# 6.3: Counting Atoms by the Gram

- Page ID
- 47489

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Use Avogadro's number to convert to moles and vice versa given the number of particles of an element.
- Use the molar mass to convert to grams and vice versa given the number of moles of an element.

When objects are very small, it is often inconvenient, inefficient, or even impossible to deal with the objects one at a time. For these reasons, we often deal with very small objects in groups, and have even invented names for various numbers of objects. The most common of these is "dozen" which refers to 12 objects. We frequently buy objects in groups of 12, like doughnuts or pencils. Even smaller objects such as straight pins or staples are usually sold in boxes of 144, or a dozen dozen. A group of 144 is called a "gross".

This problem of dealing with things that are too small to operate with as single items also occurs in chemistry. Atoms and molecules are too small to see, let alone to count or measure. Chemists needed to select a group of atoms or molecules that would be convenient to operate with.

## Avogadro's Number and Mole

In chemistry, it is impossible to deal with a single atom or molecule because we can't see them, count them, or weigh them. Chemists have selected a number of particles with which to work that is convenient. Since molecules are extremely small, you may suspect this number is going to be very large, and you are right. The number of particles in this group is \(6.02 \times 10^{23}\) particles and the name of this group is the **mole** (the abbreviation for ** mole** is \(\text{mol}\)). One mole of any object is \(6.02 \times 10^{23}\) of those objects. There is a particular reason that this number was chosen and this reason will become clear as we proceed.

When chemists are carrying out chemical reactions, it is important that the relationship between the numbers of particles of each reactant is known. 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.

The **mole** (symbol: **mol**) is the base unit of amount of substance ("number of substance") in the International System of Units or System International (SI), defined as exactly 6.02214076×10^{23} particles, e.g., atoms, molecules, ions or electrons. The current definition was adopted in November 2018, revising its old definition based on the number of atoms in 12 grams of carbon-12 (^{12}C) (the isotope of carbon with relative atomic mass 12 Daltons, by definition). For most purposes, 6.022 × 10^{23} provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 10^{23} atoms, 1 mole of eggs contains 6.022 × 10^{23} 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 × 10^{23}.

## Converting Between Number of Atoms to Moles and Vice Versa

We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. If we are given the number of atoms of an element X, we can convert it into moles by using the relationship

\[\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms}. \nonumber \]

The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is \(4.72 \times 10^{24}\) atoms of carbon?

###### Solution

Steps for Problem Solving |
The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is \(4.72 \times 10^{24}\) atoms of carbon? |
---|---|

Identify the "given" information and what the problem is asking you to "find." | Given: \(4.72 \times 10^{24}\) C atoms Find: mol C |

List other known quantities. | \(1\, mol = 6.022 \times 10^{23}\) C atoms |

Prepare a concept map and use the proper conversion factor. | |

Cancel units and calculate. | \[4.72 \times 10^{24} \: \cancel{\text{C} \: \ce{atoms}} \times \dfrac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \cancel{\text{C} \: \ce{atoms}}} = 7.84 \: \text{mol} \: \ce{C} \nonumber \] |

Think about your result. | The given number of carbon atoms was greater than Avogadro's number, so the number of moles of \(\ce{C}\) atoms is greater than 1 mole. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures. |

## Molar Mass

**Molar mass** is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of the element lithium is 6.94g, the molar mass of zinc is 65.38g, and the molar mass of gold is 196.97g. Each of these quantities contains \(6.022 \times 10^{23}\) atoms of that particular element. The units for molar mass are grams per mole or g/mol. \(1.00 \: \text{mol}\) of carbon-12 atoms has a mass of \(12.0 \: \text{g}\) and contains \(6.022 \times 10^{23}\) atoms. 1.00 mole of any element has a mass numerically equal to its atomic mass in grams and contains \(6.022 \times 10^{23}\) particles. The mass, in grams, of 1 mole of particles of a substance is now called the **molar mass** (mass of 1.00 mole).

## Converting Grams to Moles of an Element and Vice Versa

We can also convert back and forth between grams of an element and moles. The conversion factor for this is the molar mass of the substance. The **molar mass** is the ratio giving the number of grams for each one mole of the substance. This ratio is easily found by referring to the atomic mass of the element using the periodic table. This ratio has units of grams per mole or \(\text{g/mol}\).

Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure 6.4.1 illustrates what conversion factor is needed and two examples are given below.

Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.

###### Solution

Steps for Problem Solving |
Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium. |
---|---|

Identify the "given" information and what the problem is asking you to "find." | Given: 0.560 mol Cr Find: g Cr |

List other known quantities. | 1 mol Cr = 52.00g Cr |

Prepare a concept map and use the proper conversion factor. | |

Cancel units and calculate. | \[0.560 \: \cancel{\text{mol} \: \ce{Cr}} \times \dfrac{52.00 \: \text{g} \: \ce{Cr}}{1 \: \cancel{\text{mol} \: \ce{Cr}}} = 29.1 \: \text{g} \: \ce{Cr} \nonumber \] |

Think about your result. | Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass. The answer has three significant figures because of the \(0.560 \: \text{mol}\) |

How many moles are in 107.6g of Si?

###### Solution

Steps for Problem Solving |
How many moles are in 107.6g of Si. |
---|---|

Identify the "given" information and what the problem is asking you to "find." | Given: 107.6g Si Find: mol Si |

List other known quantities. | 1 mol Si = 28.09g Si |

Prepare a concept map and use the proper conversion factor. | |

Cancel units and calculate. | \[107.6 \: \cancel{\text{g} \: \ce{Si}} \times \dfrac{1 \: \text{mol} \: \ce{Si}}{28.09 \: \cancel{\text{g} \: \ce{Si}}} = 3.83 \: \text{mol} \: \ce{Si} \nonumber \] |

Think about your result. | Since 1 mol of Si is 28.09g, 107.6 should be about 4 moles. |

- How many moles are present in 100.0 g of Al?
- What is the mass of 0.552 mol of Ag metal?

**Answer a:**- 3.706 mol Al

**Answer b:**- 59.5 g Ag

## Summary

- A mole is defined as exactly 6.02214076×10
^{23}particles, e.g., atoms, molecules, ions or electrons. - There are \(6.02214076 \times 10^{23}\) particles in 1.00 mole. This number is called Avogadro's number.
- The molar mass of an element can be found by referring to the atomic mass on a periodic table with units of g/mol.
- Using dimensional analysis, it is possible to convert between grams, moles, and the number of atoms or molecules.

## Further Reading/Supplemental Links

__learner.org/resources/series61.html__- Thewebsite allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos, but there is no charge. The website has one video that relates to this lesson called*learner.org*.*The Mole*- Using Avogadro's law, the mass of a substance can be related to the number of particles contained in that mass.
__The Mole: (www.learner.org/vod/vod_window.html?pid=803)__ - Vision Learning tutorial: The Mole
__http://visionlearning.com/library/mo...p?mid-53&1=&c3__=

## Contributions & Attributions

- Wikipedia