Question

prepwork part 1: avg. atomic mass calculations
examine the mass spectrum data for element y below and use it to answer questions 1 - 6.

1. **how many isotopes are depicted in the mass spectrum data above? ____
2. **which peak depicts the isotope with the greatest abundance in nature? ____
3. **which peak depicts the isotope with the greatest number of neutrons? ____
4. **the average atomic mass of this element is probably closest to ____ amu because:
5. **what is the likely identity of element y? ____________
6. **what is the atomic number of element y? ____

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