |
62 Sm 150.36 |
63 Eu 151.96 |
64 Gd 157.25 |
65 Tb 158.93 |
66 Dy 162.50 |
67 Ho 164.93 |
68 Er 167.26 |
69 Tm 168.93 |
70 Yb 173.04 |
71 Lu 174.97 |
|||

89 Ac (227) |
90 Th 232.04 |
91 Pa 231.04 |
92 U 238.03 |
93 Np (237) |
94 Pu (244) |
95 Am (243) |
96 Cm (247) |
97 Bk (247) |
98 Cf (251) |
99 Es (252) |
100 Fm (257) |
101 Md (258) |
102 No (259) |
103 Lr (262) |

Problem Seven

Which of the following isotopes has 6 protons and 7 neutrons in the nucleus?

Problem Eight

Naturally occurring chlorine is made up of two isotopes, one with two more neutrons that the other. If the lighter isotopes has a mass number of 35, what is the correct symbol of the other isotope?

Problem Nine

An isotope of some element has 70 neutrons in its nuclues and a mass number of 120. What is the element?

After the discovery of isotopes and a means for accurate measurement of the relative mass of these isotopes, it was decided that the mass of the ^{12}C isotope should be defined as exactly 12.0000 atomic mass units (amu).

Problem Ten

How would you define one amu?

a) 12 g b) one-twelfth of the mass of one 12C atom c) 1 x 10^{-12} g

Protons and neutrons have very nearly the same mass. The mass of the electron is about 2000 times less than that of a proton or neutron.

Problem Eleven

What is the approximate mass of a proton (and a neutron) in amu units?

Problem Twelve

What is the approximate mass of one ^{13}C atom?

Problem Thirteen

What is the approximate mass of two atoms of ^{29}Si?

Problem Fourteen

Suppose that it was decided to design a new system of atomic weights in which a ^{12}C atom no longer weighed 12 amu, but rather the mass of a ^{12}C atom would be defined as 1.00 amu. In this system, what would be the mass of a ^{16}O atom?

Problem Fifteen

Naturally-occuring carbon is 99% ^{12}C and 1% ^{13}C. If you had 99 ^{12}C atoms and one ^{13}C atom, what would be the mass of the 99 ^{12}C atoms and the one ^{13}C atom?

a) 1188 amu and 13 amu b) 99 amu and 1 amu c) 1000 amu and 10 amu

Problem Sixteen

Now, determine the averageAn average, also called a mean, is calculated by dividing the sum of all of the data points in a set of values by the number of values. Suppose you had the following set of numbers that represent the length of time, in seconds, that you can hold your breath under water: 24, 29, 26, 28, and 28. The sum of these 5 numbers is 135, which we divide by 5 to get 27 s. This number is a better representation of the amount of time you are able to hold your breath than any of the individual numbers. mass of these one hundred atoms.

Sometimes, we may want to take a weighted average. Actually, all averages are weighted, but for an ordinary average every value in the set has the same weighting. For example, in the set above, each value counts 20% (one-fifth) toward the average. Indeed, we could determine the average by doing the calculation as follows:

0.20 x 24 + 0.20 x 29 + 0.20 x 26 + 0.20 x 28 + 0.20 x 28 =

4.8 + 5.8 + 5.2 + 5.6 + 5.6 = 27 s

Suppose that while you were measuring the times, you felt that the 24 second value was not a particularly good value. Perhaps this value was obtained while your brother did a cannonball in front of you. Likewise, one of the 28 second values was obtained when your chemistry teacher was walking by and probably was not a really representative value (you were trying to stay under water as long as possible). Because you don't want to exclude these values entirely, you decide to weight them differently. Perhaps you decide to weight the 24 second value only 5% (0.05) and one of the 28 second values you weight only 10% (0.10). These two points therefore contribute 15% (0.15) to the average. You want the weighting of the other values to be equal to one another so each of the other 3 must have a weighting of (100-15)/3 = 28.3%. Now your calculation will look like this:

0.05 x 24 + 0.283 x 29 + 0.283 x 26 + 0.15 x 28 + 0.283 x 28 =

1.2 + 8.2 + 7.4 + 4.2 + 7.9 = 28.9 s

Certainly, the different weightings have affected the mean.

A glance at a periodic chart will show you that the atomic mass of carbon is given as 12.01. Notice also, that there are no units on the atomic masses given on the chart. You now realize that every atomic mass shown on the periodic chart is an average for a large number of isotopes.

