Question

Using the average atomic masses given inside the front cover of this text, calculate how many moles of each element the following masses represent. a. \(1.5 \mathrm{mg}\) of chromium b. \(2.0 \times 10^{-3} \mathrm{~g}\) of strontium c. \(4.84 \times 10^{4} \mathrm{~g}\) of boron d. \(3.6 \times 10^{-6} \mu \mathrm{g}\) of californium e. 1.0 ton \((2000 \mathrm{lb})\) of iron f. \(20.4 \mathrm{~g}\) of barium g. 62.8 g of cobalt

Short answer

a. \(2.885 \times 10^{-5} \mathrm{mol}\) of Cr b. \(2.282 \times 10^{-5} \mathrm{mol}\) of Sr c. \(4476.6 \mathrm{mol}\) of B d. \(1.429 \times 10^{-11} \mathrm{mol}\) of Cf e. \(16247.7 \mathrm{mol}\) of Fe f. \(0.1485 \mathrm{mol}\) of Ba g. \(1.0655 \mathrm{mol}\) of Co

Step-by-step solution

a. Moles of chromium (Cr)

First, we need to convert the mass of chromium given in milligrams (mg) to grams (g): \(1.5 \mathrm{~mg} = 1.5 \times 10^{-3} \mathrm{~g}\) The average atomic mass of Cr is approximately 51.996 g/mol. Using the formula, the number of moles of Cr is: \(n = \frac{1.5 \times 10^{-3} \mathrm{~g}}{51.996 \mathrm{~g/mol}} = 2.885 \times 10^{-5} \mathrm{mol}\)

b. Moles of strontium (Sr)

The average atomic mass of Sr is approximately 87.62 g/mol. Using the formula, the number of moles of Sr is: \(n = \frac{2.0 \times 10^{-3} \mathrm{~g}}{87.62 \mathrm{~g/mol}} = 2.282 \times 10^{-5} \mathrm{mol}\)

c. Moles of boron (B)

The average atomic mass of B is approximately 10.81 g/mol. Using the formula, the number of moles of B is: \(n = \frac{4.84 \times 10^{4} \mathrm{~g}}{10.81 \mathrm{~g/mol}} = 4476.6 \mathrm{mol}\)

d. Moles of californium (Cf)

First, we need to convert the mass of californium given in micrograms (µg) to grams (g): \(3.6 \times 10^{-6} \mu \mathrm{g} = 3.6 \times 10^{-9} \mathrm{~g}\) The average atomic mass of Cf (californium-252 isotope) is approximately 252.08 g/mol. Using the formula, the number of moles of Cf is: \(n = \frac{3.6 \times 10^{-9} \mathrm{~g}}{252.08 \mathrm{~g/mol}} = 1.429 \times 10^{-11} \mathrm{mol}\)

e. Moles of iron (Fe)

First, we need to convert the mass of iron given in tons to grams (g): \(1.0 \mathrm{ton} = 2000 \mathrm{lb} = 907185 \mathrm{~g}\) (1 ton = 2000 lb and 1 lb ≈ 453.592 g) The average atomic mass of Fe is approximately 55.845 g/mol. Using the formula, the number of moles of Fe is: \(n = \frac{907185 \mathrm{~g}}{55.845 \mathrm{~g/mol}} = 16247.7 \mathrm{mol}\)

f. Moles of barium (Ba)

The average atomic mass of Ba is approximately 137.33 g/mol. Using the formula, the number of moles of Ba is: \(n = \frac{20.4 \mathrm{~g}}{137.33 \mathrm{~g/mol}} = 0.1485 \mathrm{mol}\)

g. Moles of cobalt (Co)

The average atomic mass of Co is approximately 58.933 g/mol. Using the formula, the number of moles of Co is: \(n = \frac{62.8 \mathrm{~g}}{58.933 \mathrm{~g/mol}} = 1.0655 \mathrm{mol}\)

Key concepts

Average Atomic Mass

Understanding the average atomic mass is pivotal when embarking on the journey of chemical calculations. It represents the weighted average of the masses of all isotopes of an element, measured in atomic mass units (amu). Each isotope's mass is multiplied by its natural abundance, and the sum provides the average atomic mass that we use in everyday calculations. For instance, the average atomic mass of chromium is 51.996 amu, which factors in the various naturally occurring isotopes of chromium and their respective abundances.

When you're tasked with determining the moles of an element given its mass in grams, the average atomic mass serves as a conversion factor, allowing you to translate grams into moles. This is the crux of many stoichiometric calculations in chemistry, essentially serving as a bridge between the macroscopic world of lab-scale masses and the microscopic scale of atoms and molecules. Thus, a clear grasp of this concept is vital for proficiency in chemistry.

Mole Concept

The mole concept is a cornerstone of chemistry – it's akin to a baker's dozen but for chemists. One mole is defined as the amount of substance that contains exactly 6.022 x 10²³ particles, be it atoms, molecules, or ions. This number is known as Avogadro's number, and it helps chemists deal with the incredibly large quantities of tiny entities that they work with.

The significance of the mole lies in its universality – one mole of carbon atoms is as numerous as one mole of oxygen molecules. When you calculate the moles of an element, such as in the provided exercise, you're effectively counting how many Avogadro's numbers of particles you have, by dividing the given mass by the element's average atomic mass. For instance, converting micrograms of californium to grams and then to moles allows us to know exactly how many atoms we're discussing in terms relatable to macroscopic measurements like weight.

Chemical Calculations

Chemical calculations are integral to the application of chemistry, allowing us to predict and quantify the outcomes of chemical reactions. Through the mole concept and an understanding of the average atomic mass, you can perform a variety of essential calculations. These can range from determining the mass of a reagent needed for a reaction to calculating the yield of a product.

In the context of the provided exercise, chemical calculations involve converting masses of elements to moles. This process usually entails converting the mass from one unit to another (like milligrams to grams), using the average atomic mass as a conversion factor, and applying the concept of the mole to arrive at the solution. By understanding these concepts and their interplay, one can tackle diverse problems in chemistry with confidence and accuracy.