Learning Objectives

- Express the masses of atoms and molecules.

Because matter is defined as anything that has mass and takes up space, it should not be surprising to learn that atoms and molecules have mass.

Individual atoms and molecules, however, are very small, and the masses of individual atoms and molecules are also very small. For macroscopic objects, we use units such as grams and kilograms to state their masses, but these units are much too big to comfortably describe the masses of individual atoms and molecules. Another scale is needed.

The atomic mass unit (u; some texts use amu, but this older style is no longer accepted) is defined as one-twelfth of the mass of a carbon-12 atom, an isotope of carbon that has six protons and six neutrons in its nucleus. By this scale, the mass of a proton is 1.00728 u, the mass of a neutron is 1.00866 u, and the mass of an electron is 0.000549 u. There will not be much error if you estimate the mass of an atom by simply counting the total number of protons and neutrons in the nucleus (i.e., identify its mass number) and ignore the electrons. Thus, the mass of carbon-12 is about 12 u, the mass of oxygen-16 is about 16 u, and the mass of uranium-238 is about 238 u. More exact masses are found in scientific references—for example, the exact mass of uranium-238 is 238.050788 u, so you can see that we are not far off by using the whole-number value as the mass of the atom.

What is the mass of an element? This is somewhat more complicated because most elements exist as a mixture of isotopes, each of which has its own mass. Thus, although it is easy to speak of the mass of an atom, when talking about the mass of an element, we must take the isotopic mixture into account.

The atomic massof an element is a weighted average of the masses of the isotopes that compose an element. What do we mean by a weighted average? Well, consider an element that consists of two isotopes, 50.% with mass 10. u and 50.% with mass 11 u. A weighted average is found by multiplying each mass by its fractional occurrence (in decimal form) and then adding all the products. The sum is the weighted average and serves as the formal atomic mass of the element. In this example, we have the following:

[latex]\begin{array}{rrrl} &0.50\times 10.\text{ u}&=&\phantom{1}5.0\text{ u} \\ +&0.50\times 11\text{ u}&=&\phantom{1}5.5\text{ u} \\ \hline &\text{sum}&=&10.5\text{ u}=\text{the atomic mass of our element} \end{array}[/latex]

Note that no atom in our hypothetical element has a mass of 10.5 u; rather, that is the average mass of the atoms, weighted by their percent occurrence.

This example is similar to a real element. Boron exists as about 20.% boron-10 (five protons and five neutrons in the nuclei) and about 80.% boron-11 (five protons and six neutrons in the nuclei). The atomic mass of boron is calculated similarly to what we did for our hypothetical example, but the percentages are different:

[latex]\begin{array}{rrrl} &0.20\times 10.\text{ u}&=&\phantom{1}2.0\text{ u} \\ +&0.80\times 11\text{ u}&=&\phantom{1}8.8\text{ u} \\ \hline &\text{sum}&=&10.8\text{ u}=\text{the atomic mass of boron} \end{array}[/latex]

Thus, we use 10.8 u for the atomic mass of boron.

Virtually all elements exist as mixtures of isotopes, so atomic masses may vary significantly from whole numbers. Table 2.4 “Selected Atomic Masses of Some Elements” lists the atomic masses of some elements. The atomic masses in Table 2.4 “Selected Atomic Masses of Some Elements” are listed to three decimal places where possible, but in most cases, only one or two decimal places are needed. Note that many of the atomic masses, especially the larger ones, are not very close to whole numbers. This is, in part, the effect of an increasing number of isotopes as the atoms increase in size. (The record number is 10 isotopes for tin.)

Element Name | Atomic Mass (u) |
---|---|

Aluminum | 26.981 |

Argon | 39.948 |

Arsenic | 74.922 |

Barium | 137.327 |

Beryllium | 9.012 |

Bismuth | 208.980 |

Boron | 10.811 |

Bromine | 79.904 |

Calcium | 40.078 |

Carbon | 12.011 |

Chlorine | 35.453 |

Cobalt | 58.933 |

Copper | 63.546 |

Fluorine | 18.998 |

Gallium | 69.723 |

Germanium | 72.64 |

Gold | 196.967 |

Helium | 4.003 |

Hydrogen | 1.008 |

Iodine | 126.904 |

Iridium | 192.217 |

Iron | 55.845 |

Krypton | 83.798 |

Lead | 207.2 |

Lithium | 6.941 |

Magnesium | 24.305 |

Manganese | 54.938 |

Mercury | 200.59 |

Molybdenum | 95.94 |

Neon | 20.180 |

Nickel | 58.693 |

Nitrogen | 14.007 |

Oxygen | 15.999 |

Palladium | 106.42 |

Phosphorus | 30.974 |

Platinum | 195.084 |

Potassium | 39.098 |

Radium | n/a |

Radon | n/a |

Rubidium | 85.468 |

Scandium | 44.956 |

Selenium | 78.96 |

Silicon | 28.086 |

Silver | 107.868 |

Sodium | 22.990 |

Strontium | 87.62 |

Sulfur | 32.065 |

Tantalum | 180.948 |

Tin | 118.710 |

Titanium | 47.867 |

Tungsten | 183.84 |

Uranium | 238.029 |

Xenon | 131.293 |

Zinc | 65.409 |

Zirconium | 91.224 |

Key Takeaways

- The atomic mass unit (u) is a unit that describes the masses of individual atoms and molecules.
- The atomic mass is the weighted average of the masses of all isotopes of an element.

Exercises

# Questions

- Define
*atomic mass unit.*What is its abbreviation? - Define
*atomic mass.*What is its unit? - Estimate the mass, in whole numbers, of each isotope.
- hydrogen-1
- hydrogen-3
- iron-56

- Estimate the mass, in whole numbers, of each isotope.
- phosphorus-31
- carbon-14
- americium-241

- Determine the atomic mass of each element, given the isotopic composition.
- lithium, which is 92.4% lithium-7 (mass 7.016 u) and 7.60% lithium-6 (mass 6.015 u)
- oxygen, which is 99.76% oxygen-16 (mass 15.995 u), 0.038% oxygen-17 (mass 16.999 u), and 0.205% oxygen-18 (mass 17.999 u)

- Determine the atomic mass of each element, given the isotopic composition.
- neon, which is 90.48% neon-20 (mass 19.992 u), 0.27% neon-21 (mass 20.994 u), and 9.25% neon-22 (mass 21.991 u)
- uranium, which is 99.27% uranium-238 (mass 238.051 u) and 0.720% uranium-235 (mass 235.044 u)

- How far off would your answer be from Exercise 5a if you used whole-number masses for individual isotopes of lithium?
- How far off would your answer be from Exercise 6b if you used whole-number masses for individual isotopes of uranium?

# Answers

1. The atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom. Its abbreviation is u.

3 a. 1

3 b.3

3 c. 56

5 a. 6.940 u

5 b. 16.000 u

7. We would get 6.924 u.