Problem Seventeen

Let's try the calculation of the atomic mass of argon for which there are three naturally occuring isotopes: ^{36}Ar, ^{38}Ar, and ^{40}Ar. The percentages of these isotopes are: 0.34%, 0.07%, and 99.58%, respectively. Calculate the atomic mass of argon. *Hint: a) You can choose the correct answer without having to use your calculator, and b) This is where you use the weighted average.*

Problem Eighteen

Chlorine exists as two isotopes: ^{35}Cl with a mass of 34.97 and ^{37}Cl with a mass of 36.95. The atomic weight of chlorine is 35.453. What is the percentage of each of the isotopes?

We now have two different interpretations of the atomic masses: 1) they represent the mass of the average atom of an element in amu's, given the convention that the mass of 12C is exactly 12 amu, and 2) they represent the relative combining masses of the elements.

Perhaps we should expand on the second interpretation. If you wanted to make the compound carbon dioxide from carbon and oxygen you first need to know that the formula of this compound is CO_{2}. The subscripts after the symbol for each element give the ratio of atoms present in the compound. In the case of CO_{2}, the subscript 2 for oxygen tells us that two oxygen atoms are present for every one atom of carbon (the absence of a subscript for carbon is equivalent to a subscript of 1). Next, you must find the atomic masses of carbon and oxygen--12 and 16, respectively. If the formula of the compound were CO, then you would merely need to combine 12 g of carbon with 16 g of oxygen in order to get the same number of atoms of each element. Or, you could combine 1.2 g of carbon with 1.6 g of oxygen, or you could combine 1.0 pound of carbon with 16/12 = 1.33 pound of oxygen, or 3.5 tons of carbon with 3.5 x 16/12 = 4.7 tons of oxygen. But, the formula of carbon dioxide is CO_{2}, which tells us that we need twice as many oxygen atoms as carbon atoms. Therefore, if we want to use 12 g of carbon we must use 2 x 16 = 32 g of oxygen. If we want to start with 4 g of carbon then we must use 4 x (16/12) x 2 = 10.7 g of oxygen. Notice that the ratio of the atomic masses and the number of atoms of each element always play a role in the calculation.

Problem Nineteen

If you happen to have one pound of carbon, how many pounds of oxygen will be required to make CO_{2}?

Problem Twenty

If the compound SO_{2} contains 32 g of sulfur, how many grams of oxygen does it contain?

Problem Twenty One

What is the ratio of the number of atoms in 16 ounces of sulfur to the number in 1 ounce of hydrogen?

Problem Twenty Two

If the compound NH3 contains 6 g hydrogen how many grams of nitrogen are also present?

Problem Twenty Three

How many grams of hydrogen contain the same number of atoms as 12.01 grams of carbon?

It should now be clear that the atomic weights in the periodic chart are numbers that can be used to determine the ratio of the number of atoms in various amounts of the elements. The weights (masses) represent the average mass of an atom of the element in atomic mass units. So far, we have no idea of the value of an atomic mass unit in a unit of mass, such as grams, with which we are familiar.

**The Mole**

Chemists usually weigh elements and compounds in grams, and it is, therefore, of particular interest to know the actual number of atoms in an amount of an element equal to its atomic mass in grams. In other words, how many atoms are present in 12.01 grams of carbon, in 1.008 grams of hydrogen, or in 32.04 grams of sulfur? We know, of course, that the number of atoms in each of these quantities is the same. But what is the number? This number has been determined experimentally in a number of different ways and turns out to be an extremely large number-- **6.022 x 10 ^{23}**. It is known as

**Avogadro's number**, in honor of the Italian chemist.

Problem Twenty Four

Which of the following statements expresses the significance of the number 6.02 x 10^{23}?

a) There are 6.022 x 10^{23} amu in a gram b) There are 6.022 x 10^{23} atoms in a gram c) There are 6.022 x 10^{23} atoms in a mass of an element equal to its atomic mass in grams.

Problem Twenty Five

How many atoms are present in 24 grams of carbon?

Avogadros's number, 6.022 x 10^{23}, is also called a **mole**, in the same way that 500 sheets of paper is called a ream, or 144 is referred to as a gross.

Problem Twenty Six

How many ping-pong balls are there in a mole of ping-pong balls and what would their molar mass be if one ping-pong ball weighs 1.0 g?

Are you saying that we can have a mole of anything? Sure, can't you have a dozen pencils, a dozen ducks, a dozen bricks? However, we normally use the term mole in dealing with atomic particles--atoms, ions, molecules.

