Chapter 7.1: Again the Mole
 Page ID
 18854
Learning Objectives
 To review how to calculate the molecular mass of a covalent compound and the formula mass of an ionic compound and to calculate the number of atoms, molecules, or formula units in a sample of a substance.
This section is a review of Chapter 1 Section 1.7. In the rest of Chapter 7 we will use the mole to calculate the outcomes of reactions, how much of one substance reacts with another, how many molecules a mass of a substance contains, what are chemical reactions and how do we account for the number of molecules of one kind that react with another and how many different products molecules the reaction produces. The name that chemists use for such things is stoichiometry.
We have been using the mole concept in previous chapters to do problems, you have been using it in ALeKs problems and you have been also using it in General Chemistry Laboratory. What follows is a bare bones review. More problems and a more thorough review can be found in Section 1.7
We start on the atomic scale. The mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass with the difference being accounted for by binding energy. Although the mass number is unitless, it is assigned units called atomic mass units (amu). Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, we can calculate the average atomic mass of any molecule or polyatomic ion from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations. We must differentiate between the mass in amu of a single isotope and the atomic weight of an element given in the periodic table chart. The latter is an average over the naturally occuring isotopes accounting for the mass percentage of each as they occur in nature.
Molecular and Formula Masses  Review
The molecular massThe sum of the average masses of the atoms in one molecule of a substance, each multiplied by its subscript. of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example 1.
Example 1
Calculate the molecular mass of ethanol, whose condensed structural formula is CH_{3}CH_{2}OH. Among its many uses, ethanol is a fuel for internal combustion engines.
Given: molecule
Asked for: molecular mass
Strategy:
A Determine the number of atoms of each element in the molecule.
B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
C Add together the masses to give the molecular mass.
Solution:
A The molecular formula of ethanol may be written in three different ways: CH_{3}CH_{2}OH (which illustrates the presence of an ethyl group, CH_{3}CH_{2}−, and an −OH group), C_{2}H_{5}OH, and C_{2}H_{6}O; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.
B Taking the atomic masses from the periodic table, we obtain
\[ \begin{matrix}
2\times\; atomic\; mass\; of\; carbon = 2\; \cancel{atoms}\frac{12.011\; amu}{\cancel{atom}}=24.022\; amu\\
6\times\; atomic\; mass\; of\; hydrogen = 6\; \cancel{atoms}\frac{1.0079\; amu}{\cancel{atom}}=6.0474\; amu\\
1\times\; atomic\; mass\; of\; oxygen = 1\; \cancel{atoms}\frac{15.994\; amu}{\cancel{atom}}=15.994\; amu
\end{matrix} \notag \]
C Adding together the masses gives the molecular mass:
24.022 amu + 6.0474 amu + 15.9994 amu = 46.069 amu
Alternatively, we could have used unit conversions to reach the result in one step, as described in Essential Skills 2 (Section 7.7 ):
The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
2C  (2 atoms)(12.011 amu/atom) = 24.022 amu 
6H  (6 atoms)(1.0079 amu/atom) = 6.0474 amu 
1O  (1 atoms)(15.9994 amu/atom) = 15.9994 amu 
C_{2}H_{6}O  molecular mass of ethanol = 46.069 amu 
Unlike molecules, which have covalent bonds, ionic compounds do not have a readily identifiable molecular unit. So for ionic compounds we use the formula mass (also called the empirical formula massAnother name for formula mass.) of the compound rather than the molecular mass. The formula massThe sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript. is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. Once again, the units are atomic mass units.
Example 2
Calculate the formula mass of Ca_{3}(PO_{4})_{2}, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.
Given: ionic compound
Asked for: formula mass
Strategy:
A Determine the number of atoms of each element in the empirical formula.
