3.3: Avogadro's Number and the Mole

Learning Objectives

• To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound.
• To calculate the number of atoms, molecules, or formula units in a sample of a substance.
• To interpret a chemical equation in terms of mole-to-mole ratios.

Molecular and Formula Masses

The molecular mass 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 \(\PageIndex{1}\).

Example \(\PageIndex{1}\): Molecular Mass of Ethanol

Calculate the molecular mass of ethanol, whose condensed structural formula is \(\ce{CH3CH2OH}\). Among its many uses, ethanol is a fuel for internal combustion engines.

Given: molecule

Asked for: molecular mass

Strategy:

1. Determine the number of atoms of each element in the molecule.
2. 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.
3. Add together the masses to give the molecular mass.

Solution:

A The molecular formula of ethanol may be written in three different ways: \(\ce{CH3CH2OH}\) (which illustrates the presence of an ethyl group, CH₃CH₂₋, and an −OH group), \(\ce{C2CH5OH}\), and \(\ce{C2H6O}\); 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

\[ \begin{align*} 2 \times \text { atomic mass of carbon} &= 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) \\[4pt] &= 24.022 \,amu \end{align*}\]

\[ \begin{align*} 6 \times \text { atomic mass of hydrogen} &= 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) \\[4pt] &= 6.0474 \,amu \end{align*}\]

\[ \begin{align*} 1 \times \text { atomic mass of oxygen} &= 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) \\[4pt] &= 15.994 \,amu \end{align*}\]

C Adding together the masses gives the molecular mass:

\[ 24.022 \,amu + 6.0474 \,amu + 15.9994 \,amu = 46.069 \,amu \nonumber\]

Exercise \(\PageIndex{1}\): Molecular Mass of Freon

Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is \(CCl_3F\). Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:

Answer

137.368 amu

Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass 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. The units are atomic mass units.

Example \(\PageIndex{2}\): Formula Mass of Calcium Phosphate

Calculate the formula mass of Ca₃(PO₄)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:

1. Determine the number of atoms of each element in the empirical formula.
2. 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.
3. Add together the masses to give the formula mass.

Solution:

A The empirical formula—Ca₃(PO₄)₂—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca²⁺ ions and two PO₄³⁻ 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

\[ 3 \times \text {atomic mass of calcium} = 3 \, atoms \left ( {40.078 \, amu \over atom } \right ) = 120.234 \, amu \nonumber \]

\[ 2 \times \text {atomic mass of phosphorus} = 2 \, atoms \left ( {30.973761 \, amu \over atom } \right ) = 61.947522 \, amu \nonumber\]

\[ 8 \times \text {atomic mass of oxygen} = 8 \, atoms \left ( {15.9994 \, amu \over atom } \right ) = 127.9952 \, amu \nonumber\]

C Adding together the masses gives the formula mass of Ca₃(PO₄)₂:

\[120.234 \,amu + 61.947522 \, amu + 127.9952 \, amu = 310.177 \, amu \nonumber\]

Exercise \(\PageIndex{2}\): Formula Mass of Silicon Nitride

Calculate the formula mass of \(\ce{Si3N4}\), commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.

Answer

\(140.29 \,amu\)

The Mole

Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. It also describes the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10⁻²³ g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction, it is therefore essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol), from the Latin moles, meaning “pile” or “heap.”

Many familiar items are sold in numerical quantities with distinct names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them. The mole is used for this purpose.

A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 10²³ atoms, but for most purposes 6.022 × 10²³ provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 10²³ atoms, 1 mole of eggs contains 6.022 × 10²³ eggs. This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.

It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 10²³.

To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 10¹⁷ mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would have more than one trillion dollars. The mole is so large that it is useful only for measuring very small objects, such as atoms.

The concept of the mole allows scientists to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, one weighs out 12 g of isotopically pure carbon-12. 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 × 10²³). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way 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 carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, Dalton’s theory can be restated: 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 (H₂O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.