Problem Twenty Seven

How many atoms are present in 31 g of phosphorus?

Let's try some questions that compare terms like ream or gross or dozen with mole.

Problem Twenty Eight

If one ream of paper weighs 1.0 pounds, how many reams are there in 300 pounds of paper?

Problem Twenty Nine

How many moles of carbon are there in 500 g of carbon?

Problem Thirty

As the head chick in the hen-house, you have responsibility for 100 eggs, which weigh a total of 1.00 x 10^{4} g. How many dozen eggs do you have under your wings and how much does a dozen weigh?

Problem Thirty One

Your Zork particle counter has just informed you that there are 3 x 10^{20} atoms of radon in your basement. How many moles of radon is this?

Problem Thirty Two

In order to do well in your favorite course (well, chemistry, what else?) you have ordered 20 gross of pencils. How many pencils will you get?

Problem Thirty Three

Your lab partner is staring out the window as usual when he suddenly exclaims, "Gadzooks! There are 1 x 10^{-23} moles of dogs running down the street!" Could he be right for a change?

Problem Thirty Four

How many moles of atoms are there in 24 grams of carbon?

Problem Thirty Five

How many grams of calcium must be weighed out in order to obtain a sample containing 3 moles of atoms?

Problem Thirty Six

What is the weight of an average atom of calcium?

Now that we know how to determine the mass of an atom in grams, we can figure out the relationship between amu's and grams.

Problem Thirty Seven

First, calculate the mass of a ^{12}C atom in grams.

Problem Thirty Eight

An amu is defined as 1/12 of the mass of a ^{12}C atom. Determine the mass of one amu in grams.

**The Mole and Compounds**

We have already found that the formula of a compound represents the ratio of the atoms of the various elements in the compound. For example, magnesium chloride has the formula MgCl_{2}, which reveals that the compound contains two chlorine atoms for every magnesium atom.

Problem Thirty Nine

In the compound Fe_{2}O_{3}, how many atoms are present for every two iron atoms?

The subscripts in formulas can also be interpreted in terms of moles. If there are two atoms of chlorine for every atom of magnesium in MgCl_{2}, then there must also be two moles of chlorine atoms for every one mole of magnesium. Or, if we prefer, we can say that for every 12 x 10^{23} atoms of chlorine there are 6 x 10^{23} atoms of magnesium.

Problem Forty

The formula for barium hydroxide is Ba(OH)_{2}. The parentheses indicate that for every barium ion there are two hydroxide ions (OH^{-} is the hydroxide ion). How many moles of oxygen are present with one mole of barium?

We can now calculate the molar mass (sometimes also called the **formula weight**) of a compound. This is the mass of one mole of the compound and can be determined by simply adding the atomic masses of each of the elements (taking into account the number of each) in the compound. For barium hydroxide, we need to add the atomic weight of barium, two times the atomic weight of oxgyen, and two times the atomic weight of hydrogen. So, this would be 137 + (2 x 16) + (2 x 1) = 171 g. Thus, one mole of Ba(OH)_{2} has a mass of 171 g.

Problem Forty One

How much barium hydroxide must be weighed out in order to obtain 0.10 mole?

Problem Forty Two

A student weighs out a sample of 20 g of barium hydroxide. How many moles of barium hydroxide is this?

Problem Forty Three

Calculate the molar mass of Cu_{2}O.

Problem Forty Four

A 5 x 10^{-3} mole sample of Cu_{2}O is required. What weight is required?

In addition to the formulas that we have used above, which are called empirical formulas, there are several formulas that are used for molecular compounds. Before we define the molecular formula let's be sure we are clear about the meaning of the empirical formula. **The empirical formula gives the simplest whole number ratio of the number of atoms (moles) of the each of the elements in the compound**. The empirical formula is the only formula used for ionic compounds - compounds that contain charged atoms (ions). There are also compounds that do not contain ions. The atoms in these compounds are held together by covalent bonds. Many of these compounds contain **molecules**, the building blocks of a molecular compound. Benzene, an interesting molecular compound, contains molecules that have six carbon atoms attached to one another in a hexagonal ring. Each carbon has one hydrogen attached to it. The structural formula (the name tells you that the formula gives the structure of the molecule) is shown below:

[Note: this is not an electron-dot formula (yes, yet another kind of formula), which you will learn about in your discussion of bonding. This **structural formula** simply shows which atoms are connected.] As you can see, each molecule of benzene contains six carbons and six hydrogens. The molecular formula of benzene therefore is written C_{6}H_{6}.