B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
C Add together the masses to give the formula mass.
Solution:
A The empirical formula—Ca_{3}(PO_{4})_{2}—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca^{2}^{+} ions and two PO_{4}^{3−} ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.
B Taking atomic masses from the periodic table, we obtain
\( atomic\; mass\; of\; calcium=3\cancel{atoms}\left ( \dfrac{40.078\; amu}{\cancel{atom}} \right )=120.234\; amu \)
\( atomic\; mass\; of\; phosphorus=2\cancel{atoms}\left ( \dfrac{30.973761\; amu}{\cancel{atom}} \right )=61.947522\; amu \)
\( atomic\; mass\; of\; oxygen=8\cancel{atoms}\left ( \dfrac{15.9994\; amu}{\cancel{atom}} \right )=127.9952\; amu \)
C Adding together the masses gives the formula mass of Ca_{3}(PO_{4})_{2}:
\[ 120.234\; amu + 61.947522\; amu + 127.9952\; amu = 310.177\; amu \]
We could also find the formula mass of Ca_{3}(PO_{4})_{2} in one step by using unit conversions or a tabular format:
\( \left [3\; atoms Ca\left ( \dfrac{40.078\; amu}{1\; atom\; Ca} \right ) \right ] + \left [2\; atoms P\left ( \dfrac{30.973761\; amu}{1\; atom\; P} \right ) \right ]+ \left [8\; atoms O\left ( \dfrac{15.9994\; amu}{1\; atom\; O} \right ) \right ]=310.177\; amu \notag \)
3Ca  (3 atoms)(40.078 amu/atom) = 24.022 amu 
2P  (2 atoms)(30.973761 amu/atom) = 6.0474 amu 
8O  (8 atoms)(15.9994 amu/atom) = 127.9952 amu 
Ca_{3}P_{2}O_{8}  formula mass of Ca_{3}(PO_{4}) = 310.177 _{2}amu 
The Mole
Chemistry is the study of how atoms and molecules interact with each other which occurs on the atomic scale. Chemists need a way of simply determining how many molecules they have in a beaker. The mole concept bridges that gap by relating the mass of a single atom or molecule in amu to the mass of a collection of a large number of such molecules in grams. Grams rather than kilograms are used, because the mole concept was introduced at a time when the base units of the measuring system were centimeters, grams and seconds, rather than today's SI units of meters, kilograms and seconds. Also, because, well a kilogram of a compound is a rather large amount of stuff.
A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon12. According to the most recent experimental measurements, this mass of carbon12 contains 6.022142 × 10^{23} atoms, but for most purposes 6.022 × 10^{23} provides an adequate number of significant figures. The number in a mole is called Avogadro’s numberThe number of units (e.g., atoms, molecules, or formula units) in 1 mol: 6.022 x 10^{23}, after the 19thcentury Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain. Since gas in a volume can be weighted on a sensitive balance, we can determine the mass in grams (g) of a number of molecules. This, indeed, up until the beginning of the 20th century, was the most accurate way of determining the molecular weight of a molecule that could be voltatized. You will do an experiment in General Chemistry Laboratory that uses this method.
The definition of a mole—that is, the decision to base it on 12 g of carbon12—is also arbitrary. The important point is that 1 mol of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 10^{23}.
Note the Pattern
One mole always has the same number of objects: 6.022 × 10^{23}.
The concept of the mole allows us to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon12 atoms, we would weigh out 12 g of isotopically pure carbon12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 10^{23}). Because of the way in which the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about onethird that of 1 mol of carbon12. Using the concept of the mole, we can now restate Dalton’s theory: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H_{2}O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.
Molar Mass
The molar massThe mass in grams of 1 mol of a substance. of a substance is defined as the mass in grams of 1 mol of that substance. One mole of isotopically pure carbon12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 10^{23} atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 10^{23} carbon atoms—is therefore 12.011 g/mol:
Substance (formula)  Atomic, Molecular, or Formula Mass (amu)  Molar Mass (g/mol) 

