1. Name two ways that isotopes of an element differ.
2. What data must you know about the isotopes of an element to calculate the atomic mass of the element?
Atomic Mass of each isotope and % abundance of each isotope
3. The four isotopes of lead are shown below, each with its percent by mass abundance and the composition of its nucleus. Using these data, calculate the approximate atomic mass of lead.
82p 82p 82p 82p
122n 124n 125n 126n
1.37% 26.26% 20.82% 51.55%
4. Lithium has two naturally occurring isotopes. Lithium-6 has an atomic mass of 6.015 amu; lithium-7 has an atomic mass of 7.016 amu. The atomic mass of lithium is 6.941 amu. What is the percentage of naturally occurring lithium-7? (Make Li-6’s percent abundance x and Li-7’s percent abundance 1-x)
Isotope Atomic Mass % Abundance Lithium-6 6.015 amu x = 0.0749 . 7.49% Lithium-7 7.016 amu 1-x = 1 – 0.0749 = 0.9251 92.51%
6.015x + 7.016(1-x) = 6.941 x = 0.7492
5. What is the average atomic mass of an element, and how does it differ from the mass number?
The atomic mass of an element is the average of all of the masses of the naturally occurring isotopes of that element. The mass number represents the protons plus the neutrons for a certain isotope of that element.
6. Imagine you are standing on top of a boron-11 nucleus. Describe the numbers and kinds of subatomic particles you would see looking down into the nucleus, and those you would see looking out from the nucleus.
Inside the nucleus—5 protons and 6 neutrons Outside the nucleus—6 electrons Unit 3—Atomic Theory and the Periodic Table
7. The element boron, B, has an atomic mass of 10.81 amu according to the periodic table. However, no single atom of boron has a mass of exactly 10.81 amu. How can you explain this difference? The periodic table reports the average atomic mass, which is a weighted average of all isotopes of b oron. 8. A certain element has two isotopes. One isotope, which has a percent abundance of 72.15% has a mass of 84.9118 amu. The other isotope has a mass of 86.9092 amu.
a. Calculate the average atomic mass of this element to three decimal places
Isotope Atomic Mass % Abundance
1 84.9118 ∗ 72.15/100 = 61.2638637
2 86.9092 ∗ 27.85/100 = + 24.2042122
b. Identify the element.
9. Three isotopes of uranium occur in nature. If the relative atomic masses and abundances of each of these isotopes are as follows, calculate the average atomic mass of uranium to three decimal places. Isotope Atomic Mass Percent Abundance
uranium-234 234.0409amu ∗ 0.0055%/100 = 0.0128722495
uranium-235 235.0439 amu ∗ 0.7200%/100 = 1.69231608
uranium-238 238.0508 amu ∗ 99.2745%/100 = +236.3237414
238.0289298 0.0129 amu + 1.692 amu +236.324 amu = 238.029 amu