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Chapter 3: Problem 38

The element silver (Ag) has two naturally occurring isotopes: \({ }^{109} \mathrm{Ag}\) and \({ }^{107} \mathrm{Ag}\) with a mass of \(106.905\) amu. Silver consists of \(51.82 \%^{107} \mathrm{Ag}\) and has an average atomic mass of \(107.868 \mathrm{amu} .\) Calculate the mass of \({ }^{109} \mathrm{Ag}\).

Short Answer

Expert verified
The mass of \({ }^{109} \mathrm{Ag}\) is approximately \(108.905 \mathrm{amu}\).

Step by step solution

01

Gather the given information

We are given the following information: 1. Silver consists of \(51.82\% \:^{107} \mathrm{Ag}\) 2. The average atomic mass of silver is \(107.868 \mathrm{amu}\) 3. The mass of \({ }^{107} \mathrm{Ag}\) is \(106.905 \mathrm{amu}\)
02

Calculate the proportion of \({ }^{109} \mathrm{Ag}\) in natural silver

Since there are only two isotopes in natural silver, we can calculate the proportion of \({ }^{109} \mathrm{Ag}\) using the proportion of \({ }^{107} \mathrm{Ag}\). Percentage of \({ }^{109} \mathrm{Ag} = 100 - 51.82\) Percentage of \({ }^{109} \mathrm{Ag} = 48.18\%\)
03

Write down the equation for calculating average atomic mass

We can use the equation for average atomic mass which takes into account the proportion and mass of each isotope. The equation is: Average Atomic Mass = (Mass of isotope 1 × Proportion of isotope 1) + (Mass of isotope 2 × Proportion of isotope 2)
04

Substitute the given values into the equation

Substitute the given values into the equation: \(107.868 \mathrm{amu} = (106.905 \mathrm{amu}\times 51.82\%) + (m_{109} \times 48.18\%)\) Here, \(m_{109}\) is the mass of \({ }^{109} \mathrm{Ag}\).
05

Solve for the mass of \({ }^{109} \mathrm{Ag}\)

Now we can solve for the mass of \({ }^{109} \mathrm{Ag}\) by first rearranging the equation: \(m_{109} \times 48.18\% = 107.868 \mathrm{amu} - (106.905 \mathrm{amu}\times 51.82\%)\) Next, divide both sides by 48.18\%: \(m_{109} = \frac{107.868 \mathrm{amu} - (106.905 \mathrm{amu} \times 51.82\%)}{48.18\%}\) Now, calculate the mass of \({ }^{109} \mathrm{Ag}\): \(m_{109} = \frac{107.868 \mathrm{amu} - (106.905 \mathrm{amu} \times 0.5182)}{0.4818}\) \(m_{109} \approx 108.905\ \mathrm{amu}\) The mass of \({ }^{109} \mathrm{Ag}\) is approximately \(108.905 \mathrm{amu}\).

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Most popular questions from this chapter

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. A 752 g sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(\mathrm{g}) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

Commercial brass, an alloy of \(Z n\) and \(\mathrm{Cu}\), reacts with hydrochloric acid as follows: $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ (Cu does not react with HCl.) When \(0.5065 \mathrm{~g}\) of a certain brass alloy is reacted with excess \(\mathrm{HCl}, 0.0985 \mathrm{~g} \mathrm{ZnCl}_{2}\) is eventually isolated. a. What is the composition of the brass by mass? b. How could this result be checked without changing the above procedure?

Coke is an impure form of carbon that is often used in the industrial production of metals from their oxides. If a sample of coke is \(95 \%\) carbon by mass, determine the mass of coke needed to react completely with \(1.0\) ton of copper(II) oxide. $$ 2 \mathrm{CuO}(s)+\mathrm{C}(s) \longrightarrow 2 \mathrm{Cu}(s)+\mathrm{CO}_{2}(g) $$

In the production of printed circuit boards for the electronics industry, a \(0.60-\mathrm{mm}\) layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q)\) A plant needs to manufacture 10,000 printed circuit boards, each \(8.0 \times 16.0 \mathrm{~cm}\) in area. An average of \(80 . \%\) of the copper is remoyed from each board (density of copper \(=8.96 \mathrm{~g} / \mathrm{cm}^{3}\) ). What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume \(100 \%\) yield.

A compound contains only carbon, hydrogen, and oxygen. Combustion of \(10.68 \mathrm{mg}\) of the compound yields \(16.01 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(4.37 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The molar mass of the compound is \(176.1 \mathrm{~g} / \mathrm{mol}\). What are the empirical and molecular formulas of the compound?
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