Molar Mass

The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 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 × 10²³ 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 × 10²³ 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 (C2H5OH)	46.069 (molecular mass)	46.069
calcium phosphate [Ca3(PO4)2]	310.177 (formula mass)	310.177

The molar mass of naturally-occurring carbon is different from that of carbon-12, and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 10²³ carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. (For more information, see Section 1.6 "Isotopes and Atomic Masses".) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When dealing with elements such as iodine and sulfur, which occur as a diatomic molecule (I₂) and a polyatomic molecule (S₈), 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 (I₂ and S₈).

The molar mass of ethanol is the mass of ethanol (C₂H₅OH) that contains 6.022 × 10²³ ethanol molecules. As in Example \(\PageIndex{1}\), 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 [Ca₃(PO₄)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 × 10²³ 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, the following relationship is used:

\[(moles)(molar mass) \rightarrow mass \]

or, more specifically,

\[ moles \left ( {grams \over mole } \right ) = grams \]

Conversely, to convert the mass of a substance to moles:

\[ \left ( {grams \over molar mass } \right ) \rightarrow moles \]

\[ \left ( { grams \over grams/mole} \right ) = grams \left ( {mole \over grams } \right ) = moles \]

The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:

\[\ce{2H2(g) + O2(g) \rightarrow 2H2O(l)}\]

the production of two moles of water would require the consumption of 2 moles of \(H_2\) and one mole of \(O_2\). Therefore, when considering this particular reaction

• 2 moles of H₂
• 1 mole of O₂ and
• 2 moles of H₂O

would be considered to be stoichiometrically equivalent quantitites.

These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of \(H_2O\) would be produced from 1.57 moles of \(O_2\)?

\[ (1.57\; mol\; O_2) \left( \dfrac{2\; mol\;H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O\]

The ratio \( \left( \dfrac{2\; mol\; H_2O}{1\;mol\;O_2} \right)\) is the stoichiometric relationship between \(H_2O\) and \(O_2\) from the balanced equation for this reaction.

Example \(\PageIndex{3}\): Combustion of Butane

For the combustion of butane (\(C_4H_{10}\)) the balanced equation is:

\[\ce{2C4H10(l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l)}\]

Calculate the mass of \(CO_2\) that is produced in burning 1.00 gram of \(C_4H_{10}\).

Solution

Thus, the overall sequence of steps to solve this problem is:

First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:

\[ (1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}\]

Now, the stoichiometric relationship between \(C_4H_{10}\) and \(CO_2\) is:

\[\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)\]

Therefore:

\[ \left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2\]

The question called for the determination of the mass of \(CO_2\) produced, thus we have to convert moles of \(CO_2\) into grams (by using the molecular weight of \(CO_2\)):

\[ 6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2\]

Be sure to pay attention to the units when converting between mass and moles. Figure \(\PageIndex{1}\) is a flowchart 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 Examples \(\PageIndex{3}\) and \(\PageIndex{4}\).

Figure \(\PageIndex{1}\): A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units

Example \(\PageIndex{4}\): Ethylene Glycol

For 35.00 g of ethylene glycol (HOCH₂CH₂OH), which is used in inks for ballpoint pens, calculate the number of

1. moles.
2. molecules.

Given: mass and molecular formula

Asked for: number of moles and number of molecules

Strategy:

1. Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
2. Convert from mass to moles by dividing the mass given by the compound’s molar mass.
3. 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 \(\PageIndex{1}\): 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):

\[ { \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }\]

So

\[ 35.00 \, g \text {ethylene glycol} \left ( {1 \, mole \text {ethylene glycol} \over 62.068 \, g \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol} \]

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:

\[ \text {molecules of ethylene glycol} = 0.5639 \, mol \left ( {6.022 \times 10^{23} \, molecules \over 1 \, mol } \right ) \]

\[ = 3.396 \times 10^{23} \, molecules \]

Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 10²³ molecules, which is indeed the case.

Exercise \(\PageIndex{4}\): Freon-11

For 75.0 g of CCl₃F (Freon-11), calculate the number of

1. moles.
2. molecules.

Answer a

0.546 mol

Answer b

3.29 × 10²³ molecules

Example \(\PageIndex{5}\)

Calculate the mass of 1.75 mol of each compound.

