# Introductory Chemistry Online/Mole and Measurement

Import and Editing in Progress

## Contents

## **Chapter 4. The Mole and Measurement in Chemistry**[edit]

## 4.1 Measurement and Scale in Chemistry: The Mole Concept[edit]

One of the important concepts to grasp in chemistry is *scale.* Atoms and molecules are very small. A single atom of hydrogen has a mass of about 1.67 x 10^{-24} grams (that’s 0.00000000000000000000000167 grams). One cubic centimeter of water (one mL) contains about 3.3 x 10^{22} water molecules (that’s 33 sextillion molecules). Because chemists routinely use numbers that are both incredibly small and incredibly large, unique units of measurement have been developed to simplify working with these numbers. As we learned in Chapter 1, the atomic mass unit (amu) helps us talk about the mass of atoms on a scale appropriate to atoms (one grams is about 600 sextillion amu). In this chapter, we will introduce the concept of a mole to help us talk about numbers of atoms on a scale appropriate to the size of a sample we could work with in a laboratory (for example, in grams). The **mole** is defined as the number of atoms contained in *exactly* 12 grams of carbon-12 (the isotope ). Chemist have measured this number and it turns out to be 6.0221415 x 10^{23}. We can think of the term *mole* as a number, just like the word *dozen* represents the number 12. We will use the mole to represent this very large number (a *chemist’s dozen*) and we will see that there is a special relationship between a mole of a pure substance and the mass of the substance measured in amu. Figure 4.1

The origin of the mole concept is generally attributed to the Italian chemical physicist, Amadeo Avogadro Figure 4.2. In 1811, Avogadro published an important article that drew a distinction between atoms and molecules (although these terms were not in use at the time). As part of his explanation of the behavior of gasses, he suggested *that equal volumes of all gases at the same temperature and pressure contained the same number of molecules.* This statement is referred to as **Avogadro’s Hypothesis** and today we commonly refer to the number of *things* in a mole, (6.0221415 x 10^{23}) as **Avogadro’s number** (this is rounded to 6.02 x 10^{23} for most calculations). Because a mole can be thought of as a *number*, you can convert any number to moles by dividing that number by 6.02 x 10^{23}. For example, at the time of this writing, the national debt of the United States is about 7.9 trillion dollars (7.9 x 10^{12} dollars). This could be expressed as *moles of dollars* as shown below:

The ratio of any number to the number of things in a mole is often referred to as a **mole fraction**.

**Exercise 4.1 Numbers of Things in Multiple Moles and in Mole Fractions.**

a. It is estimated there are 7 x 10^{22}stars in the universe. How many moles of stars is this?

b. It is estimated there are 7.5 x 10^{18}grains of sand on the earth. How many moles of sand grains is this?

c. You have 0.0555 moles of jelly donuts. What number of donuts would that be?

d. You drink a small bottle of drinking water that contains 13 moles of water.

What is the number of molecules of water you drank?

## 4.2 Molar Mass[edit]

As we described in Section 4.1, in chemistry, the term **mole** can be used to describe a particular number. The number of things in a mole is large, *very* large (6.0221415 x 10^{23}). We are all familiar with common copy-machine paper that comes in 500 sheet reams. If you stacked up 6.02 x 10^{23} sheets of this paper, the pile would reach from the earth to the moon **80 billion times**! The mole is a *huge* number, and by appreciating this, you can also gain an understanding of how *small* molecules and atoms really are. Figure 4.3

Chemists work simultaneously on the level of individual atoms, and on the level of samples large enough to work with in the laboratory. In order to go back and forth between these two scales, they often need to know how many atoms or molecules there are in the sample they’re working with. The concept that allows us to bridge these two scales is **molar mass**. Molar mass is defined **as the mass in grams of one mole of a substance**. The units of molar mass are grams *per* mole, abbreviated as .