The molecular weight or molar mass of benzene is entirely analogous to the formula weight, which is derived from the empirical formula.

Problem Forty Five

Calculate the molecular weight of benzene.

Problem Forty Six

A 78 g sample of benzene contains how many moles of benzene molecules?

It is important to understand that the empirical formula of benzene is CH, which indicates the one to one mole ratio of carbon to hydrogen. The **molecular formula**, C_{6}H_{6}, **gives the number of each atom in one molecule of the compound**.

We sometimes need to convert a formula, be it empirical or molecular, to the weight percentageThe use of percentages is very common in science. The word comes from the Latin, per centum, meaning by the hundreds, and means per hundred parts. For example, if we have a class that is 40% women, we know that out of every hundred students, 40 of them are women. If a sample is 40% carbon then for every hundred parts (grams, pounds, tons, whatever), 40 of them are carbon. In order to determine the amount of an element from its percentage, you must first convert the percentage to its decimal equivalent (40% = 0.40) and then multiply that number by the weight. The amount of carbon in 50 grams of benzene (92% carbon) is 0.92 x 50 g = 46 g. of the elements. For example, benzene is 92% carbon and 8% hydrogen. In order to determine the percentage from the formula we first obtain the molar mass from the empirical formula. For benzene, the molar mass is 13 g. The percent of carbon in that 13 g is:

12 g/13 g x 100 = 92 %

Problem Forty Seven

Determine the percentage of oxygen in acetic acid, which has the empirical formula CH_{2}O.

**Calculation of Empirical Formula from Percent Composition**

The conversion of experimental composition data to empirical formula is an important step in determining the identity of a substance. Suppose that we have a sample of a heavy pale yellow crystalline material that has the experimentally determined composition: 59.0% Ba, 13.5% S, and 27.4% O. We need to convert this percent composition data to an empirical formula, which will tell us the relative number of moles of barium, sulfur and oxygen in the substance. The key to this conversion lies in the understanding of the phrase "relative number of moles of barium, sulfur and oxygen in the substance." If we want the number of moles of each we must know the mass of each element in some amount of the substance and then we must convert each of thoses masses to moles.

In fact, we can choose any amount of the substance and convert it to the mass of each of the elements. If we choose 100 grams of the substance it makes the problem a little easier because we can do the first part of the math in our heads.

Problem Forty Eight

For example, if we have a 100 g sample of the substance and it contains 59.0% barium, what mass of barium is present?

Likewise, in this same sample we have 13.5 g sulfur and 27.4 g oxygen. In order to convert these masses to moles we must divide each by their atomic mass.

Ba 59.0 g/ 137.33 g/mol = 0.430 mol

S 13.5 g/ 32.07 g/mol = 0.421 mol

O 27.4 g/16.00 g/mol = 1.71 mol

We could at this point write the empirical formula as Ba0_{.430}S0_{.421}O_{1.71}, but we know that the empirical formula should be expressed in whole numbers. Consequently, we divide each of the subscripts by the smallest one (0.421), which gives us subscripts of 1.02, 1.00, and 4.06. These still are not whole numbers, but because these numbers have been derived from experimental data, which contain errors, we can not expect the numbers to come out to exact whole numbers. The numbers appear close enough to 1, 1, and 4 to allow us to safely round them off. Hence, the empirical formula of the substance is BaSO_{4}.

Problem Forty Nine

Determine the empirical formula of a compound that contains 10.0% C, 0.84% H, and 89.1% Cl.

In some cases the numbers will not come close to whole numbers and we must then multiply each by some number that will produce a set of whole numbers. The following is such a case.

Problem Fifty

Determine the empirical formula of a compound with percent composition data of Fe, 69.9%, O 30.1%.

Problem Fifty One

Determine the empirical formula of a compound with the percent composition of 50.04% C, 5.59% H, and 44.37% O.

The compound in the drill question has an empirical formula of C_{3}H_{4}O_{2}. If this were also the molecular formula the compound would have a molecular weight of 3 x 12 + 4 x 1 + 2 x 16 = 72 g/mole. If, in fact, the compound has a molecular weight of 144 g/mole, then the molecular formula must be some multiple of the empirical formula; that is, the subscripts must be multiplied by some whole number. The number, in this case, is 2, and the molecular formula is C_{6}H_{8}O_{4}.