1. \(\ce{S2Cl2}\) (common name: sulfur monochloride; systematic name: disulfur dichloride)
2. \(\ce{Ca(ClO)2}\) (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 \(\ce{S2Cl2}\) and the formula mass of \(\ce{Ca(ClO)2}\).

A The molar mass of \(\ce{S2Cl2}\) is 135.036 g/mol.

B The mass of 1.75 mol of \(\ce{S2Cl2}\) is calculated as follows:

\[moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g) \]

\[ 1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2 \]

A 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:

\[ moles Ca(ClO)_2 \left [{\text {molar mass} Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2} \right ]=mass Ca(ClO)_2\]

\[ 1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2 \]

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 \(\PageIndex{5}\)

Calculate the mass of 0.0122 mol of each compound.

1. Si₃N₄ (silicon nitride), used as bearings and rollers
2. (CH₃)₃N (trimethylamine), a corrosion inhibitor

Answer a

1.71 g

Answer b

0.721 g

The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:

\[\ce{2H2(g) + O2(g) \rightarrow 2H2O(l)} \nonumber\]

the production of two moles of water would require the consumption of 2 moles of \(H_2\) and one mole of \(O_2\). Therefore, when considering this particular reaction

• 2 moles of H₂
• 1 mole of O₂ and
• 2 moles of H₂O

would be considered to be stoichiometrically equivalent quantitites.

These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of \(H_2O\) would be produced from 1.57 moles of \(O_2\)?

\[ (1.57\; mol\; O_2) \left( \dfrac{2\; mol \;H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O \nonumber\]

The ratio \( \left( \dfrac{2\; mol\;H_2O}{1\;mol\;O_2} \right)\) is the stoichiometric relationship between \(H_2O\) and \(O_2\) from the balanced equation for this reaction.

Example \(\PageIndex{6}\)

For the combustion of butane (\(C_4H_{10}\)) the balanced equation is:

\[ \ce{2C4H_{10} (l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l)} \nonumber\]

Calculate the mass of \(CO_2\) that is produced in burning 1.00 gram of \(C_4H_{10}\).

Solution

First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:

\[ (1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10} \nonumber\]

Now, the stoichiometric relationship between \(C_4H_{10}\) and \(CO_2\) is:

\[\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right) \nonumber\]

Therefore:

\[ \left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2 \nonumber\]

The question called for the determination of the mass of \(CO_2\) produced, thus we have to convert moles of \(CO_2\) into grams (by using the molecular weight of \(CO_2\)):

\[ 6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2 \nonumber\]

Thus, the overall sequence of steps to solve this problem were:

In a similar way we could determine the mass of water produced, or oxygen consumed, etc.

Interpreting Chemical Equations (Part 2)

Previously, we discussed that chemical equations give the relative amounts of reactants and products consumed or produced in a reaction. This was discussed in terms of the the number of atoms, molecules, or formula units of a reactant or a product. A chemical equation can also be interpreted interms of moles of a reactant or product. As illustrated below, the coefficients allow interpreted in any of the following ways:

• Two NH₄⁺ ions and one Cr₂O₇²⁻ ion yield 1 formula unit of Cr₂O₃, 1 N₂ molecule, and 4 H₂O molecules.
• One mole of (NH₄)₂Cr₂O₇ yields 1 mol of Cr₂O₃, 1 mol of N₂, and 4 mol of H₂O.
• A mass of 252 g of (NH₄)₂Cr₂O₇ yields 152 g of Cr₂O₃, 28 g of N₂, and 72 g of H₂O.
• A total of 6.022 × 10²³ formula units of (NH₄)₂Cr₂O₇ yields 6.022 × 10²³ formula units of Cr₂O₃, 6.022 × 10²³ molecules of N₂, and 24.09 × 10²³ molecules of H₂O.

Figure \(\PageIndex{4}\): The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced Chemical Reaction for the Ammonium Dichromate Volcano

These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of \(H_2O\) to \(N_2\) in Equation \(\ref{3.1.1}\) is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass:

\[252 \;g \;\text{of}\; \ce{(NH_4)_2Cr_2O_7} \]

yield

\[152 + 28 + 72 = 252 \; g \; \text{of products.}\]

The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters.