The mass of a single isotope of any given element (the **isotopic atomic mass**) is a value relating the mass of that isotope to the mass of the isotope carbon-12 (); a carbon atom with six proton and six neutrons in its’ nucleus, surrounded by six electrons. The **atomic mass** of an element is the relative average of *all* of the naturally occurring isotopes of that element and atomic mass is the number that appears in the periodic table. We have defined a mole based on the isotopic atomic mass of carbon-12. By definition, the *molar mass* of carbon-12 is numerically the same, and is therefore *exactly* 12 grams. Generalizing this definition, **the molar mass of any substance in grams per mole is numerically equal to the mass of that substance expressed in atomic mass units**. For example, the

*atomic mass*of an oxygen atom is 16.00 amu; that means the

*molar mass*of an oxygen atom is 16.00 g/mol. Further, if you have 16.00 grams of oxygen atoms, you know from the definition of a mole that your sample contains 6.022 x 10

^{23}oxygen atoms.

The concept of molar mass can also be applied to compounds. For a molecule (for example, nitrogen, N_{2}) the mass of molecule is the sum of the atomic masses of the two nitrogen atoms. For nitrogen, the mass of the N_{2} molecule is simply (14.01 + 14.01) = 28.02 amu. This is referred to as the **molecular mass** and the molecular mass of any molecule is simply the sum of the atomic masses of all of the elements in that molecule. The molar mass of the N_{2} molecule is therefore 28.02 g/mol. For compounds that are *not molecular* (ionic compounds), it is improper to use the term “molecular mass” and “**formula mass**” is generally substituted. This is because there are no individual molecules in ionic compounds. However when talking about a mole of an ionic compound we will still use the term molar mass. Thus, the *formula* mass of calcium hydrogen carbonate is 117.10 amu and the *molar* mass of calcium hydrogen carbonate is 117.10 grams per mole (g/mol).

**Example 4.1 Calculation of Molecular and Masses**

Calculate the molecular mass for the sugar glucose (C_{6}H_{12}O_{6}).

**Exercise 4.2: Calculation of Molar Masses.**

**Find the molar mass of each of the following compounds:**

Sand, silicon dioxide (SiO_{2})
Draino™, sodium hydroxide (NaOH)
Nutrasweet™, Aspartame (C_{14}H_{18}N_{2}O_{5})
Bone phosphate, calcium phosphate Ca_{3}(PO_{4})_{2}

## 4.3 Mole-Mass Conversions[edit]

As described in the previous section, molar mass is expressed as “grams *per* mole”. The word *per* in this context implies a mathematical relationship between grams and mole. Think of this as a ratio. The fact that a *per* relationship, ratio, exists between grams and moles implies that you can use dimensional analysis to interconvert between the two. For example, if we wanted to know the mass of 0.50 mole of molecular hydrogen (H_{2}) we could set up the following equations:

Equation The * known* molar mass of H

_{2}is .

We are * given* that we have 0.50 moles of H

_{2}and we want to

*the number of grams of H*

**find**_{2}that this represents. To perform the dimensional analysis, we arrange the

*known*and the

*given*so that the units cancel, leaving only the units of the item we want to

*find*.

**Example 4.2 Mole-to-Mass and Mass-to-Mole Conversions**

a. Determine the mass of 0.752 mol of H_{2}gas.

b. How manymolesof molecular hydrogen are present in 6.022 grams of H_{2}?

c. If you have 22.414 grams of Cl_{2}, how manymolesof molecular chlorine do you have?

We can also use what is often called a *per* relationship (really just a **ratio**) to convert between number of *moles* and the number to *things* (as in 6.02 x 10^{23} *things per mole*). For example, if we wanted to know how many *molecules* of H_{2} are there in 3.42 moles of H_{2} gas we could set up the following equations:

The * known* ratio of molecules

*per*mole is .

Equation We are * given* that we have 3.42 moles of H

_{2}and we want to

*the number of molecules of H*

**find**_{2}that this represents. To perform the dimensional analysis, we arrange the

*known*and the

*given*so that the units cancel, leaving only the units of the item we want to

*find*.

And finally, we can combine these two operations and use the *per* relationships to convert between *mass* and the *number* of atoms or molecules. For example, if we wanted to know how many *molecules* of H_{2} are there in 6.022 *grams* of H_{2} gas we could set up the following series of equations:

The * known* molar mass of H

_{2}is .

The * known* ratio of molecules

*per*mole is .

