Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

You are given a small bar of an unknown metal X. You find the density of the metal to be \(10.5 \mathrm{g} / \mathrm{cm}^{3} .\) An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as \(4.09 Å\left(1 Å=10^{-10} \mathrm{m}\right) .\) Identify X.

Short Answer

Expert verified
The unknown metal X can be identified as silver (Ag) based on its density, face-centered cubic unit cell structure, and atomic mass of approximately 108 u.

Step by step solution

01

Determine the number of atoms in an FCC unit cell

In a face-centered cubic unit cell, there is 1 atom at each corner and 1 atom at the center of each face. There are 8 corners and 6 faces in an FCC unit cell. From this information, we can calculate the total number of atoms in a FCC unit cell: Number of atoms = 8 (atoms at corners) * 1/8 + 6 (atoms at faces) * 1/2 = 1 * 8 + 3 * 6 = 1 + 18 = 4 atoms per unit cell
02

Calculate the volume of the unit cell

We are given the edge length of the unit cell as 4.09 Å, and we need to convert it to meters (1 Å = 10^{-10} m): Edge length = 4.09 Å * (10^{-10} m/Å) = 4.09 * 10^{-10} m Now, we can calculate the volume of the unit cell: Volume = edge length^3 = (4.09 * 10^{-10} m)^3 = 6.84 * 10^{-29} m^3
03

Calculate the mass of the unit cell

We are given the density of the metal as 10.5 g/cm^3, and we need to convert it to kg/m^3: Density = 10.5 g/cm^3 * (1 kg/1000 g) * (100 cm/m)^3 = 10.5 * 10^3 kg/m^3 Now, calculate the mass of the unit cell using density and volume: Mass = density * volume = 10.5 * 10^3 kg/m^3 * 6.84 * 10^{-29} m^3 = 71.82 * 10^{-25} kg
04

Calculate the mass of one atom and identify the metal

We determined that there are 4 atoms in the unit cell. To find the mass of one atom, divide the mass of the unit cell by the number of atoms: Mass of one atom = (71.82 * 10^{-25} kg) / 4 = 17.96 * 10^{-25} kg Now, we need to convert the mass of one atom to atomic mass units (u) to identify the metal using the periodic table. The conversion is 1 u = 1.66054 * 10^{-27} kg. Mass of one atom in atomic mass units (u) = 17.96 * 10^{-25} kg * (1 u / 1.66054 * 10^{-27} kg) = 107.87 u With a mass of approximately 108 u, the unknown metal X is most likely silver (Ag), which has an atomic mass of about 107.87 u.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Density calculation
The calculation of density involves understanding both mass and volume. Density is defined as the mass of an object divided by its volume. In this exercise, the density of metal X is given as 10.5 g/cm³. To use it in calculations involving the metric system, it is advantageous to convert this into kg/m³:
  • Convert grams to kilograms: 1 g = 0.001 kg, so 10.5 g/cm³ becomes 10.5 * 10³ kg/m³.
  • Convert cm³ to m³: 1 cm³ = 1×10⁻⁶ m³, thus adjusting our density to kg/m³.
This conversion facilitates the subsequent calculation of mass and volume in compatible units, ensuring that the calculations are both accurate and straightforward. Keeping the units consistent is critical in ensuring the accuracy of final results and interpretations.
X-ray diffraction
X-ray diffraction is a powerful technique used in determining the structure of crystalline materials. In this method, X-rays are directed at a crystal, and the way they scatter provides valuable information about the crystal's atomic structure. The resulting diffraction pattern helps us find out distances within the unit cell of the crystal.
In the context of our problem, X-ray diffraction provides the edge length of the face-centered cubic (FCC) unit cell. The measurements from the diffraction experiment allow us to calculate the actual dimensions of the cell in meters, essential for further volume and density calculations. This is achieved by using the edge length, given as 4.09 Å, and converting it into meters, acknowledging that 1 Å equals 10⁻¹⁰ meters. Thus, the precise measurement helps in the accurate identification of the crystal's atomic arrangement and properties.
Atomic mass units
Atomic mass units (AMU) are a standard unit of mass used to express atomic and molecular weights. The conversion to AMUs is pivotal in identifying the unknown metal based on its atomic properties. After determining the mass of one atom of metal X in kilograms, the conversion to AMU is accomplished by the relation:
  • 1 AMU = 1.66054 × 10⁻²⁷ kg
Using this conversion factor, we determine that the approximate mass of one atom of the metal is 107.87 AMUs. This value is essential for comparing it to known atomic masses of elements, thereby leading to the identification of the unknown metal. AMU acts as a bridge between laboratory measurements and the well-known masses from the periodic table.
Periodic table identification
The periodic table is an invaluable tool in identifying elements by their atomic mass. In this exercise, we derived the atomic mass in atomic mass units, approximately 107.87 AMUs, from the given data on density and unit cell volume.
Upon consulting the periodic table, we find that this atomic mass closely corresponds to that of silver (Ag), which has an atomic mass of about 107.87 AMUs. This comparison allows us to confidently identify the unknown metal as silver.
Periodic table identification uses fundamental chemical data such as atomic mass, which, when paired with structural data obtained from X-ray diffraction and density calculations, provides a comprehensive picture that facilitates the accurate determination of chemical identity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A common response to hearing that the temperature in New Mexico is \(105^{\circ} \mathrm{F}\) is, "It's not that bad; it's a dry heat," whereas at the same time the summers in Atlanta, Georgia, are characterized as "dreadful," even though the air temperature is typically lower. What role does humidity play in how our bodies regulate temperature?

Cake mixes and other packaged foods that require cooking often contain special directions for use at high elevations. Typically these directions indicate that the food should be cooked longer above 5000 ft. Explain why it takes longer to cook something at higher elevations.

One method of preparing elemental mercury involves roasting cinnabar (HgS) in quicklime (CaO) at 600.^ C followed by condensation of the mercury vapor. Given the heat of vaporization of mercury (296 J/g) and the vapor pressure of mercury at \(25.0^{\circ} \mathrm{C}\left(2.56 \times 10^{-3} \text {torr }\right),\) what is the vapor pressure of the condensed mercury at \(300 .^{\circ} \mathrm{C} ?\) How many atoms of mercury are present in the mercury vapor at \(300 .^{\circ} \mathrm{C}\) if the reaction is conducted in a closed 15.0 -L container?

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69 ). Given that the density of cesium chloride is \(3.97 \mathrm{g} / \mathrm{cm}^{3},\) and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm},\) and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

The molar enthalpy of vaporization of water at \(373 \mathrm{K}\) and 1.00 atm is \(40.7 \mathrm{kJ} / \mathrm{mol} .\) What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free