carbon (C)  12.011 (atomic mass)  12.011 
ethanol (C_{2}H_{5}OH)  46.069 (molecular mass)  46.069 
calcium phosphate [Ca_{3}(PO_{4})_{2}]  310.177 (formula mass)  310.177 
The molar mass of naturally occurring carbon is different from that of carbon12 and is not an integer because carbon occurs as a mixture of carbon12, carbon13, and carbon14. One mole of carbon still has 6.022 × 10^{23} carbon atoms, but 98.89% of those atoms are carbon12, 1.11% are carbon13, and a trace (about 1 atom in 10^{12}) are carbon14. (For more information, see Section 1.6) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When we deal with elements such as iodine and sulfur, which occur as a diatomic molecule (I_{2}) and a polyatomic molecule (S_{8}), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I_{2} and S_{8}).
The molar mass of ethanol is the mass of ethanol (C_{2}H_{5}OH) that contains 6.022 × 10^{23} ethanol molecules. As you calculated in Example 1, the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca_{3}(PO_{4})_{2}] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 10^{23} formula units.
The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, we use the relationship we discussed in Section 1.7
\[ (moles)(molar mass) → mass \tag{7.1.1} \]
or, more specifically,
\( \cancel{moles}\left ( \dfrac{grams}{\cancel{mole}} \right )=grams \)
Conversely, to convert the mass of a substance to moles, we use
\( moles\left ( \dfrac{grams}{mole} \right ) = grams \)
\( \left ( \dfrac{mass}{molar\; mass} \right )\rightarrow moles \tag{7.2.2}\)
\( \left ( \dfrac{grams}{grams/mole} \right )=grams\left ( \dfrac{mole}{grams} \right )=moles \)
Be sure to pay attention to the units when converting between mass and moles.
Figure 1.7.2 below is a copy from Chapter 1 of theflowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example 3 and Example 4.
Figure 1.7.2 A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units
Example 3
For 35.00 g of ethylene glycol (HOCH_{2}CH_{2}OH), which is used in inks for ballpoint pens, calculate the number of
 moles.
 molecules.
Given: mass and molecular formula
Asked for: number of moles and number of molecules
Strategy:
A Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
B Convert from mass to moles by dividing the mass given by the compound’s molar mass.
C Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.
Solution:
A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example 1:
2C  (2 atoms)(12.011 amu/atom) = 24.022 amu 
6H  (6 atoms)(1.0079 amu/atom) = 6.0474 amu 
2O  (2 atoms)(15.9994 amu/atom) = 31.9988 amu 
C2H6O 
molecular mass of ethanol = 62.068 amu 
The molar mass of ethylene glycol is 62.068 g/mol.
B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):
\( 35.00\; g\; ethylene\; glycol\left ( \dfrac{1\; mol\; ethylene\; glycol\;}{62.068\; g\; ethylene\; glycol} \right )=0.5639\; mol\; ethylene\; glycol \)
It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.
C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:
\( molecules\; of\; ethylene\; glycol=0.5639\; mol\left ( \dfrac{6.022\times 10^{23}}{1\; mol} \right ) \)
\( = 3.396\times 10^{23}\; molecules \)
Example 4
Calculate the mass of 1.75 mol of each compound.
 S_{2}Cl_{2} (common name: sulfur monochloride; systematic name: disulfur dichloride)
 Ca(ClO)_{2} (calcium hypochlorite)
Given: number of moles and molecular or empirical formula
Asked for: mass
Strategy:
A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).
B Convert from moles to mass by multiplying the moles of the compound given by its molar mass.
Solution:
We begin by calculating the molecular mass of S_{2}Cl_{2} and the formula mass of Ca(ClO)_{2}.
A The molar mass of S_{2}Cl_{2} is obtained from its molecular mass as follows:
2S  (2 atoms)(32.065 amu/atom) = 64.130 amu 
2Cl  (2 atoms)(35.353 amu/atom) = 70.906 amu 
S_{2}Cl_{2}  molecular mass of S_{2}Cl_{2} = 135.036 amu 
The molar mass of S_{2}Cl_{2} is 135.036 g/mol.
B The mass of 1.75 mol of S_{2}Cl_{2} is calculated as follows:
\( moles\; S{_{2}}Cl_{2} \left [molar\; mass \dfrac{g}{mol} \right ]= mass\; S{_{2}}Cl_{2} \)
\( 1.75\; mol\; S{_{2}}Cl_{2}\left ( \dfrac{135.036\; g\; S{_{2}}Cl_{2}}{1\;mol\;S{_{2}}Cl_{2}} \right )=236\;g\; S{_{2}}Cl_{2} \)
A The formula mass of Ca(ClO)_{2} is obtained as follows:
1Ca  (1 atom )(40.078 amu/atom) = 40.078 amu 
2Cl  (2 atoms)(35.453 amu/atom) = 70.906 amu 
2O  (2 atoms)(15.9994 amu/atom) = 31.9988 amu 
Ca(ClO)_{2}  formula mass of Ca(ClO)_{2} = 142.983 amu 
The molar mass of Ca(ClO)_{2} 142.983 g/mol.
B The mass of 1.75 mol of Ca(ClO)_{2} is calculated as follows:
\( moles\; Ca\left ( ClO \right )_{2}\left [ \dfrac{molar\; mass\; Ca\left ( ClO \right )_{2}}{1\; mol\; Ca\left ( ClO \right )_{2}} \right ]=mass\; Ca\left ( ClO \right )_{2} \)
\( 1.75\; mol\; Ca\left ( ClO \right )_{2}\left [ \dfrac{142.983\; g Ca\left ( ClO \right )_{2}}{1\; mol\; Ca\left ( ClO \right )_{2}} \right ]=250.\; g\; Ca\left ( ClO \right )_{2} \)
Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.
Exercise
Calculate the mass of 0.0122 mol of each compound.
 Si_{3}N_{4} (silicon nitride), used as bearings and rollers
 (CH_{3})_{3}N (trimethylamine), a corrosion inhibitor
Answer:
 1.71 g
 0.721 g
Summary
The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon12 and consists of Avogadro’s number (6.022 × 10^{23}) of atoms of carbon12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 10^{23} atoms, molecules, or formula units of that substance.
Key Takeaway
 To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole.
Conceptual Problems