Equation We are * given* that we have 6.022 grams of H

_{2}and we want to

*the number of molecules of H*

**find**_{2}that this represents. As always, to perform the dimensional analysis, we arrange the

*known*ratios and the

*given*so that the units cancel, leaving only the units of the item we want to

*find*.

**Example 4.5 Number-to-Mass Conversion**

A sample of molecular chlorine is found to contain 1.0 x 10^{20}molecules of Cl_{2}.

What is themass(in grams) of this sample?

**Exercise 4.3: Mole, Mass and Number Conversions.**

a. How many moles of sand, silicon dioxide (SiO_{2}), and how many molecules of sand are found in 1.00 pound (454g)

of sand?

b. You add 2.64 x 10^{23}molecules of sodium hydroxide (Drano™; NaOH), to your drain. How many moles are this

and how many grams?Figure 4.4

## 4.4 Percentage Composition[edit]

We are all familiar with the term *percentage*. We take an exam that is worth 100 points, we get 96 correct and we describe our score as 96%. We arrive at 96% by first taking our score and dividing it by the total number of points to get the *fraction* that we got correct. To convert the fraction to a *percentage*, we multiply the fraction by 100.

Applying this concept to molecules, we could describe HCl as consisting of one atom of hydrogen and one atom of chlorine. Likewise, we could use *mole* nomenclature and say that one *mole* of the molecule consists of one *mole* of hydrogen and one *mole* of chlorine, and that one *mole* of HCl has a mass of 36.46 grams. These descriptions, however, tell us nothing about how much of this mass is attributable to the hydrogen and how much comes from the chlorine. In order to do this, we need to speak of the **percentage composition** of a molecule, that is, what percent of the total mass arises from each element. These calculations are simple and involve taking the atomic mass of the element in question and dividing by the molar mass of the molecule.

For HCl, the *fraction* of hydrogen in HCl is given by the molar mass of hydrogen divided by the molar mass of HCl:

and the *percentage* of hydrogen in HCl is obtained by multiplying the *fraction* by 100:

Combining the steps, the *percentage* of chlorine in HCl can be calculated by dividing the molar mass of chlorine by the molar mass of HCl and multiplying by 100:

**Example 4.6 The Percentage of an Element in a Compound**

Find the percentage of fluorine in calcium fluoride (CaF_{2}).

These types of problems can also be presented as mass calculations. For example, determine the mass of calcium in 423.6 grams of CaF_{2}. Collecting our known, given and find values:

The known molar mass of Ca is .

The known molar mass of CaF_{2} is .

Equation We are given that we have a sample of CaF_{2} with a mass of 423.6 grams and we want to find the mass of Ca in this sample. We could find the mass of Ca if we knew the fraction of Ca in CaF_{2}. We could then multiply this fraction by the known mass of CaF_{2} to obtain the mass of calcium in the sample. The fraction of Ca in CaF_{2} is the ratio of the two known molar masses (the percentage, before you multiply by 100). As always, to perform the dimensional analysis, we arrange the known and the given so that the units cancel, leaving only the units of the item we want to find.

Note that in this example, you want to use the fraction of Ca in CaF2, not the percentage; if you used percentage, you would have to divide your answer by 100 to get the proper number of grams of Ca in the sample.

**Example 4.7 Percentage-to-Mass Conversions**

A sample of CaF_{2}is known to contain 18.00 grams of calcium; what mass of fluorine is contained in this sample?

**Exercise 4.4: Percentage Composition Calculations.**

a. Carbon dioxide is a green house gas produced in combustion. What is the percentage of oxygen in CO_{2}?

b. Barium sulfate is use when x-raying the gastrointestinal track. Determine the mass of barium in 523 grams

of BaSO_{4}.

c. Sulfuric acid, H_{2}SO_{4}is the most used chemical in industrial processes. If a sample of sulfuric acid

contained 3.67 g of hydrogen how many grams of sulfur would it contain?

## 4.5 Empirical and Molecular Formulas[edit]

In Chapter 2, we introduced the concept of a chemical compound as a substance that results from the combination of two or more atoms, in such a way that the atoms are bonded together in a constant ratio. We represented that ratio using the symbols for the atoms in the molecule, with subscripts to indicate the fixed ratios of the various atoms. The result is a molecular formula, and in Chapter 3, we used molecular formulas to devise chemical names for both molecular and ionic compounds. As we have seen in this chapter, molecular formulas can be used to directly calculate the molar mass of a compound, as shown in Figure 4.5.

