Question

Mass spectrometric analysis showed that there are four isotopes of an unknown element having the following masses and abundances: $$\begin{array}{llll} \hline \text { Isotope } & \text { Mass } \text { Number } & \text { Isotope Mass } & \text { Abundance }(\%) \\ \hline 1 & 136 & 135.9090 & 0.193 \\ 2 & 138 & 137.9057 & 0.250 \\ 3 & 140 & 139.9053 & 88.48 \\ 4 & 142 & 141.9090 & 11.07 \\ \hline \end{array}$$ Three elements in the periodic table that have atomic weights near these values are lanthanum (La), atomic number 57 , atomic weight \(139.9055 ;\) cerium (Ce), atomic number \(58,\) atomic weight \(140.115 ;\) and praeseodymium (Pr), atomic number 59 , atomic weight \(140.9076 .\) Using the data above, calculate the atomic weight and identify the element if possible.

Short answer

The element is cerium (Ce) with an atomic weight of 140.1157.

Step-by-step solution

Understand Atomic Weight Calculation

The atomic weight of an element is the weighted average of masses of all its isotopes, calculated using their natural abundances. This is given by the formula: \( \text{Atomic Weight} = \sum (\text{Isotope Mass} \times \text{Abundance}) \). Abundance must be converted from percentage to decimal form.

Convert Abundances to Decimal

Convert each abundance percentage to a decimal by dividing by 100. Thus, for isotope 1: \(0.193/100 = 0.00193\), for isotope 2: \(0.250/100 = 0.00250\), for isotope 3: \(88.48/100 = 0.8848\), and for isotope 4: \(11.07/100 = 0.1107\).

Calculate Contribution of Each Isotope

Multiply the mass of each isotope by its decimal abundance:- Isotope 1: \(135.9090 \times 0.00193 = 0.2623\)- Isotope 2: \(137.9057 \times 0.00250 = 0.3448\)- Isotope 3: \(139.9053 \times 0.8848 = 123.8029\)- Isotope 4: \(141.9090 \times 0.1107 = 15.7057\).

Sum Contributions to Find Atomic Weight

Add up all the contributions calculated in Step 3 to get the atomic weight:\[ 0.2623 + 0.3448 + 123.8029 + 15.7057 = 140.1157 \]

Identify the Element

Compare the calculated atomic weight (140.1157) with the atomic weights of lanthanum, cerium, and praeseodymium: The atomic weight 140.1157 is closest to cerium (Ce) which has an atomic weight of 140.115.

Key concepts

Mass Spectrometry

Mass spectrometry is a powerful technique used to identify the masses and abundance of different isotopes of an element. It works by ionizing chemical compounds to generate charged particles, or ions, which are then measured by their mass-to-charge ratio. This technique allows scientists to accurately determine the composition and molecular weight of a substance.

In the context of atomic weight calculation, mass spectrometry provides crucial data on the specific isotopes present in a sample and their relative abundances. This is important because it helps in identifying the elementary composition of unknown substances and can lead to the determination of an element by its unique isotopic signature.

Isotopes

Isotopes are variants of the same chemical element that have the same number of protons but different numbers of neutrons. This means they have the same atomic number but different mass numbers. Each isotope of an element can have a different natural abundance or proportion in which it occurs in a sample.

For example, the element in the exercise has four isotopes, each with different mass numbers and abundances. The isotopic distribution reflects on the ultimate atomic weight of the element measured. Despite having different masses, isotopes generally exhibit similar chemical properties since their electron configuration remains the same.

Abundance

Natural abundance refers to the relative amounts of each isotope of an element found in a natural source. It is an important factor in calculating the atomic weight of an element because the atomic weight is determined by averaging the masses of an element’s isotopes, weighted by their abundance.

In the exercise data, abundances are expressed as percentages, which need to be converted into decimal form for calculation purposes. For instance, an abundance of 0.193% means that in a given sample, this isotope makes up a very small proportion. When calculating atomic weight, each isotopic mass is multiplied by its decimal abundance, and then summed up to get the final result, reflecting each isotope’s contribution relative to its prevalence.

Periodic Table

The periodic table is an organized arrangement of all known elements. It allows for a clear representation of atomic numbers, atomic weights, and the electronic configuration of elements, facilitating the understanding and identification of elements based on their properties.

In this exercise, the calculated atomic weight of the unknown element, found to be closest to that of cerium, is used to identify it. By comparing this weight to known atomic weights in the periodic table, scientists can determine which element an unknown sample corresponds to. This highlights the periodic table's essential role in correlating empirical data with known elemental properties, helping both in education and scientific research.