Describe the relationship between an atomic mass unit and a gram.

Is it correct to say that ethanol has a formula mass of 46? Why or why not?

If 2 mol of sodium react completely with 1 mol of chlorine to produce sodium chloride, does this mean that 2 g of sodium reacts completely with 1 g of chlorine to give the same product? Explain your answer.

Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi_{3}O_{8}), a mineral used in the manufacture of porcelain.

Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass.
Answer

A = %N by mass, expressed as a decimal
\( B=\dfrac{1}{molar\; mass\; of\; nitrogen\; in\; g} \)
\( g\; nitroglycerine \overset{\times A}{\rightarrow} gN\overset{\times B}{\rightarrow} mol N \)
Numerical Problems
These are the same problems as in Section 1.7
Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass.

Calculate the molecular mass or formula mass of each compound.
 KCl (potassium chloride)
 NaCN (sodium cyanide)
 H_{2}S (hydrogen sulfide)
 NaN_{3} (sodium azide)
 H_{2}CO_{3} (carbonic acid)
 K_{2}O (potassium oxide)
 Al(NO_{3})_{3} (aluminum nitrate)
 Cu(ClO_{4})_{2} [copper(II) perchlorate]

Calculate the molecular mass or formula mass of each compound.
 V_{2}O_{4} (vanadium(IV) oxide)
 CaSiO_{3} (calcium silicate)
 BiOCl (bismuth oxychloride)
 CH_{3}COOH (acetic acid)
 Ag_{2}SO_{4} (silver sulfate)
 Na_{2}CO_{3} (sodium carbonate)
 (CH_{3})_{2}CHOH (isopropyl alcohol)

Calculate the molar mass of each compound.


Calculate the molar mass of each compound.


For each compound, write the condensed formula, name the compound, and give its molar mass.


For each compound, write the condensed formula, name the compound, and give its molar mass.


Calculate the number of moles in 5.00 × 10^{2} g of each substance. How many molecules or formula units are present in each sample?
 CaO (lime)
 CaCO_{3} (chalk)
 C_{12}H_{22}O_{11} [sucrose (cane sugar)]
 NaOCl (bleach)
 CO_{2} (dry ice)

Calculate the mass in grams of each sample.
 0.520 mol of N_{2}O_{4}
 1.63 mol of C_{6}H_{4}Br_{2}
 4.62 mol of (NH_{4})_{2}SO_{3}

Give the number of molecules or formula units in each sample.
 1.30 × 10^{−2} mol of SCl_{2}
 1.03 mol of N_{2}O_{5}
 0.265 mol of Ag_{2}Cr_{2}O_{7}

Give the number of moles in each sample.
 9.58 × 10^{26} molecules of Cl_{2}
 3.62 × 10^{27} formula units of KCl
 6.94 × 10^{28} formula units of Fe(OH)_{2}

Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I_{2})?

What is the total number of atoms in each sample?
 0.431 mol of Li
 2.783 mol of methanol (CH_{3}OH)
 0.0361 mol of CoCO_{3}
 1.002 mol of SeBr_{2}O

What is the total number of atoms in each sample?
 0.980 mol of Na
 2.35 mol of O_{2}
 1.83 mol of Ag_{2}S
 1.23 mol of propane (C_{3}H_{8})

What is the total number of atoms in each sample?
 2.48 g of HBr
 4.77 g of CS_{2}
 1.89 g of NaOH
 1.46 g of SrC_{2}O_{4}

Decide whether each statement is true or false and explain your reasoning.
 There are more molecules in 0.5 mol of Cl_{2} than in 0.5 mol of H_{2}.
 One mole of H_{2} has 6.022 × 10^{23} hydrogen atoms.
 The molecular mass of H_{2}O is 18.0 amu.
 The formula mass of benzene is 78 amu.

Complete the following table.
Substance Mass (g) Number of Moles Number of Molecules or Formula Units Number of Atoms or Ions MgCl_{2} 37.62 AgNO_{3} 2.84 BH_{4}Cl 8.93 × 10^{25} K_{2}S 7.69 × 10^{26} H_{2}SO_{4} 1.29 C_{6}H_{14} 11.84 HClO_{3} 2.45 × 10^{26} 
Give the formula mass or the molecular mass of each substance.
 PbClF
 Cu_{2}P_{2}O_{7}
 BiONO_{3}
 Tl_{2}SeO_{4}

Give the formula mass or the molecular mass of each substance.
 MoCl_{5}
 B_{2}O_{3}
 UO_{2}CO_{3}
 NH_{4}UO_{2}AsO_{4}
Contributors
 Anonymous
Modified by Joshua Halpern