Question

Naturally occurring titanium exists as five stable isotopes. Four of the isotopes are \({ }^{46} \mathrm{Ti}\) with a mass of \(45.953 \mathrm{u}(8.0 \%){ }^{47} \mathrm{Ti}\) with a mass of \(46.952 \mathrm{u}(7.3 \%)\); \({ }^{48} \mathrm{Ti}\) with a mass of \(47.948 \mathrm{u}\) \((73.8 \%)\); and \({ }^{49} \mathrm{Ti}\) with a mass of \(48.948 \mathrm{u}(5.5 \%)\). The average mass of an atom of titanium is \(47.9 \mathrm{u}\). Determine the mass of the fifth isotope of titanium.

Short answer

The mass of the fifth isotope of titanium is 50.3 u.

Step-by-step solution

- Understand the Problem

You need to find the mass of the fifth isotope of titanium, given the masses and percentages of four isotopes, and the overall average atomic mass of titanium.

- Formula for Average Atomic Mass

The average atomic mass is calculated by the sum of the masses of all isotopes, each multiplied by their relative abundance.

- Write the Equation

Denote the mass of the fifth isotope as \(m_5\). The equation for the average atomic mass is: \[47.9 = (45.953 \times 0.08) + (46.952 \times 0.073) + (47.948 \times 0.738) + (48.948 \times 0.055) + m_5 \times x\], where \(x\) is the relative abundance of the fifth isotope.

- Calculate Abundance of Fifth Isotope

First, calculate the total abundance of the given four isotopes: \[0.08 + 0.073 + 0.738 + 0.055 = 0.946\]. Thus, the relative abundance of the fifth isotope is \[1 - 0.946 = 0.054\].

- Substitute Known Values

Substitute chemical isotopes values into the average atomic mass equation: \[47.9 = (45.953 \times 0.08) + (46.952 \times 0.073) + (47.948 \times 0.738) + (48.948 \times 0.055) + m_5 \times 0.054\].

- Perform Calculations

Calculate each term: \[(45.953 \times 0.08) = 3.67624\], \[(46.952 \times 0.073) = 3.427496\], \[(47.948 \times 0.738) = 35.388024\], \[(48.948 \times 0.055) = 2.69214\]. Substitute these values: \[47.9 = 3.67624 + 3.427496 + 35.388024 + 2.69214 + (m_5 \times 0.054)\].

- Combine and Solve for \(m_5\)

Sum up the constants: \[47.9 = 45.1839 + (m_5 \times 0.054)\]. Isolate \(m_5\): \[47.9 - 45.1839 = m_5 \times 0.054\], \[2.7161 = m_5 \times 0.054\], \[m_5 = \frac{2.7161}{0.054} = 50.3\].

Key concepts

average atomic mass

The average atomic mass of an element is a weighted mean based on the masses and relative abundances of its isotopes. It is calculated by summing the products of each isotope's mass and its fractional abundance. For example, to compute the average atomic mass of titanium, we use the formula:
\text{Average atomic mass} = \frac{(m_1 \times a_1) + (m_2 \times a_2) + (m_3 \times a_3) + (m_4 \times a_4) + (m_5 \times a_5)}{1}
Where:

• \(m_1, m_2, ... m_5\) are the masses of the isotopes
• \(a_1, a_2, ... a_5\) are the relative abundances of the isotopes

The concept of an average atomic mass is crucial because it helps us understand the overall mass of an element as it occurs naturally. This value is typically what you'll see listed on the periodic table.

chemical isotopes

Chemical isotopes are atoms of the same element with different numbers of neutrons, which results in different mass numbers. Despite these mass differences, isotopes exhibit the same chemical behavior because they possess identical numbers of protons and electrons. Isotopes are generally written in the form: \text{Element-Mass Number}. For instance, the isotopes of titanium in the exercise are denoted as \({ }^{46} \text{Ti}, { }^{47} \text{Ti}, { }^{48} \text{Ti}, { }^{49} \text{Ti}\), and the fifth isotope we need to find.
These isotopes have the same atomic number (22 for titanium), meaning they each have 22 protons, but differ in their neutron count:

• \({ }^{46} \text{Ti}\) has 24 neutrons (46 - 22)
• \({ }^{47} \text{Ti}\) has 25 neutrons (47 - 22)
• \({ }^{48} \text{Ti}\) has 26 neutrons (48 - 22)
• \({ }^{49} \text{Ti}\) has 27 neutrons (49 - 22)

Understanding isotopes helps explain differing atomic weights for the same element and predicts how isotopes behave in nuclear reactions.

relative abundance

Relative abundance refers to how common or rare an isotope is in nature compared to other isotopes of the same element. This metric is typically expressed as a percentage or fraction. It's a crucial component of calculating the average atomic mass. In problems like the one involving titanium isotopes, relative abundance is used to weigh each isotope's mass contribution to the overall atomic mass.
For instance, in the exercise:

• \({ }^{46} \text{Ti}\) has a relative abundance of 8.0% (or 0.08)
• \({ }^{47} \text{Ti}\) has a relative abundance of 7.3% (or 0.073)
• \({ }^{48} \text{Ti}\) has a relative abundance of 73.8% (or 0.738)
• \({ }^{49} \text{Ti}\) has a relative abundance of 5.5% (or 0.055)

To determine the relative abundance of any remaining isotopes, sum the given abundances and subtract from one:
For the fifth isotope: \(1 - (0.08 + 0.073 + 0.738 + 0.055) = 0.054\) or 5.4%. This calculation ensures that all isotopes account for 100% of the element's occurrence in nature.