The atomic masses in the periodic table are relative masses and average masses. Explain.

Short Answer

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The atomic masses in the periodic table are both relative and average masses. They are relative because they are compared to the mass of the reference element, carbon-12, with an atomic mass unit (amu) defined as 1/12 of the mass of a carbon-12 atom. This allows for convenient expression and comparison of atomic masses. They are average masses because they represent the weighted average of the masses of all isotopes of an element, factoring in the relative abundance of each isotope. This provides a single atomic mass value for each element, rather than masses for individual isotopes.

Step by step solution

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1. Overview of Atomic Structure

Atoms are the basic building blocks of matter. They consist of a nucleus, which contains protons and neutrons, and electrons that orbit the nucleus. Protons and electrons have a charge, with protons being positively charged and electrons being negatively charged. Neutrons have no charge. The number of protons in the nucleus determines the element to which the atom belongs.
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2. Understanding Isotopes and Their Masses

Isotopes are atoms of the same element that have different numbers of neutrons in their nucleus. This means they have the same number of protons and electrons, but different masses because the number of neutrons varies. Each isotope has a specific mass, and the atomic mass for a given element is generally different for varying isotopes.
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3. Averaging Isotope Masses to Determine Atomic Mass

Atomic mass in the periodic table is an average of the masses of all isotopes of an element, weighted by the relative abundance of each isotope. The abundant isotopes have a higher contribution to the average mass, while the less abundant isotopes contribute less to the average. As a result, the atomic mass listed for an element in the periodic table represents an average value and not the mass of a single isotope.
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4. Atomic Masses are Relative Masses

Atomic masses measured in atomic mass units (amu) are relative, meaning they are compared to the mass of a reference element. The reference element used is carbon-12, with a mass of exactly 12 amu. An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom. Relative masses are used because the actual masses of atoms are extremely small, making it difficult to work with them directly. By comparing them to a reference element, it becomes more convenient to express and compare the atomic masses of different elements. In summary, atomic masses in the periodic table are relative masses because they are compared to the mass of a reference element, carbon-12, and they are average masses because they represent the weighted average of the masses of all isotopes for a given element, taking into account the relative abundance of each isotope.

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Most popular questions from this chapter

Balance the following equations: a. $\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)$ b. $\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ c. $\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{HNO}_{3}(a q)$

Consider a mixture of potassium chloride and potassium nitrate that is 43.2\(\%\) potassium by mass. What is the percent \(\mathrm{KCl}\) by mass of the original mixture?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$ \begin{array}{l}{\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\\ {\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)}\end{array} $$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ 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.

A \(0.4230-\) g sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

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