An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose (\(C_6H_{12}O_6\)), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.

Example \(\PageIndex{2}\): Combustion of Glucose

The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:

\[ \ce{C_6H_{12}O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)}\]

Construct a table showing how to interpret the information in this equation in terms of

1. a single molecule of glucose.
2. moles of reactants and products.
3. grams of reactants and products represented by 1 mol of glucose.
4. numbers of molecules of reactants and products represented by 1 mol of glucose.

Given: balanced chemical equation

Asked for: molecule, mole, and mass relationships

Strategy:

1. Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.
2. Use the molar masses of the reactants and products to convert from moles to grams.
3. Use Avogadro’s number to convert from moles to the number of molecules.

Solution:

This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.

1. One molecule of glucose reacts with 6 molecules of O₂ to yield 6 molecules of CO₂ and 6 molecules of H₂O.
2. One mole of glucose reacts with 6 mol of O₂ to yield 6 mol of CO₂ and 6 mol of H₂O.
3. To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O₂, 31.9988; CO₂, 44.010; and H₂O, 18.015.

\[ \begin{align*} \text{mass of reactants} &= \text{mass of products} \\[4pt] g \, glucose + g \, O_2 &= g \, CO_2 + g \, H_2O \end{align*}\]

\[ 1\,mol\,glucose \left ( {180.16 \, g \over 1 \, mol \, glucose } \right ) + 6 \, mol \, O_2 \left ( { 31.9988 \, g \over 1 \, mol \, O_2} \right ) \]

\[= 6 \, mol \, CO_2 \left ( {44.010 \, g \over 1 \, mol \, CO_2} \right ) + 6 \, mol \, H_2O \left ( {18.015 \, g \over 1 \, mol \, H_2O} \right ) \]

\[ 372.15 \, g = 372.15 \, g \]

C One mole of glucose contains Avogadro’s number (6.022 × 10²³) of glucose molecules. Thus 6.022 × 10²³ glucose molecules react with (6 × 6.022 × 10²³) = 3.613 × 10²⁴ oxygen molecules to yield (6 × 6.022 × 10²³) = 3.613 × 10²⁴ molecules each of CO₂ and H₂O.

In tabular form:

	\(C_6H_{12}O_{6\;(s)}\)	+	\(6O_{2\;(g)}\)	→	\(6CO_{2\;(g)}\)		\(6H_2O_{(l)}\)
a.	1 molecule		6 molecules		6 molecules		6 molecules
b.	1 mol		6 mol		6 mol		6 mol
c.	180.16 g		191.9928 g		264.06 g		108.09 g
d.	6.022 × 10^23 molecules		3.613 × 10^24 molecules		3.613 × 10^24 molecules		3.613 × 10^24molecule

Exercise \(\PageIndex{2}\): Ammonium Nitrate Explosion

Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas. The explosion resulted from the following reaction:

\[ 2NH_4NO_{3\;(s)} \rightarrow 2N_{2\;(g)} + 4H_2O_{(g)} + O_{2\;(g)} \]

Construct a table showing how to interpret the information in the equation in terms of

1. individual molecules and ions.
2. moles of reactants and products.
3. grams of reactants and products given 2 mol of ammonium nitrate.
4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.

Answer:

	\(2NH_4NO_{3\;(s)}\)	→	\(2N_{2\;(g)}\)	+	\(4H_2O_{(g)}\)	+	\(O_{2\;(g)}\)
a.	2NH4^+ ions and 2NO3^− ions		2 molecules		4 molecules		1 molecule
b.	2 mol		2 mol		4 mol		1 mol
c.	160.0864 g		56.0268 g		72.0608 g		31.9988 g
d.	1.204 × 10^24 formula units		1.204 × 10^24 molecules		2.409 × 10^24 molecules		6.022 × 10^23 molecules

Summary

To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole. 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 carbon-12, Avogadro’s number (6.022 × 10²³) of atoms of carbon-12. 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 × 10²³ atoms, molecules, or formula units of that substance.