Problem Fifty Two

How many moles of the molecular compound C_{6}H_{8}O_{4} are there in a 10 g sample of the compound?

Problem Fifty Three

How many molecules are there in this sample?

Problem Fifty Four

How many moles of carbon are present in this sample?

Problem Fifty Five

The empiricial formula of chloroform is CHCl_{3} and its molar mass is 119. What is the molecular formula of chloroform?

Problem Fifty Six

What is the weight of one molecule of chloroform?

Problem Fifty Seven

How many moles of chloroform are present in 1 mg of the sample [mg = milligram = 10^{-3} g]?

Problem Fifty Eight

How many moles of chlorine are present in this 1 mg sample of CHCl_{3}?

Problem Fifty Nine

What weight of chlorine is present in this 1 mg sample?

Problem Sixty

What mass of chloroform is required to provide 1 x 10^{25} atoms of hydrogen?

**Moles in Solution**

**Solutions are homogeneous mixtures of two or more substances**. We will restrict our discussion to solutions that contain only two substances: a **solute**, the substance present in the smallest amount, and the **solvent**, the substance present in the greatest quantity. One of the important characteristics of a solution is its concentration: the relative amount of solute in the solution. There are a number of ways to express concentration--weight percent, molarity, molality, and mole fraction--but we will concentrate (no pun intended) on molarity.

**Molarity** is defined as the **number of moles of solute per liter of solution**. If we dissolve 1.0 mole of NaCl in enough water to make a total of one liter of solution, we will have a 1.0 molar solution of NaCl in water. Molarity is usually designated with a capital M.

Problem Sixty One

Which of the following describes how you would make up a 1.0 M NaCl solution in water?

- a) Add 58 g of NaCl to a liter of water and then mix carefully
- b) Add 58 g of NaCl to a beaker, add water and mix, and then continue to add water (and mix) until a total volume of 1.0 liter as been reached.
- c) Add 5.8 g of NaCl to a beaker, add enough water to dissolve the NaCl. Then transfer the solution to a graduated cylinder and add enough water, while mixing, to reach a volume of 100 mL.

Problem Sixty Two

If 50 mL of a solution contains 5.8 g NaCl, what is the molarity of the solution?

Problem Sixty Three

George prepares one liter of a 0.05 M NaCl solution. Jack, a good friend of Mary, came to lab late and needs 50 mL of a 0.05 M NaCl solution for a titration. Mary suggests that Jack ask George for 50 mL of his solution. Jack hesitates to use George's solution because he says that if he takes 50 mL he will not have a 0.05 M solution. He will have 0.05 mole of NaCl in 0.050 L, says Jack, which will make the solution a 1 M solution. What should Mary say?

The previous question emphasizes the fact that no matter how much of a solution you have, the concentration remains the same. Let's go through Mary's reasoning (answer b) again with a different amount of the 0.05 M solution. If we start with 1 liter of the solution and take from it 100 mL, we have taken a tenth of the solution. That tenth will contain a tenth of the original amount of solute (0.05 moles); that is, the 100 mL will contain 0.005 moles of solute. But, that 0.005 moles will be present in a total of 100 ml (0.1 L). Thus, the molarity, which is just moles per liter, will be:

0.005 moles/0.1 L = 0.05 M

Problem Sixty Four

How many moles of solute will be present in 310 mL of the 0.05 M NaCl solution?

Problem Sixty Five

What volume of a 0.05 M NaCl solution will be required to obtain 2 x 10^{-3} moles of NaCl?

Problem Sixty Six

How many moles of solute will be present in 310 mL of the 0.05 M NaCl solution?

Problem Sixty Seven

How many moles of acetic acid are there in 100 mL of a 0.60 M solution?

Problem Sixty Eight

If 10 mL of a 0.60 M acetic acid solution is diluted to 55 mL, what is the molarity of the resulting solution?

Problem Sixty Nine

How many mL of a 12 M stock solution of HCl must be used to prepare 100 mL of a 0.10 M HCl solution?

**The Mole and Chemical Reactions**

Although the quantitative aspects of chemical reactions will be studied in the module on *Stoichiometry*, we will introduce the idea that substances react in certain mole ratios. For every reaction there is an equation that expresses the mole ratio between the reactants and products. We have looked at a few simple equations early in this module; let's look at a few more from our new vantage point of the mole.

