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An element " \(\mathrm{X}\) " has five major isotopes, which are listed below along with their abundances. What is the element? $$ \begin{array}{|ccc|} \hline \text { Isotope } & \text { Percent Natural Abundance } & \text { Mass (amu) } \\ \hline{ }^{46} \mathrm{X} & 8.00 \% & 45.95269 \\ { }^{47} \mathrm{X} & 7.30 \% & 46.951764 \\ { }^{45} \mathrm{X} & 73.80 \% & 47.947947 \\ { }^{49} \mathrm{X} & 5.50 \% & 48.947841 \\ { }^{50} \mathrm{X} & 5.40 \% & 49.944792 \\ \hline \end{array} $$

Short Answer

Expert verified
The element X is \(\textbf{Titanium}\), with an average atomic mass of approximately \(47.88amu\).

Step by step solution

01

Calculate the mass contribution of each isotope

Multiply the percent abundance of each isotope by its mass to find the contribution to the average atomic mass. 1. \(46X: \: 8.00\% * 45.95269 = 3.6762152\) 2. \(47X: \: 7.30\% * 46.951764 = 3.427286472\) 3. \(45X: \: 73.80\% * 47.947947 = 35.39210566\) 4. \(49X: \: 5.50\% * 48.947841 = 2.69153125\) 5. \(50X: \: 5.40\% * 49.944792 = 2.69677896\)
02

Calculate the average atomic mass

Add up the mass contributions calculated in Step 1: Average atomic mass = \(3.6762152 + 3.427286472 + 35.39210566 + 2.69153125 + 2.69677896 = 47.884ยข01054\)
03

Identify the element using the periodic table

Using the average atomic mass calculated in Step 2, refer to the periodic table and find the element with atomic mass close to the calculated value, which is \(47.88401054amu\). The element with an atomic mass close to \(47.88401054amu\) is \(\textbf{Titanium}\) with an atomic mass of approximately \(47.87amu\). Therefore, the element X is \(\textbf{Titanium}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Percent Natural Abundance
The concept of percent natural abundance is pivotal when understanding how elements behave and exist in nature. A specific element can have different isotopes, which are atoms that contain the same number of protons but a different number of neutrons. Each isotope of an element can exist in a certain percentage compared to others. This percentage is called the percent natural abundance, indicating how common an isotope is on Earth compared to other isotopes of the same element.
For example, if you have an isotope with 50% natural abundance, this means that out of all the occurrences of that element, half of them will be that specific isotope. Calculating the average atomic mass of an element relies heavily on understanding these abundances because they show how each isotope contributes to the element's overall atomic mass.
Isotopes
Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of an element have the same number of protons, meaning their identity as a specific element remains unchanged, but the number of neutrons can vary. This variance leads to differences in atomic mass, even though they have similar chemical properties.
In our example, element "X" has five isotopes listed, each with a unique atomic mass and percent natural abundance. Even though they are forms of the same element, they provide us with different properties and make up different portions of the total abundance of element "X" in nature.
Element Identification
The task of identifying an element involves understanding the isotopes and their contributions to its average atomic mass. All elements are identified by the sum of the contributions of their isotopes' masses based on their natural abundances.
Once the average atomic mass is calculated, we can compare it to known atomic masses presented in the periodic table. This comparison helps us pinpoint which element is being dealt with. In this example, once the average atomic mass of approximately 47.884 amu was calculated, it matched closely with the atomic mass of Titanium, around 47.87 amu, helping us identify element "X" as Titanium.
Atomic Mass Calculation
Calculating atomic mass accurately is crucial for identifying elements and understanding their properties. The average atomic mass is computed by taking a weighted average of all the isotopes of an element, considering both their isotopic masses and percent abundances.
The formula for calculating the average atomic mass is:
  • For each isotope:
    • Multiply the mass of the isotope by its percent abundance (as a fraction, not a percentage).
  • Sum the results from all isotopes.
This calculation gives a single value representing the atom's mass, which helps chemists and physicists identify the element and predict how it might behave in reactions and compounds.

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Most popular questions from this chapter

Give the balanced equation for each of the following chemical reactions: a. Glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{t}\right)\) reacts with oxygen gas to produce gaseous carbon dioxide and water vapor. b. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. c. Carbon disulfide liquid reacts with ammonia gas to produce hydrogen sulfide gas and solid ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\).

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

A substance \(\mathrm{X}_{2} Z\) has the composition (by mass) of \(40.0 \% \mathrm{X}\) and \(60.0 \% \mathrm{Z}\). What is the composition (by mass) of the compound \(\mathrm{XZ}_{2}\) ?

Boron consists of two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\). Chlorine also has two isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\). How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

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