Many of the methods, however, that chemists use in the laboratory to determine the composition of compounds do not give the molecular formula of the compound directly, but instead simply yield the *lowest whole-number ratio* of the elements in the compound. A formula such as this is called an **empirical formula**. For example, the molecular formula for glucose is C_{6}H_{12}O_{6}, but the simplest whole-number ratio of the elements in glucose is CH_{2}O; if you multiply each element in (CH_{2}O) by *six*, you obtain the molecular formula for glucose. An empirical formula cannot be converted into a molecular formula unless you know the molar mass of the compound. For example, the empirical formula for acetic acid (the acidic component in vinegar) is identical to that for glucose (CH_{2}O). If you analyzed these two compounds and determined only an empirical formula, you could not identify which compound you had.

Conversion of an empirical formula into a molecular formula requires that you know the molar mass of the compound in question. Knowing this, you can calculate a molecular formula based on the fact that an empirical formula can always be multiplied by an integer *n* to yield a molecular formula. Thus, some value of *n*, multiplying each element in CH_{2}O will yield the molecular formula of acetic acid. The value of *n* can be determined as follows:

*n* x (CH_{2}O ), where

For acetic acid, the molar mass is 60.05 g/mol, and the molar mass of the empirical formula CH_{2}O is 30.02 g/mol. The value of the integer *n* for acetic acid is therefore,

; and the molecular formula is 2 x (CH</nowiki>_{2}O), or C_{2}H_{4}O_{2}.

Note that *n* must be an integer and that your calculation should always yield a whole number (or very close to one).

**Example 4.7 Empirical and Molecular Formula Calculations**

A compound is determined to have a molar mass of 58.12 g/mol and an empirical formula of C_{2}H_{5};

determine the molecular formula for this compound.

**Exercise 4.5: Empirical and Molecular Formula Calculations.**

Benzene is an intermediate in the production of many important chemicals used in the manufacture of plastics,

drugs, dyes, detergents and insecticides. Benzene has an empirical formula of CH. It has a molar mass of

78.11 g/mol.What is the molecular formula?

## Study Points[edit]

- The
**mole**is defined as the number of atoms contained in*exactly*12 grams of carbon-12 (the isotope ). There are 6.0221415 x 10^{23}particles in a mole. Remember, a*mole*is just a number (like*dozen*) and you can have a mole of*anything.* - The concept of a mole is based on
**Avogadro’s Hypothesis**(*equal volumes of all gases at the same temperature and pressure contained the same number of molecules*) and the number of particles in a mole (6.0221415 x 10^{23}) is commonly referred to as**Avogadro’s number**(typically rounded to 6.02 x 10^{23}for most calculations). - Because atomic masses, and the number of particles in a mole, are both based on the isotopic atomic mass of the isotope carbon-12,
**the****mass of any substance expressed in atomic mass units is numerically equal to the**. Thus,*molar mass*of the substance in grams*per*mole*exactly*12 grams of carbon-12 contains*exactly*a mole of carbon atoms; likewise, 31.9988 grams of O_{2}contains 6.02214 x 10^{23}oxygen molecules (note,*six*significant figures), etc.

- To convert the number of
**moles**of a substance into the**mass**of a substance, you simply need to multiply (**moles**x**molar mass**).

- To convert the
**mass**of a substance into the number of**moles**, you simply need to*divide*the**mass**by the**molar mass**. - To convert the number of
**moles**of a substance into the number of**particles**of that substance, you simply need to multiply (**moles**x**Avogadro’s number**).

- The
**percentage composition**of a compound, simply tells us what*percent*of the total mass arises from each element in the compound. To do the calculation, simply take the atomic mass of the element in question and divide it by the molar mass of the molecule. - The
**empirical formula**for a compound is the*lowest whole-number ratio*of the elements in that compound. For example, the molecular formula for glucose is C_{6}H_{12}O_{6}, but the simplest whole-number ratio of the elements in glucose is CH_{2}O.

## **Supplementary Problems**[edit]

Please follow this link for End-of-Chapter Problems