# Introductory Chemistry Online/Mole and Measurement

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## 4.1 Measurement and Scale in Chemistry: The Mole Concept

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 $_{6}^{12}{\text{C}}$). 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:
$\left(7.9\times 10^{12}dollars\right)\left({\frac {1{\text{ }}mol{\text{ }}dollars}{6.02\times 10^{23}dollars}}\right)=mol{\text{ }}dollars$
$=1.3\times 10^{-11}mol{\text{ }}dollars$
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.

## 4.2 Molar Mass

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 $\left(^{g}\!\!\diagup \!\!_{mol}\;\right)$.
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 ($_{6}^{12}{\text{C}}$); 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
Exercise 4.2: Calculation of Molar Masses.
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

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 $\left({\frac {2.{\text{016 }}g{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)$.
We are given that we have 0.50 moles of H$_2$ and we want to find the number of grams of H$_2$ 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.
$\left(0.50{\text{ }}mol{\text{ H}}_{\text{2}}\right)\times \left({\frac {2.{\text{016 }}g{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)=x{\text{ }}g{\text{ H}}_{\text{2}}$
$={\text{1}}{\text{.0 }}g{\text{ H}}_{\text{2}}$
Example 4.2 Mole-to-Mass and Mass-to-Mole Conversions
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 $\left({\frac {6.{\text{02 }}\times {\text{ 10}}^{\text{23}}{\text{ }}molecules{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)$.
Equation We are given that we have 3.42 moles of H$_2$ and we want to find the number of molecules of H$_2$ 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.
$\left(3.{\text{42 }}mol{\text{ H}}_{\text{2}}\right)\times \left({\frac {6.{\text{02 }}\times {\text{ 10}}^{\text{23}}{\text{ }}molecules{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)=x{\text{ }}molecules{\text{ H}}_{\text{2}}$
$={\text{2}}.06\times {\text{10}}^{\text{24}}molecules{\text{ H}}_{\text{2}}$
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 $\left({\frac {2.{\text{016 }}g{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)$.
The known ratio of molecules per mole is $\left({\frac {6.{\text{02 }}\times {\text{ 10}}^{\text{23}}{\text{ }}molecules{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)$.
Equation We are given that we have 6.022 grams of H$_2$ and we want to find the number of molecules of H$_2$ 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.
$\left(6.{\text{022 }}g{\text{ H}}_{\text{2}}\right)\times \left({\frac {1{\text{ }}mol{\text{ H}}_{\text{2}}}{2.{\text{016 }}g{\text{ H}}_{\text{2}}}}\right)\times \left({\frac {6.{\text{02 }}\times {\text{ 10}}^{\text{23}}{\text{ }}molecules{\text{ H}}_{\text{2}}}{1{\text{ }}mol{\text{ H}}_{\text{2}}}}\right)=x{\text{ }}molecules{\text{ H}}_{\text{2}}$
$={\text{1}}.80\times {\text{10}}^{\text{24}}{\text{ }}molecules{\text{ H}}_{\text{2}}$
Example 4.5 Number-to-Mass Conversion
Exercise 4.3: Mole, Mass and Number Conversions.

## 4.4 Percentage Composition

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.
$\left({\frac {\text{96 points}}{100{\text{ points total}}}}\right)\times 100=96\%$
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:
$^{\left(1.008{\text{ }}^{\text{g}}\!\!\diagup \!\!_{\text{mol}}\;\right)}\!\!\diagup \!\!_{\left({\text{36}}{\text{.46 }}^{\text{g}}\!\!\diagup \!\!_{\text{mol}}\;\right)}$
and the percentage of hydrogen in HCl is obtained by multiplying the fraction by 100:
$0.02765\times 100=2.765\%$
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:
$^{\left(1.008{\text{ }}^{\text{g}}\!\!\diagup \!\!_{\text{mol}}\;\right)}\!\!\diagup \!\!_{\left({\text{36}}{\text{.46 }}^{\text{g}}\!\!\diagup \!\!_{\text{mol}}\;\right)}\;=0.02765$
Example 4.6 The Percentage of an Element in a Compound
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 $\left({\frac {4{\text{0}}{\text{.08 }}g{\text{ Ca}}}{1{\text{ }}mol{\text{ Ca}}}}\right)$.
The known molar mass of CaF$_2$ is $\left({\frac {7{\text{8}}{\text{.08 }}g{\text{ CaF}}_{\text{2}}}{1{\text{ }}mol{\text{ Cl}}_{\text{2}}}}\right)$.
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.
$\left(4{\text{23}}{\text{.6 }}g{\text{ CaF}}_{\text{2}}\right)\times \left({\frac {1{\text{ }}mol{\text{ }}}{7{\text{8}}{\text{.08 }}g{\text{ CaF}}_{\text{2}}}}\right)\times \left({\frac {4{\text{0}}{\text{.08 }}g{\text{ Ca}}}{1{\text{ }}mol{\text{ }}}}\right)=x{\text{ }}g{\text{ Ca}}$
$=217.4{\text{ }}g{\text{ Ca}}$
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
Exercise 4.4: Percentage Composition Calculations.

## 4.5 Empirical and Molecular Formulas

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 $n={\frac {\text{molar mass of the compound}}{\text{molar mass of the empirical formula}}}$
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,
$n={\frac {{\text{60}}{\text{.05 }}^{g}\!\!\diagup \!\!_{mol}\;}{{\text{30}}{\text{.02 }}^{g}\!\!\diagup \!\!_{mol}\;}}=2$; 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
Exercise 4.5: Empirical and Molecular Formula Calculations.

## Study Points

* 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 molar mass of the substance in grams per mole. Thus, 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.