1.5: Molecules and Moles
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)So far we have talked only about individual atoms or molecules, and about masses measured in atomic mass units. But individual molecules are hard to manipulate in the laboratory, and chemists weigh their materials in grams, not in atomic mass units. To scale up from the molecular level to the laboratory level, we use a unit called a mole. A mole of a substance is equal to as many molecules of that substance as there are atoms of carbon-12 in exactly 12 g of carbon-12. This means that 1 mole of any substance is a weight, in grams, equal to that substance's molecular weight expressed in atomic mass units. Most important of all, by this definition, 1 mole of any substance contains the same number of molecules. The chemist can count atoms and molecules in the laboratory simply by weighing them. The word mole applies not just to molecules but also to atoms; in practice, we speak of a mole of helium atoms as well as of a mole of water molecules. The term gram-atom applied to a mole of atoms is no longer widely used.
How many grams of each of the following substances are there in 1 mole of that substance: H2, H20 , CH3OH, octane (C8H18), and neon gas (Ne)?
Solution
The molecular weights (in atomic mass units) of most of these substances have been given in previous examples, and the atomic weight of neon is listed on the inside back cover. One mole of each substance is therefore:
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H2 2.0160 g break C8H18 114.23 g H2O 18.0154 g break Ne 20.179 g CH3OH 32.04 g break
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Because the weights listed in Example 9 give the correct relative weights of the molecules that are being weighed out, each of the quantities of material will contain the same number of molecules. This is what makes the concept of moles useful. It is not even necessary to know what that number is, although we know it to be 6.022 X 1023 ; it is called Avogadro's number and is given the symbol N Going from molecules to moles means a scale-up of 6.022 X 1023 times. Avogadro's number is also the conversion factor between atomic mass units and grams as units of mass: 1 g = 6.022 X 1023 amu. If we think of the molecular weight as being the mass of a mole of substance, the units for molecular weight are grams per mole; if we think of it as the actual weight of one molecule, the numerical value is unchanged but the units become atomic mass units per molecule. Both are correct.
One molecule of H2 reacts with one molecule of Cl2 to form two molecules of hydrogen chloride gas, HCl. What weight of chlorine gas should be used in order to react completely with 1 kilogram (kg) of hydrogen gas?
Solution
The molecular weights of H2 and Cl2 are 2.0160 g mole-1 and 70.906 g mole-1, respectively. * Hence 1000 g of H2 contains:
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= 496.0 moles of H2 molecules
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- 496.0 moles X 70.906 g mole-1 = 35,170 g of Cl2
* The expression "g mole-1" should be read as "grams per mole." In this notation, a speed in miles per hour is written with units of "miles hr-I."
How many molecules of \(H_2\) and \(Cl_2\) would be present in the experiment of Example 1.5.2?
Solution
In 496.0 moles of any substance, there will be
\[(496.0\; \cancel{mol}) \left(6.022 \times 10^{23}\; \dfrac{molecules}{\cancel{mole}}\right) = 2.99 \times 10^{26}\; molecules\]
As a sobering example of just how large Avogadro's number is, 1 mole of coconuts, each 14 centimeters (cm) in diameter, would fill a volume as large as the entire planet earth. The use of moles in chemical calculations is the subject of the next chapter, but the idea has been introduced here because we need to know how to scale up from the molecular to the laboratory level.
Contributors and Attributions
R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors.