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Chapter 2: Problem 83

Deuterium, \(^{2} \mathrm{H}(2.0140 \mathrm{u}),\) is sometimes used to replace the principal hydrogen isotope \(^{1} \mathrm{H}\) in chemical studies. The percent natural abundance of deuterium is 0.015\%. If it can be done with 100\% efficiency, what mass of naturally occurring hydrogen gas would have to be processed to obtain a sample containing \(2.50 \times 10^{21}^{2} \mathrm{H}\) atoms?

Short Answer

Expert verified
Substitution and calculation with the above steps will yield the mass of naturally occurring hydrogen gas that needs to be processed to obtain a specific number of Deuterium atoms.

Step by step solution

01

Calculate moles of Deuterium

First, convert the number of deuterium atoms needed (\(2.50 \times 10^{21}\)) to moles using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole). The calculation is given by: \n\n\[ \frac{2.50 \times 10^{21} \text{ atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \]
02

Convert moles of Deuterium to moles of Hydrogen

Given the natural abundance of Deuterium (0.015%), convert the moles of Deuterium to moles of Hydrogen using the percent abundance as follows:\n\n\[ \text{moles of hydrogen} = \frac{\text{moles of deuterium}}{\text{percent abundance of deuterium} / 100} \]\n\nNote that the percent abundance needs to be converted to decimal form by dividing by 100.
03

Convert moles of Hydrogen to grams

Now, convert the moles of hydrogen to grams using the molar mass of hydrogen (1.008 u). The molar mass is the mass of one mole of a substance. It is used as a conversion factor to go from moles to grams (and vice versa). The formula is:\n\n\[ \text{grams of hydrogen} = \text{moles of hydrogen} \times \text{molar mass of hydrogen} \]

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Key Concepts

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

Isotopes
Isotopes are variants of a particular chemical element that share the same number of protons but differ in their number of neutrons. This results in isotopes having different atomic masses. However, because they have the same number of protons, these isotopes maintain the same chemical behavior.
For example, hydrogen has three main isotopes: protium ( ^{1}H) with zero neutrons, deuterium ( ^{2}H) with one neutron, and tritium ( ^{3}H) with two neutrons. In this exercise, the focus is on deuterium, an isotope that is relatively rare with only 0.015% abundance in nature.
Understanding isotopes is crucial in chemical calculations as they affect the mass and sometimes the behavior of the substances involved. They are used in various scientific fields, from nuclear energy to medicine, due to their unique properties.
Mole Concept
The mole is a fundamental concept in chemistry that provides a bridge between the atomic scale and macroscopic measurements. A mole is defined as the amount of substance that contains the same number of entities (such as atoms, ions, or molecules) as there are in 12 grams of pure carbon-12 ( ^{12}C), which is about 6.022 x 10^{23} entities. This value is known as Avogadro's number. The mole allows chemists to convert between the mass of a sample and the number of atoms or molecules it contains. In the provided solution, the mole concept helps quantify the number of deuterium atoms by converting them into moles. This conversion is essential since chemical reactions and calculations are often more conveniently expressed in terms of moles rather than individual atoms.
Avogadro's Number
Avogadro's number ( 6.022 imes 10^{23}) is a constant used to describe the number of atoms, ions, or molecules in one mole of a substance. This number is incredibly large because atoms and molecules are extremely small. In chemical calculations, Avogadro's number acts as a critical conversion factor. It connects the microscale (atoms or molecules) to the macroscale (grams or liters that are measurable). For instance, as seen in the step-by-step solution, converting the number of deuterium atoms to moles requires division by Avogadro's number. By employing this constant, the number of particles in a given sample can be appreciated in terms of measurable quantities, making it fundamental for calculating reactions, determining material properties, and understanding the structure of substances at a molecular level.

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Most popular questions from this chapter

Without doing detailed calculations, determine which of the following samples occupies the largest volume: (a) 25.5 mol of sodium metal \(\left(d=0.971 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) 0.725 L of liquid bromine \((d=3.12 \mathrm{g} / \mathrm{mL})\) (c) \(1.25 \times 10^{25}\) atoms of chromium metal \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right)\) (d) \(2.15 \mathrm{kg}\) of plumber's solder \(\left(d=9.4 \mathrm{g} / \mathrm{cm}^{3}\right), \mathrm{a}\) lead-tin alloy with a 2: 1 atom ratio of lead to tin

Refer to the periodic table inside the front cover and identify (a) the element that is in group 11 and the sixth period (b) an element with atomic number greater than 50 that has properties similar to the element with atomic number 18 (c) the group number of an element \(\mathrm{E}\) that forms an ion \(\mathrm{E}^{2-}\) (d) an element \(M\) that you would expect to form the ion \(\mathrm{M}^{3+}\)

Atoms are spherical and so when silver atoms pack together to form silver metal, they cannot fill all the available space. In a sample of silver metal, approximately \(26.0 \%\) of the sample is empty space. Given that the density of silver metal is \(10.5 \mathrm{g} / \mathrm{cm}^{3}\), what is the radius of a silver atom? Express your answer in picometers.

For the ion \(^{228} \mathrm{Ra}^{2+}\) with a mass of 228.030 u, determine (a) the numbers of protons, neutrons, and electrons in the ion; (b) the ratio of the mass of this ion to that of an atom of \(^{16} \mathrm{O}\) (refer to page 47 ).

To four significant figures, all of the following masses are possible for an individual titanium atom except one. The exception is (a) 45.95 u; (b) 46.95 u; (c) 47.87 u; (d) 47.95 u; (e) 48.95 u; (f) 49.94 u.
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