Question

worksheet - isotopes and average atomic masses name: jaliah d period: 6 date: 01 - 11 - 25

1. four isotopes of lead include 204 - lead, 206 - lead, 207 - lead and 208 - lead. the average atomic mass of a lead atom is 207.2 amu. which isotope of lead is likely to be the most abundant?
2. what do all isotopes of an element have in common?
3. explain why carbon - 14 and nitrogen - 14 are not considered isotopes.
4. write the atomic symbol (symbol notation) for the two isotopes of uranium (u), whose atomic number is 92. one isotope has 142 neutrons, and the other isotope has 146 neutrons.
5. calculate the average atomic mass of the element iron (fe) using the following data:

isotope abundance
iron - 54 6%
iron - 56 92%
iron - 57 2%

1. calculate the average atomic mass of the element nitrogen (n) using the following data:

isotope abundance
nitrogen - 14 95%
nitrogen - 15 3%
nitrogen - 16 2%

1. calculate the average atomic mass of the element iodine (i) using the following data:

isotope abundance
iodine - 127 86%
iodine - 126 11%
iodine - 128 3%

Answer:

1. Question: Four isotopes of lead include 204 - lead, 206 - lead, 207 - lead and 208 - lead. The average atomic mass of a lead atom is 207.2 amu. Which isotope of lead is likely to be the most abundant?

• Explanation:
• ## Step1: Recall the concept of average atomic mass
• The average atomic mass is a weighted - average of the masses of all the isotopes of an element. The isotope with the mass closest to the average atomic mass will be the most abundant.
• ## Step2: Compare isotope masses to average atomic mass
• The mass of lead - 204 is 204 amu, lead - 206 is 206 amu, lead - 207 is 207 amu, and lead - 208 is 208 amu. The average atomic mass of lead is 207.2 amu. Since 208 amu is closest to 207.2 amu, lead - 208 is likely to be the most abundant.
• Answer: Lead - 208

1. Question: What do all isotopes of an element have in common?

• Brief Explanations:
• Isotopes of an element have the same number of protons. Protons determine the atomic number and the identity of the element.
• Answer: The same number of protons

1. Question: Explain why carbon - 14 and nitrogen - 14 are not considered isotopes.

• Brief Explanations:
• Isotopes are atoms of the same element with different numbers of neutrons. Carbon - 14 has 6 protons and 8 neutrons, while nitrogen - 14 has 7 protons and 7 neutrons. Since they are atoms of different elements (different atomic numbers due to different numbers of protons), they are not isotopes.
• Answer: They are atoms of different elements (different number of protons)

1. Question: Write the atomic symbol (symbol notation) for the two isotopes of uranium (U), whose atomic number is 92. One isotope has 142 neutrons, and the other isotope has 146 neutrons.

• Explanation:
• ## Step1: Calculate the mass number for the first isotope
• The mass number ($A$) is the sum of the number of protons ($Z$) and neutrons ($N$). Given $Z = 92$ and $N_1=142$, then $A_1=Z + N_1=92 + 142=234$. The atomic symbol is $^{234}_{92}U$.
• ## Step2: Calculate the mass number for the second isotope
• Given $Z = 92$ and $N_2 = 146$, then $A_2=Z+N_2=92 + 146=238$. The atomic symbol is $^{238}_{92}U$.
• Answer: $^{234}_{92}U$, $^{238}_{92}U$

1. Question: Calculate the average atomic mass of the element iron (Fe) using the following data:

• Isotope: Iron - 54, Abundance: 6%
• Isotope: Iron - 56, Abundance: 92%
• Isotope: Iron - 57, Abundance: 2%
• Explanation:
• ## Step1: Convert percentages to decimals
• The abundance of iron - 54 as a decimal is $0.06$, iron - 56 is $0.92$, and iron - 57 is $0.02$.
• ## Step2: Use the average - atomic - mass formula
• The average atomic mass ($A_{avg}$) formula is $A_{avg}=\sum_{i}(A_i\times x_i)$, where $A_i$ is the mass of the $i$ - th isotope and $x_i$ is its abundance.
• $A_{avg}=(54\times0.06)+(56\times0.92)+(57\times0.02)$
• $A_{avg}=3.24 + 51.52+1.14$
• $A_{avg}=55.9$ amu
• Answer: 55.9 amu

1. Question: Calculate the average atomic mass of the element nitrogen (N) using the following data:

• Isotope: Nitrogen - 14, Abundance: 95%
• Isotope: Nitrogen - 15, Abundance: 3%
• Isotope: Nitrogen - 16, Abundance: 2%
• Explanation:
• ## Step1: Convert percentages to decimals
• The abundance of nitrogen - 14 as a decimal is $0.95$, nitrogen - 15 is $0.03$, and nitrogen - 16 is $0.02$.
• ## Step2: Use the average - atomic - mass formula
• $A_{avg}=(14\times0.95)+(15\times0.03)+(16\times0.02)$
• $A_{avg}=13.3+0.45 + 0.32$
• $A_{avg}=14.07$ amu
• Answer: 14.07 amu

1. Question: Calculate the average atomic mass of the element iodine (I) using the following data:

• Isotope: Iodine - 127, Abundance: 86%
• Isotope: Iodine - 126, Abundance: 11%
• Isotope: Iodine - 128, Abundance: 3%
• Explanation:
• ## Step1: Convert percentages to decimals
• The abundance of iodine - 127 as a decimal is $0.86$, iodine - 126 is $0.11$, and iodine - 128 is $0.03$.
• ## Step2: Use the average - atomic - mass formula
• $A_{avg}=(127\times0.86)+(126\times0.11)+(128\times0.03)$
• $A_{avg}=109.22+13.86 + 3.84$
• $A_{avg}=126.92$ amu
• Answer: 126.92 amu