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

The element lead (Pb) consists of four naturally occurring isotopes with atomic masses 203.97302, 205.97444, 206.97587, and 207.97663 amu. The relative abundances of these four isotopes are \(1.4,24.1,22.1\), and \(52.4 \%\), respectively. From these data, calculate the atomic weight of lead.

Short Answer

Expert verified
The atomic weight of lead is approximately \(207.21\) amu.

Step by step solution

01

Convert percentages to decimals

First, we need to convert the relative abundances given in percentages to decimals. We will do this by dividing each percentage value by 100. This will help us in the further calculations. Relative abundance of isotope 1: \(1.4 \% = \frac{1.4}{100} = 0.014\) Relative abundance of isotope 2: \(24.1 \% = \frac{24.1}{100} = 0.241\) Relative abundance of isotope 3: \(22.1 \% = \frac{22.1}{100} = 0.221\) Relative abundance of isotope 4: \(52.4 \% = \frac{52.4}{100} = 0.524\)
02

Multiply each isotope's atomic mass with its relative abundance

Next, we will multiply the atomic mass of each isotope with its relative abundance. Isotope 1: \(203.97302 * 0.014 = 2.855622\) Isotope 2: \(205.97444 * 0.241 = 49.6383904\) Isotope 3: \(206.97587 * 0.221 = 45.7424757\) Isotope 4: \(207.97663 * 0.524 = 108.97833412\)
03

Sum the values obtained in Step 2

Finally, we will add the values obtained in the previous step to find the atomic weight of lead. Atomic weight = \(2.855622 + 49.6383904 + 45.7424757 + 108.97833412 = 207.21482222\)
04

Round the result

The atomic weight of lead can be rounded to a reasonable number of decimal places, such as two decimal places: Atomic weight of lead (rounded) = \(207.21\) So, the atomic weight of lead is approximately \(207.21\) amu.

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.

Isotopes
Isotopes are different forms of the same chemical element. They have the same number of protons but different numbers of neutrons in their nuclei. This difference in neutron number causes isotopes to have differing atomic masses. However, chemically, they behave almost identically due to having the same number of electrons.
Lead, for instance, has four naturally occurring isotopes, which means there are four types of lead atoms with slightly differing masses. These isotopes are expressed based on their atomic mass units (amu). Understanding isotopes is important because it helps explain why elements like lead have an atomic mass that isn't a whole number. The presence of isotopes contributes to the averaged atomic weight of the element seen on the periodic table.
Relative Abundance
Relative abundance refers to how common or rare an isotope is in nature compared to the total abundance of all isotopes of that element. It is often expressed as a percentage. In calculations like atomic weight, we need to convert these percentages into decimal form by dividing by 100.
For example, lead's isotopes vary in abundance. The isotope with an atomic mass of 203.97302 has a relative abundance of 1.4%, whereas the isotope with an atomic mass of 207.97663 has the highest abundance at 52.4%. These values need to be converted into decimals before they can be used to calculate the average atomic weight of the element. Understanding relative abundance is crucial because it impacts the atomic weight more significantly than isotopes that are less abundant.
Atomic Mass
Atomic mass is the mass of a single atom of a chemical element. It primarily comes from the protons and neutrons in the nucleus, with electrons having negligible mass. In practice, the atomic mass of an element is not a single value but a weighted average of all its isotopes' atomic masses, factoring in their relative abundances.
To find an element's atomic mass, we multiply each isotope's atomic mass by its relative abundance (in decimal form) and add all these values up. This gives us the element's atomic weight, reflecting the average mass one would find in a sample of the element. So, while each isotope might have a precise atomic mass, the atomic weight (calculated mass) accounts for how nature balances these isotopes in everyday existence.
Lead
Lead is a heavy metal known by the symbol Pb, from its Latin name plumbum. It is a dense, soft, and malleable element that has been utilized by humans for thousands of years. The understanding of its isotopic composition is essential for various applications, including scientific research and industrial processes.
The atomic weight of lead is an average value calculated from its naturally occurring isotopes' atomic masses and their relative abundances. This calculation gives us a value that reflects the typical mass of lead found in nature, which, as calculated, is approximately 207.21 amu.
Despite its beneficial uses, it's important to handle lead with care due to its toxicity to human health and the environment. This makes understanding its composition and behavior both practically and environmentally crucial.

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

Give the name or chemical formula, as appropriate, for each of the following binary molecular substances: (a) \(\mathrm{SF}_{6},(\mathrm{~b}) \mathrm{IF}_{5},(\mathrm{c}) \mathrm{XeO}_{3}\), (d) dinitrogen tetroxide, \((\mathrm{e}) \mathrm{hy}\) - drogen cyanide, (f) tetraphosphorus hexasulfide.

(a) What is a hydrocarbon? (b) Butane is the alkane with a chain of four carbon atoms. Write a structural formula for this compound, and determine its molecular and empirical formulas.

Hydrogen sulfide is composed of two elements: hydrogen and sulfur. In an experiment, \(6.500 \mathrm{~g}\) of hydrogen sulfide is fully decomposed into its elements. (a) If \(0.384 \mathrm{~g}\) of hydrogen is obtained in this experiment, how many grams of sulfur must be obtained? (b) What fundamental law does this experiment demonstrate? (c) How is this law explained by Dalton's atomic theory?

Fill in the gaps in the following table, assuming each column represents a neutral atom: $$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Symbol } & { }^{52} \mathrm{Cr} & & & & \\ \hline \text { Protons } & & 25 & & & 82 \\ \hline \text { Neutrons } & & 30 & 64 & & \\ \hline \text { Electrons } & & & 48 & 86 & \\ \hline \text { Mass no. } & & & & 222 & 207 \\ \hline \end{array} $$

Determine whether each of the following statements is true or false. If false, correct the statement to make it true: (a) The nucleus has most of the mass and comprises most of the volume of an atom; (b) every atom of a given element has the same number of protons; (c) the number of electrons in an atom equals the number of neutrons in the atom; (d) the protons in the nucleus of the helium atom are held together by a force called the strong nuclear force.

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