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Chapter 18: Problem 18

Find the required ratios. The atomic mass of an atom of carbon is defined to be 12 u. The ratio of the atomic mass of an atom of oxygen to that of an atom of carbon is \(\frac{4}{3} .\) What is the atomic mass of an atom of oxygen? (The symbol u represents the unified atomic mass unit, where \(\left.1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg} .\right)\)

Short Answer

Expert verified
The atomic mass of an atom of oxygen is 16 u.

Step by step solution

01

Understand the Given Ratio

The problem states that the ratio of the atomic mass of an atom of oxygen to that of carbon is \( \frac{4}{3} \). This means if the atomic mass of carbon is set at 12 u, then the atomic mass of oxygen should be \( \frac{4}{3} \times 12 \).
02

Calculate the Oxygen Atomic Mass

Multiply the atomic mass of carbon, which is 12 u, by the ratio \( \frac{4}{3} \): \[ \text{Atomic mass of oxygen} = 12 \times \frac{4}{3} = 16 \text{ u} \].
03

Conclude with the Result

Based on the calculation, the atomic mass of an atom of oxygen is found to be 16 u.

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Key Concepts

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

Atomic Mass Unit
The concept of the atomic mass unit (amu), often symbolized as "u", is fundamental in understanding atomic masses and scales of atomic particles. An atomic mass unit is defined as one-twelfth of the mass of an atom of carbon-12, which is a natural form of carbon. This standardization allows for a uniform way of measuring atomic masses across different elements.
  • An atomic mass unit provides a simple scale where the masses of atoms are often close to whole numbers. For example, carbon has an atomic mass of 12 u, while oxygen, as calculated in the problem, has an atomic mass of 16 u.
  • The atomic mass unit also allows scientists to convert atomic masses into more perceivable scales. For instance, 1 u corresponds to approximately \(1.66 \times 10^{-27} \) kilograms. This conversion bridges atomic-scale measurements with the macroscopic scales more familiar in everyday experience.
Thus, the atomic mass unit serves as a practical tool that standardizes and simplifies calculations relating to atomic and molecular masses, making it easier to communicate and understand chemical phenomena.
Carbon Atomic Mass
Carbon is a cornerstone element in chemistry, given its significant role in organic molecules. Its atomic mass of 12 u underlines its importance and use as the standard for atomic mass units. This mass stems from the fact that carbon-12, an isotope of carbon, was chosen as the reference for defining the atomic mass unit.
Carbon's atomic mass plays a crucial role in relative calculations involving other elements:
  • Since carbon's atomic mass is set to 12 u, it provides a baseline that allows chemists to compare and calculate the masses of other elements relative to carbon.
  • This standardization is what enables the straightforward calculation of ratios between atomic masses, such as in the problem where the atomic mass of oxygen is calculated based on its relationship to carbon.
Thus, the atomic mass of carbon is not just a number but a foundational reference point for both theoretical and practical chemical computations.
Oxygen Atomic Mass
Oxygen, an element essential for life, is often found in nature bonded with other elements like hydrogen in water or carbon in carbonates. Its atomic mass is calculated to be 16 u, based on the given ratio in the exercise:
  • The atomic mass of carbon is 12 u. Given the ratio \(\frac{4}{3}\), the calculation \(12 \times \frac{4}{3}\) yields 16 u for oxygen.
  • Having an atomic mass of 16 u aligns oxygen with its position on the periodic table and reflects its weight relative to other key elements like carbon.
Understanding oxygen's atomic mass is crucial because it affects how we calculate molecules like water (\(H_2O\)), where each oxygen molecule contributes significantly to the overall molecular mass. This understanding helps in various applications, from chemical reactions to industrial processes and biological systems.

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Most popular questions from this chapter

Find the required ratios. For a particular acute angle in a right triangle, the ratio of the opposite side to the hypotenuse is \(12 / 13 .\) Find the ratio of the adjacent side to the hypotenuse.

Find the required ratios. The efficiency of a power amplifier is defined as the ratio of the power output to the power input. Find the efficiency of an amplifier for which the power output is \(2.6 \mathrm{W}\) and the power input is \(9.6 \mathrm{W}\).

Find the required ratios. The percent error in a measurement is the ratio of the error in the measurement to the actual correct value, expressed as a percent. When writing a computer program, the memory remaining is determined as \(2450 \mathrm{MB}\) and then it is correctly found to be \(2540 \mathrm{MB}\). What is the percent error in the first reading?

Find the required ratios. The coefficient of friction for two contacting surfaces is the ratio of the frictional force between them to the perpendicular force that presses them together. If it takes \(450 \mathrm{N}\) to overcome friction to move a \(1.10-\mathrm{kN}\) crate along the floor, what is the coefficient of friction between the crate and the floor? See Fig. \(18.2 .\)

Solve the given applied problems involving variation. The velocity \(v\) of a pulse traveling in a string varies directly as the square root of the tension \(T\) in the string. If the velocity of a pulse in a string is \(450 \mathrm{ft} / \mathrm{s}\) when the tension is \(20.0 \mathrm{lb}\), find the velocity when the tension is 30.0 lb.
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