prepwork part 1: avg. atomic mass calculations\nexamine the mass spectrum data for element y below and use…

prepwork part 1: avg. atomic mass calculations\nexamine the mass spectrum data for element y below and use it to answer questions 1 - 6.\n1. **how many isotopes are depicted in the mass spectrum data above? ____\n2. **which peak depicts the isotope with the greatest abundance in nature? ____\n3. **which peak depicts the isotope with the greatest number of neutrons? ____\n4. **the average atomic mass of this element is probably closest to ____ amu because:\n5. **what is the likely identity of element y? ____________\n6. **what is the atomic number of element y? ____

prepwork part 1: avg. atomic mass calculations\nexamine the mass spectrum data for element y below and use it to answer questions 1 - 6.\n1. **how many isotopes are depicted in the mass spectrum data above? ____\n2. **which peak depicts the isotope with the greatest abundance in nature? ____\n3. **which peak depicts the isotope with the greatest number of neutrons? ____\n4. **the average atomic mass of this element is probably closest to ____ amu because:\n5. **what is the likely identity of element y? ____________\n6. **what is the atomic number of element y? ____

Answer

Explanation:

Step1: Identify number of isotopes

Each peak represents an isotope. There are 2 peaks (Peak A and Peak B), so 2 isotopes.

Step2: Determine most abundant isotope

Peak A has a relative abundance of 75% and Peak B has 25%. So Peak A represents the most - abundant isotope.

Step3: Find isotope with most neutrons

Neutron number is related to mass number. Higher mass number means more neutrons. Peak B has a relative mass of 37 compared to Peak A's 35, so Peak B represents the isotope with the most neutrons.

Step4: Calculate average atomic mass

Average atomic mass (=\sum_{i}(m_i\times a_i)), where (m_i) is the mass of isotope (i) and (a_i) is its relative abundance. (m_1 = 35), (a_1=0.75), (m_2 = 37), (a_2 = 0.25). So, average atomic mass (=35\times0.75 + 37\times0.25=35\times\frac{3}{4}+37\times\frac{1}{4}=\frac{105 + 37}{4}=\frac{142}{4}=35.5) amu.

Step5: Identify element

Chlorine has two main isotopes (^{35}Cl) and (^{37}Cl) with similar abundances as in the given data. So element Y is likely Chlorine.

Step6: Determine atomic number

The atomic number of Chlorine is 17.

Answer:

  1. 2
  2. Peak A
  3. Peak B
  4. 35.5; calculated using the formula for average atomic mass with given relative abundances and masses of isotopes.
  5. Chlorine
  6. 17