Reference Section \(3.2\) to find the atomic masses of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\), the relative abundance of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) in natural carbon, and the average mass (in u) of a carbon atom. If you had a sample of natural carbon containing exactly 10,000 atoms, determine the number of \({ }^{12} \mathrm{C}\) and \({ }^{1.3} \mathrm{C}\) atoms present. What would be the average mass (in u) and the total mass (in u) of the carbon atoms in this 10,000 -atom sample? If you had a sample of natural carbon containing \(6.0221 \times 10^{23}\) atoms, determine the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms present. What would be the average mass (in u) and the total mass (in u) of this \(6.0221 \times 10^{23}\) atom sample? Given that \(1 \mathrm{~g}=6.0221 \times 10^{23} \mathrm{u}\), what is the total mass of 1 mole of natural carbon in units of grams?

Short Answer

Expert verified
In summary, the average mass of a carbon atom is \(11.86762 \text{ u}\). In the 10,000-atom sample, there are 9893 \({ }^{12} \mathrm{C}\) atoms and 107 \({ }^{13} \mathrm{C}\) atoms, with a total mass of \(118,676.2 \mathrm{~u}\). In the \(6.0221 \times 10^{23}\)-atom sample, there are \(5.9584 \times 10^{23}\) \({ }^{12} \mathrm{C}\) atoms and \(6.4413 \times 10^{21}\) \({ }^{13} \mathrm{C}\) atoms, with a total mass of \(7.142 \times 10^{24}\mathrm{~u}\). The total mass of 1 mole of natural carbon is \(11.867 \mathrm{g}\).

Step by step solution

01

Find the atomic masses and relative abundance of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\)

Based on section \(3.2\), we have the following information for \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\): Atomic mass of \({ }^{12} \mathrm{C}\): 12 u Relative abundance of \({ }^{12} \mathrm{C}\): 98.93 % Atomic mass of \({ }^{13} \mathrm{C}\): 13.00335 u Relative abundance of \({ }^{13} \mathrm{C}\): 1.07 %
02

Calculate the average mass of a carbon atom

To find the average mass, we need to multiply the atomic mass of each isotope by its relative abundance, and then add these values. Average mass of \(C\) = (12 u)(0.9893) + (13.00335 u)(0.0107) Average mass of \(C\) = \(11.86762 \text{ u}\)
03

Calculate the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms in the 10,000-atom sample

We can find the number of each isotope by multiplying their relative abundance with the total number of atoms present in the sample which is 10,000. Number of \({ }^{12} \mathrm{C}\) atoms = (10,000)(0.9893) = 9893 Number of \({ }^{13} \mathrm{C}\) atoms = (10,000)(0.0107) = 107
04

Calculate the average mass and total mass of carbon atoms in the 10,000-atom sample

Using the average mass calculated in step 2 (11.86762 u), we can find the total mass: Total mass = (Number of atoms) x (Average mass of each atom) Total mass = (10,000) x (11.86762 u) = \(118676.2 \text{ u}\)
05

Calculate the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms in the \(6.0221 \times 10^{23}\)-atom sample

Similarly as in Step 3, we can calculate the number of each isotope: Number of \({ }^{12} \mathrm{C}\) atoms = \((6.0221 \times 10^{23})(0.9893) = 5.9584 \times 10^{23}\) Number of \({ }^{13} \mathrm{C}\) atoms = \((6.0221 \times 10^{23})(0.0107) = 6.4413 \times 10^{21}\)
06

Calculate the average mass and total mass of carbon atoms in the \(6.0221 \times 10^{23}\)-atom sample

Average mass of carbon atom remains the same as in Step 2. To find the total mass, we would multiply the number of atoms by the average mass: Total mass = \((6.0221 \times 10^{23})(11.86762\text{ u}) = 7.142 \times 10^{24}\text{ u}\)
07

Calculate the total mass of 1 mole of natural carbon in grams

As given in the exercise, \(1 \mathrm{g}=6.0221 \times 10^{23} \mathrm{u}\) We already found the total mass of 1 mole of natural carbon in u, we just need to convert it to grams: Total mass in grams = \((7.142 \times 10^{24}\text{ u}) \times \frac{1\text{ g}}{6.0221 \times 10^{23}\text{ u}} = 11.867 \text{g}\) So, the total mass of 1 mole of natural carbon is \(11.867 \text{g}\).

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Key Concepts

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

Understanding Atomic Mass
In chemistry, the atomic mass is a fundamental concept that represents the mass of an atom. It's measured in atomic mass units (u), where one atomic mass unit is defined as one-twelfth the mass of a single carbon-12 atom. The importance of understanding atomic mass lies in its role in various calculations, including the determination of the molar mass of substances and the composition of mixtures of elements.

When it comes to isotopes, which are atoms of the same element with different numbers of neutrons, they have different atomic masses. The atomic mass listed in the periodic table for an element is the weighted average of the masses of its naturally occurring isotopes. This takes into account both the mass and the relative abundance of each isotope.
For example, carbon has two stable isotopes, carbon-12 and carbon-13. Their atomic masses and relative abundances must be used to calculate the average atomic mass of natural carbon. The concept becomes especially crucial when dealing with more complex elements that have a large number of isotopes, each contributing to the weighted average.
The Role of Isotope Relative Abundance
The relative abundance of isotopes is an essential piece of information in isotope abundance calculations. It indicates the percentage of each isotope present in a naturally occurring sample of an element. This data, when combined with the atomic masses of the isotopes, determines the average atomic mass of an element.

To illustrate, let's consider a hypothetical element with two isotopes: Isotope A and Isotope B. If Isotope A has an atomic mass of 10 u and a relative abundance of 70%, and Isotope B has an atomic mass of 12 u and a relative abundance of 30%, the average atomic mass would be calculated by multiplying the mass of each isotope by its relative abundance and adding the results together.
This process plays a critical role in understanding the properties of elements and in calculating the molar mass required for stoichiometric computations in chemical reactions.
Demystifying the Mole Concept
The mole concept is a bridge between the micro world of atoms and the macro world we experience. A mole is a unit in chemistry used to express amounts of a chemical substance, defined as exactly 6.02214076×10^23 (Avogadro's number) particles, which can be atoms, molecules, ions, or electrons. This vast number is used because atoms and molecules are so tiny that we need a lot of them to make up even the smallest visible amount of material.

When we say that the 'molar mass' of an element is the mass of one mole of its particles, we refer to the mass of Avogadro’s number of particles of that element. For instance, the molar mass of carbon is based on the mass of a mole of carbon atoms, which, thanks to the isotope abundance calculations, we understand as the weighted average of the atomic masses of carbon's isotopes.
The practicality of the mole concept becomes evident when converting between the mass of a substance and the number of atoms or molecules it contains, an essential step in many chemical calculations such as reacting masses and gas volumes.

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Most popular questions from this chapter

An element \(\mathrm{X}\) forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right)\). Treatment of \(10.00 \mathrm{~g} \mathrm{XCl}_{2}\) with excess chlorine forms \(12.55 \mathrm{~g} \mathrm{XCl}_{4}\). Calculate the atomic mass of \(\mathrm{X}\), and identify \(\mathrm{X}\).

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What amount (moles) is represented by each of these samples? a. \(20.0 \mathrm{mg}\) caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\) b. \(2.72 \times 10^{21}\) molecules of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) c. \(1.50 \mathrm{~g}\) of dry ice, \(\mathrm{CO}_{2}\)

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