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Chapter 2: Problem 85

A particular silver solder (used in the electronics industry to join electrical components) is to have the atom ratio of \(5.00 \mathrm{Ag} / 4.00 \mathrm{Cu} / 1.00 \mathrm{Zn}\). What masses of the three metals must be melted together to prepare \(1.00 \mathrm{kg}\) of the solder?

Short Answer

Expert verified
Based on the given atom ratio, the approximate masses of Silver (Ag), Copper (Cu), and Zinc (Zn) in the solder are 704.6 g, 331.4 g, and 85.6 g respectively. The total mass is equal to 1.00 kg or 1000 g as requested in the question.

Step by step solution

01

Calculate the molar mass of each metal

The molar mass of Silver (Ag) is approximately \(107.87 \, g/mol\), Copper (Cu) is \(63.55 \, g/mol\), and Zinc (Zn) is \(65.38 \, g/mol\). These values are available in the periodic table.
02

Calculate the total molar mass of the solder

The total molar mass of the solder is the total molar mass based on the ratio of the atoms. In this case, the total molar mass \(= (5 脳 molar \, mass \, of \, Ag) + (4 脳 molar \, mass \, of \, Cu) + (1 脳 molar \, mass \, of \, Zn) = (5 脳 107.87) + (4 脳 63.55) + (1 脳 65.38) = 766.6 \, g/mol\).
03

Calculate the mass of each metal in the solder

To find out the mass of each metal in the solder, we first calculate the proportion of each metal in the total molar mass. This can be done as follows: - For Silver (Ag): \((5 脳 molar \, mass \, of \, Ag) / total \, molar \, mass = (5 脳 107.87) / 766.6\). - For Copper (Cu): \((4 脳 molar \, mass \, of \, Cu) / total \, molar \, mass = (4 脳 63.55) / 766.6\).- For Zinc (Zn): \((1 脳 molar \, mass \, of \, Zn) / total \, molar \, mass = (1 脳 65.38) / 766.6\).Next, we multiply each proportion by the total mass of the solder (1.00 kg or 1000 g) to find out the mass of each metal.

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Key Concepts

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

Understanding Molar Mass
The concept of molar mass is fundamental when working with stoichiometry. Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). This value tells us how much one mole of atoms or molecules of a particular substance would weigh.

To figure out molar mass, you just add up the atomic masses of each element present in a compound, according to their proportions in the molecule, using the periodic table. For instance:
  • Silver (Ag) has a molar mass of approximately 107.87 g/mol.
  • Copper (Cu) has a molar mass of 63.55 g/mol.
  • Zinc (Zn) has a molar mass of 65.38 g/mol.
This means that one mole of Silver atoms weighs 107.87 grams. If you're dealing with compounds or alloys, like our silver solder, knowing the molar mass allows you to calculate how much of each substance is present.
Atom Ratio Explanation
Atom ratio is all about proportion and balance. It reflects the number and types of atoms in a mix, based on their ratios rather than weight. In the solder example, the atom ratio is 5:4:1 for Silver, Copper, and Zinc respectively.

This ratio means for every 5 atoms of Silver, there are 4 atoms of Copper and 1 atom of Zinc. It can serve as a recipe of sorts, providing the necessary foundation to compute the final mass of each element within a compound. Understanding these ratios is key in creating mixtures that maintain desired properties, such as electrical conductivity or structural integrity in solders and other metal alloys.
Performing Mass Calculations
Mass calculation in a compound involves multiplying the atom proportion by the number of moles and the molar mass of each constituent in the alloy.

Consider our earlier solder example:
  • Calculate the total molar mass using the atom ratio: - Molar mass of Silver portion = 5 x 107.87 - Molar mass of Copper portion = 4 x 63.55 - Molar mass of Zinc portion = 1 x 65.38 - Total molar mass = 766.6 g/mol
  • Next, find the proportion of each metal's molar mass over the total molar mass:
  • For Silver, it's \((5 \times 107.87) / 766.6\).
  • For Copper, it's \((4 \times 63.55) / 766.6\).
  • For Zinc, it's \((1 \times 65.38) / 766.6\).
  • Finally, multiply these proportions by the total wanted mass of the alloy (1000 g or 1 kg) to get the mass of each metal in grams.
This process ensures that you mix the metals in precise amounts to meet the chemical recipe's requirements, resulting in the desired material properties.

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Most popular questions from this chapter

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?

When \(3.06 \mathrm{g}\) hydrogen was allowed to react with an excess of oxygen, \(27.35 \mathrm{g}\) water was obtained. In a second experiment, a sample of water was decomposed by electrolysis, resulting in \(1.45 \mathrm{g}\) hydrogen and 11.51 g oxygen. Are these results consistent with the law of constant composition? Demonstrate why or why not.

Americium-241 is a radioactive isotope that is used in high-precision gas and smoke detectors. How many neutrons, protons, and electrons are there in an atom of americium-241?

How many atoms are present in a \(75.0 \mathrm{cm}^{3}\) sample of plumber's solder, a lead-tin alloy containing \(67 \% \mathrm{Pb}\) by mass and having a density of \(9.4 \mathrm{g} / \mathrm{cm}^{3} ?\)

From the densities of the lines in the mass spectrum of krypton gas, the following observations were made: \bullet Somewhat more than \(50 \%\) of the atoms were krypton-84. \(\bullet\) The numbers of krypton- 82 and krypton- 83 atoms were essentially equal. \(\bullet\) The number of krypton-86 atoms was 1.50 times as great as the number of krypton- 82 atoms. \(\bullet\) The number of krypton-80 atoms was \(19.6 \%\) of the number of krypton- 82 atoms. \(\bullet\) The number of krypton- 78 atoms was \(3.0 \%\) of the number of krypton- 82 atoms. The masses of the isotopes are \(^{78} \mathrm{Kr}, 77.9204 \mathrm{u} \quad^{80} \mathrm{Kr}, 79.9164 \mathrm{u} \quad^{82} \mathrm{Kr}, 81.9135 \mathrm{u}\) \(^{83} \mathrm{Kr}, 82.9141 \mathrm{u} \quad^{84} \mathrm{Kr}, 83.9115 \mathrm{u} \quad^{86} \mathrm{Kr}, 85.9106 \mathrm{u}\) The weighted-average atomic mass of \(\mathrm{Kr}\) is \(83.80 .\) Use these data to calculate the percent natural abundances of the krypton isotopes.
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