Question

1. what is the atomic mass of phosphorous if phosphorous-29 has a percent abundance of 35.5%, phosphorous-30 has a percent abundance of 42.6%, and phosphorous-31 has a percent abundance of 21.9%?
2. calculate the average atomic mass for the following element given information about the relevant isotopes:

(isotope 1) mass = 20 amu; isotopic abundance = 90.92%
(isotope 2) mass = 21 amu; isotopic abundance = 0.257%
(isotope 3) mass = 22 amu; isotopic abundance = 8.82%
looking at the periodic table, which element is this most likely?

Explanation:

Problem 3

Step 1: Recall the formula for atomic mass

The atomic mass of an element is the weighted average of the masses of its isotopes, calculated as \( \text{Atomic Mass} = \sum (\text{Isotope Mass} \times \text{Abundance Fraction}) \).

Step 2: Convert abundances to fractions

• Phosphorous - 29: Abundance = \( 35.5\% = 0.355 \), Mass = \( 29 \) amu
• Phosphorous - 30: Abundance = \( 42.6\% = 0.426 \), Mass = \( 30 \) amu
• Phosphorous - 31: Abundance = \( 21.9\% = 0.219 \), Mass = \( 31 \) amu

Step 3: Calculate the contribution of each isotope

• Contribution of P - 29: \( 29 \times 0.355 = 10.295 \)
• Contribution of P - 30: \( 30 \times 0.426 = 12.78 \)
• Contribution of P - 31: \( 31 \times 0.219 = 6.789 \)

Step 4: Sum the contributions

\( 10.295 + 12.78 + 6.789 = 29.864 \)

Step 1: Use the weighted average formula

The formula for average atomic mass is \( \text{Average Atomic Mass} = (m_1 \times f_1) + (m_2 \times f_2) + (m_3 \times f_3) \), where \( m \) is the mass of the isotope and \( f \) is the fractional abundance.

Step 2: Convert abundances to fractions

• Isotope 1: \( f_1 = 90.92\% = 0.9092 \), \( m_1 = 20 \) amu
• Isotope 2: \( f_2 = 0.257\% = 0.00257 \), \( m_2 = 21 \) amu
• Isotope 3: \( f_3 = 8.82\% = 0.0882 \), \( m_3 = 22 \) amu

Step 3: Calculate each isotope's contribution

• Contribution of Isotope 1: \( 20 \times 0.9092 = 18.184 \)
• Contribution of Isotope 2: \( 21 \times 0.00257 = 0.05397 \)
• Contribution of Isotope 3: \( 22 \times 0.0882 = 1.9404 \)

Step 4: Sum the contributions

\( 18.184 + 0.05397 + 1.9404 = 20.17837 \)

To identify the element, we use the average atomic mass calculated (from Problem 4, ~20.18 amu). Looking at the periodic table, the element with an atomic mass close to 20.18 amu is Neon (Ne)? Wait, no—wait, the isotopes here have masses 20, 21, 22. Wait, actually, the element with isotopes 20, 21, 22 (with those abundances) is Neon? Wait, no, Neon has isotopes, but let's check: The average atomic mass we calculated is ~20.18, which matches Neon's atomic mass (Neon has atomic mass ~20.18). Wait, but let's confirm: Isotope 1 (20 amu, 90.92%) is Ne - 20, Isotope 2 (21 amu, 0.257%) is Ne - 21, Isotope 3 (22 amu, 8.82%) is Ne - 22. Yes, Neon's isotopes have these abundances and masses. So the element is Neon.

Answer:

The atomic mass of phosphorous is approximately \( 29.86 \) amu (or rounded as needed, often close to 30 amu considering significant figures, but the calculated value is ~29.86).

Problem 4