Aluminum reacts with sulfur to form aluminum sulfide as shown below:

2 Al + 3 S → Al_{2}S_{3}

The substances on the left side of the arrow are the reactants, those on the right side are the products. Every equation must obey the **Law of Conservation of Mass**, which means that all of the aluminum atoms on the left must be converted to aluminum atoms in some form on the right if the reaction goes to completion. The same is true, of course, of the sulfur. In order to balance an equation, we must, therefore, simply make certain that the number of atoms of each element is the same on both sides. In this sense, a chemical equation is like a mathematical equation; there must be an equality between the number of atoms.

Problem Seventy

Methane (CH_{4}) reacts with elemental oxygen, which exists as O_{2} molecules, to form CO_{2} and water. Which of the following is a balanced equation for this reaction?

a) CH_{4} + 4 O → CO_{2} + 2 H_{2}O b) CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O c) CH_{4} + 3 O_{2} → CO_{2} + 4 H_{2}O

Problem Seventy One

Aluminum metal reacts with HCl to form elemental hydrogen (H_{2}) and AlCl_{3}. Write a balanced equation for this reaction.

Problem Seventy Two

Balance the following "equation."

Na_{2}CO_{3} + HCl → H_{2}CO_{3} + NaCl

What do the coefficients in from of the substances in the balanced equation mean? For example, what do the 2, the 3 and the 1 (there is an implied 1 in front of Al_{2}S_{3}) mean in the following equation?

2 Al + 3 S → Al_{2}S_{3}

We have already encountered one interpretation: Two atoms of aluminum combine with 3 atoms of aluminum to make aluminum sulfide. Notice that in this sentence we did not specify how much aluminum sufide was produced because we can not say one molecule of aluminum sulfide - the compound does not exist in molecular form. It is very convenient to use the mole to interpret equations. Thus, for this equation we can say: Two moles of aluminum combine with three moles of sulfur to make one mole of aluminum sulfide.

Problem Seventy Three

Verbalize the following equation:

H_{2} + Cl_{2} → 2 HCl

The language in our statements above should make it clear that the coefficients in front of each substance in a chemical equation represent the relative number of moles of each involved in the reaction. For example, in the equation:

H_{2} + Cl_{2} → 2 HCl

The coefficients, which are 1, 1, and 2, tell us that if we start the reaction with one mole of H_{2} and one mole of Cl_{2}, we will get two moles of HCl when the reaction is complete. Or, for the reaction:

Fe_{2}(CO_{3})_{3} → Fe_{2}O_{3} + 3 CO_{2}

we know that if we start the reaction with one mole of Fe_{2}(CO_{3})_{3} we will get one mole of Fe_{2}O_{3} and three moles of CO_{2} when the reaction is complete.

But, what if we start with some other amount, say 0.15 mole of Fe_{2}(CO_{3})_{3} ? Certainly, the same mole ratio applies no matter how much we start with. That is, we will get three times as many moles of CO_{2} as the number of moles of Fe_{2}(CO_{3})_{3} that we start with. Thus, if we start with 0.15 mole of Fe_{2}(CO_{3})_{3}, we will get:

0.15 mol Fe_{2}(CO_{3})_{3} x 3 mole CO_{2} /1 mole Fe_{2}(CO_{3})_{3} = 0.45 mole CO_{2}

Problem Seventy Four

How many moles of Fe_{2}(CO_{3})_{3} must we start the reaction with if we want 1.8 mol of CO_{2} when the reaction has gone to completion?

Elemental nitrogen and hydrogen, both of which exist as diatomic molecules (N_{2} and H_{2}), react to form ammonia, according to the equation:

N_{2} + 3 H_{2} → 2 NH_{3}

Problem Seventy Five

If 1.0 mole of H_{2} is consumed during the reaction, how many moles of ammonia are formed?

In the problem above, and in many other such problems, it is convenient to divide all of the coefficients by the coefficient of the substance in question. In the above case, we could divide all coefficients by the 3 in front of H_{2} to get the equation:

1/3 N_{2} + H_{2} → 2/3 NH_{3}

This equation makes it clear that every mole of H_{2} will be converted to two-thirds mole of NH_{3}.

Try the same approach with the reaction of ethane with O_{2} as shown below:

2 C_{2}H_{6} + 7 O_{2} → 4 CO_{2} + 6 H_{2}O

Problem Seventy Six

How many moles of CO_{2} are formed as a result of the reaction of 2 moles of O_{2}?

Problem Seventy Seven

If 1.0 mole ethane is mixed with 5.0 mole O_{2} and the 1.0 mole ethane is completely consumed during the reaction, how many moles of O_{2} will be left at the end of the reaction?