Question

1. what is the atomic mass of an element with 3 isotopes with the following information:

isotope 1 – 42 amu (75% abundance)
isotope 2 – 44 amu (10% abundance)
isotope 3 – 45 amu (____% abundance)

1. sulfur-32 95.0% abundance

sulfur-33 0.76% abundance
sulfur-34 4.22% abundance
sulfur-36 0.014% abundance

1. (isotope 1) mass = 35 amu 75.53%

(isotope 2) mass = 37 amu 24.47%
which element is this?

Explanation:

Problem 5:

Step1: Find abundance of Isotope 3

The total abundance of all isotopes should be 100%. So, abundance of Isotope 3 = \(100\% - 75\% - 10\% = 15\%\)

Step2: Calculate atomic mass

Atomic mass is the weighted average of the masses of the isotopes. The formula is:
\(\text{Atomic Mass} = (m_1 \times a_1) + (m_2 \times a_2) + (m_3 \times a_3)\)
where \(m\) is the mass of the isotope and \(a\) is the abundance (in decimal form).

Convert abundances to decimals:
\(a_1 = 0.75\), \(a_2 = 0.10\), \(a_3 = 0.15\)

Substitute the values:
\(\text{Atomic Mass} = (42 \times 0.75) + (44 \times 0.10) + (45 \times 0.15)\)
\(= 31.5 + 4.4 + 6.75\)
\(= 42.65\) amu

Problem 6:

Step1: Convert abundances to decimals

\(a_1 = 0.95\), \(a_2 = 0.0076\), \(a_3 = 0.0422\), \(a_4 = 0.00014\)

Step2: Calculate atomic mass

Using the weighted average formula:
\(\text{Atomic Mass} = (32 \times 0.95) + (33 \times 0.0076) + (34 \times 0.0422) + (36 \times 0.00014)\)
\(= 30.4 + 0.2508 + 1.4348 + 0.00504\)
\(= 32.09064\) amu (≈ 32.09 amu)

Problem 7:

Step1: Calculate atomic mass

Using the weighted average formula:
\(\text{Atomic Mass} = (35 \times 0.7553) + (37 \times 0.2447)\)
\(= 26.4355 + 9.0539\)
\(= 35.4894\) amu (≈ 35.49 amu)

Step2: Identify the element

The atomic mass of chlorine (Cl) is approximately 35.45 amu, which is close to our calculated value (due to rounding differences in abundance or mass). So the element is Chlorine.

Answer:

s:

1. Atomic mass: \(\boldsymbol{42.65}\) amu
2. Atomic mass: \(\boldsymbol{\approx 32.09}\) amu
3. Element: \(\boldsymbol{\text{Chlorine (Cl)}}\)