Ch.2 - Atoms, Molecules & Ions

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

McMurry 8th Edition

Ch.2 - Atoms, Molecules & Ions

Problem 7

Chapter 2, Problem 7

What is the atomic weight of an element that consists of two naturally occurring isotopes? The first isotope has a mass of 84.911 and an abundance of 72.17% and the second isotope has a mass of 86.909 and an abundance of 27.83%. (LO 2.15)(a) 85.47 (b) 86.35 (c) 85.91 (d) 85.17

Verified step by step guidance

Convert the percentage abundance of each isotope into a decimal by dividing by 100. For the first isotope, 72.17% becomes 0.7217, and for the second isotope, 27.83% becomes 0.2783.

Calculate the contribution of each isotope to the atomic weight by multiplying the mass of each isotope by its decimal abundance. For the first isotope, multiply 84.911 by 0.7217. For the second isotope, multiply 86.909 by 0.2783.

Add the contributions from each isotope to find the average atomic weight of the element.

Compare the calculated average atomic weight with the given options to determine the correct answer.

Ensure that the calculated atomic weight is reasonable and matches one of the provided options.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Isotopes

Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons, resulting in different atomic masses. For example, carbon-12 and carbon-14 are isotopes of carbon. Understanding isotopes is crucial for calculating atomic weights, as the atomic weight of an element is a weighted average of the masses of its isotopes based on their natural abundances.

Atomic Weight Calculation

Atomic weight is calculated by taking the weighted average of the masses of an element's isotopes, factoring in their relative abundances. The formula used is: Atomic Weight = (mass of isotope 1 × abundance of isotope 1) + (mass of isotope 2 × abundance of isotope 2). This calculation allows chemists to determine the average mass of an atom of an element as it occurs in nature.

Percentage Abundance

Percentage abundance refers to the proportion of each isotope of an element present in a natural sample, expressed as a percentage. It is essential for atomic weight calculations, as it indicates how much each isotope contributes to the overall atomic weight. In the given question, the abundances of 72.17% and 27.83% are used to weight the respective isotopes' masses in the calculation.

Recommended video:

Guided course

5:42

Calculating Abundance Example

Related Practice

Textbook Question

A compound containing sulfur and fluorine contains 8.00 g of S and 9.50 g of F. Which combination of S and F masses represents a different compound that obeys the Law of Mul-tiple Proportions? (LO 2.8) (a) 32.0 g of S and 38.0 g of F (b) 4.00 g of S and 4.75 g of F (c) 8.00 g of S and 10.5 g of F (d) 16.0 g of S and 57.0 g of F

710

views

Textbook Question

Which experiment and subsequent observation led to the discovery that atoms contain negatively charged particles, now known as electrons? (LO 2.10–2.12)(a) Oil is sprayed into a chamber and the speed at which the oil droplets fall is measured with and without an applied voltage. X rays in the chamber knock electrons out of air molecules. The electrons stick to the oil pro-ducing an overall negative charge on the drops. Adjusting the voltage changes the speed at which the negatively charged oil droplets fall. (b) When a high voltage is applied across metal electrodes at opposite ends of a sealed glass tube, a cathode ray is produced. The cathode ray is repelled by a negatively charged plate. (c) A radioactive substance emits alpha particles, which are directed at a thin gold foil. Most of the alpha par-ticles pass through the foil, but a few alpha particles are slightly deflected and some even bounce back toward the radioactive source. (d) The mass of different elements in a pure chemical com-pound are measured. Different samples of the compound always contains the same proportion of elements by mass.

451

views

Textbook Question

How many protons, neutrons, and electrons are present in an atom of 2 206Pb? (LO 2.14)(a) 82 protons, 206 neutrons, 82 electrons (b) 124 protons, 82 neutrons, 124 electrons (c) 82 protons, 124 neutrons, 82 electrons (d) 82 protons, 82 neutrons, 124 electrons

Which sample has the greatest mass? (LO 2.16)(a) 5.5 mol of C (b) 2.1 mol of S (c) 4.2 mol of Be (d) 0.52 mol of Ag

How many atoms are present in 1.2 g of gold? (LO 2.16)(a) 2.5 * 1021 (b) 1.4 * 1026 (c) 7.2 * 1023 (d) 3.7 * 1021

1005

views

Textbook Question

Bromine has two naturally occurring isotopes; 79Br (mass of 78.918) and 81Br (mass of 80.916). If the atomic weight of bromine is 79.904, predict the mass spectrum of a sample of bromine atoms. (LO 2.17) (a)

(b)

(c)

(d)

538

views