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Naturally occurring lead has the following isotopic abundances: $$ \begin{array}{lcc} \hline \text { Isotope } & \text { Abundance (\%) } & \text { Atomic mass (u) } \\ \hline{ }^{204} \mathrm{~Pb} & 1.4 & 203.9730 \\ { }^{206} \mathrm{~Pb} & 24.1 & 205.9744 \\ { }^{207} \mathrm{~Pb} & 22.1 & 206.9759 \\ { }^{208} \mathrm{~Pb} & 52.4 & 207.9766 \\ \hline \end{array} $$ (a) What is the average atomic mass of \(\mathrm{Pb}\) ? (b) Sketch the mass spectrum of \(\mathrm{Pb}\).

Short Answer

Expert verified
The average atomic mass of Pb is 206.14 u. The mass spectrum shows peaks at 204, 206, 207, and 208 u with heights proportional to their abundances.

Step by step solution

01

Convert Abundance Percentages to Decimal Form

To calculate the average atomic mass, we first need to convert the percentage abundances into decimal form by dividing each percentage by 100. For each isotope:- \(^{204}\mathrm{Pb}: \frac{1.4}{100} = 0.014\)- \(^{206}\mathrm{Pb}: \frac{24.1}{100} = 0.241\)- \(^{207}\mathrm{Pb}: \frac{22.1}{100} = 0.221\)- \(^{208}\mathrm{Pb}: \frac{52.4}{100} = 0.524\)
02

Calculate the Weighted Atomic Mass

Multiply the atomic mass of each isotope by its decimal abundance to get the weighted contribution to the average atomic mass:- \(^{204}\mathrm{Pb}: 203.9730 \times 0.014 = 2.855622\)- \(^{206}\mathrm{Pb}: 205.9744 \times 0.241 = 49.6402944\)- \(^{207}\mathrm{Pb}: 206.9759 \times 0.221 = 45.7426639\)- \(^{208}\mathrm{Pb}: 207.9766 \times 0.524 = 108.5776884\)
03

Sum the Weighted Contributions

Add the weighted contributions from each isotope to find the average atomic mass of lead:\[2.855622 + 49.6402944 + 45.7426639 + 108.5776884 = 206.1402687\]uThus, the average atomic mass of lead is approximately 206.14 u.
04

Graph the Mass Spectrum

To sketch the mass spectrum, plot peaks on a graph with the x-axis representing the mass (in atomic mass units) and the y-axis representing the relative abundance. - Plot a peak at 203.9730 u with a relative height of 1.4. - Plot a peak at 205.9744 u with a relative height of 24.1. - Plot a peak at 206.9759 u with a relative height of 22.1. - Plot a peak at 207.9766 u with a relative height of 52.4. The height of each peak corresponds to the abundance percentage of each isotope.

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Key Concepts

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

Isotopic Abundances
Isotopic abundance provides insight into how much of each type of isotope exists within a natural sample of an element. Each isotope of an element shares the same number of protons but differs in the number of neutrons, giving each a unique atomic mass. When calculating the average atomic mass of an element like lead, it is important to consider these abundances as each isotope contributes differently based on its frequency. To work with isotopic abundances, we first convert the percentages to decimal form by dividing by 100. This helps in determining the proportion of each isotope in the element's average atomic mass calculation. Key takeaways: - Isotopic abundance helps to understand the composition of an element. - Conversion to decimal allows for easier mathematical calculations. - This data is crucial for calculating the weighted average atomic mass, ensuring accurate representation of an element's mass in nature.
Mass Spectrum
A mass spectrum is a graphical representation used to display the results of a mass spectrometry analysis. This technique identifies and quantifies the isotopes present in a sample by measuring the mass-to-charge ratio of particles.In a mass spectrum, the x-axis represents the atomic masses of the isotopes, while the y-axis shows their relative abundances. Each isotope is illustrated as a peak, where the height relates to its isotopic abundance percentage. For example, in a mass spectrum of lead:
  • There is a peak at 203.9730 u for \(^{204}\mathrm{Pb}\) with a height of 1.4%.
  • Another peak at 205.9744 u for \(^{206}\mathrm{Pb}\) reaching 24.1%.
  • The next peak at 206.9759 u for \(^{207}\mathrm{Pb}\) at 22.1% height.
  • Lastly, the tallest peak at 207.9766 u for \(^{208}\mathrm{Pb}\) achieving 52.4% relative height.
The information provided by the mass spectrum helps in calculating the average atomic mass by visualizing how much each isotope contributes.
Lead Isotopes
Lead, a heavy metal, naturally occurs in multiple isotopic forms, each represented in terms of its nucleon number, or mass number. These isotopes are written as \(^{204}Pb\), \(^{206}Pb\), \(^{207}Pb\), and \(^{208}Pb\). Each isotope of lead has a different abundance and atomic mass, contributing variably to the average atomic mass of lead.By analyzing the isotopic abundances and atomic masses, scientists can determine the overall average atomic mass of lead through weighted calculations. For instance, by multiplying each isotope’s atomic mass by its abundance (in decimal format), and summing these values, we derive an accurate average atomic mass of lead.Important Points About Lead Isotopes:- Lead has four naturally occurring isotopes, with \(^{204}Pb\) being the least abundant and \(^{208}Pb\) the most.- Each isotope slightly differs in atomic mass due to varying numbers of neutrons, impacting the calculation of lead's average atomic mass.- Understanding these isotopes is crucial in fields requiring precise knowledge of elemental composition, such as chemistry and geoscience.

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Most popular questions from this chapter

Complete the table by filling in the formula for the ionic compound formed by each pair of cations and anions, as shown for the first pair. $$ \begin{array}{c|c|c|c|c} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Ca}^{2+} & \mathrm{Fe}^{2+} & \mathrm{Al}^{3+} \\ \hline \mathrm{O}^{2-} & \mathrm{Na}_{2} \mathrm{O} & & & \\ \mathrm{NO}_{3}^{-} & & & & \\ \mathrm{SO}_{4}^{2-} & & & & \\ \mathrm{AsO}_{4}{ }^{3-} & & & & \\ \hline \end{array} $$

(a) What isotope is used as the standard in establishing the atomic mass scale? (b) The atomic weight of boron is reported as 10.81 , yet no atom of boron has the mass of \(10.81 \mathrm{u}\). Explain.

Name the following ionic compounds: \((\mathbf{a}) \mathrm{KCN},(\mathbf{b}) \mathrm{NaBrO}_{2},\) (c) \(\mathrm{Sr}(\mathrm{OH})_{2}\), (d) CoTe, (e) \(\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3},\), (f) \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\), (h) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\), (i) \(\mathrm{KMnO}_{4}\), (g) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{3},\), , (j) \(\mathrm{Ag}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\).

Elements in the same group of the periodic table often form oxyanions with the same general formula. The anions are also named in a similar fashion. Based on these observations, suggest a chemical formula or name, as appropriate, for each of the following ions: \((\mathbf{a}) \mathrm{BrO}_{4}^{-},(\mathbf{b}) \mathrm{SeO}_{3}^{2-},(\mathbf{c})\) arsenate ion, (d) hydrogen tellurate ion.

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