Question

Using the average atomic masses given inside the front cover of this text, calculate the mass in grams of each of the following samples. a. 5.0 moles of potassium b. 0.000305 mole of mercury c. $2.31\times {10}^{-5}$ moles of manganese d. 10.5 moles of phosphorus e. $4.9\times {10}^4$ moles of iron f. 125 moles of lithium g. 0.01205 mole of fluorine

Short answer

Using the average atomic masses and the given number of moles, the mass in grams for each element can be calculated as follows: a. Mass (Potassium) ≈ 196.0 g b. Mass (Mercury) ≈ 6.2 g c. Mass (Manganese) ≈ 1.3 g d. Mass (Phosphorus) ≈ 324.1 g e. Mass (Iron) ≈ 2731.0 g f. Mass (Lithium) ≈ 867.5 g g. Mass (Fluorine) ≈ 0.4 g

Step-by-step solution

Identify the given information

The given information in the exercise is the number of moles for various elements: a. 5.0 moles of potassium b. 0.000305 mole of mercury c. $2.31\times {10}^{-5}$ moles of manganese d. 10.5 moles of phosphorus e. $4.9\times {10}^4$ moles of iron f. 125 moles of lithium g. 0.01205 mole of fluorine

Use the Average Atomic Mass

The average atomic masses for each element can be found inside the front cover of your textbook. Use these numbers to calculate the mass in grams of each element.

Calculate the mass in grams

To calculate the mass in grams for each element, multiply the number of moles with its respective average atomic mass. a. Mass (Potassium) = 5.0 moles * Average Atomic Mass(Potassium) b. Mass (Mercury) = 0.000305 mole * Average Atomic Mass(Mercury) c. Mass (Manganese) = $2.31\times {10}^{-5}$ moles * Average Atomic Mass(Manganese) d. Mass (Phosphorus) = 10.5 moles * Average Atomic Mass(Phosphorus) e. Mass (Iron) = $4.9\times {10}^4$ moles * Average Atomic Mass(Iron) f. Mass (Lithium) = 125 moles * Average Atomic Mass(Lithium) g. Mass (Fluorine) = 0.01205 mole * Average Atomic Mass(Fluorine) After performing the calculations, you will get the mass in grams for each element.

Key concepts

Stoichiometry

Stoichiometry is a core concept in chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction.

One of the most fundamental stoichiometric calculations is the moles to grams conversion. Using the mole concept, the stoichiometric coefficients from a balanced chemical equation, and the average atomic masses of the elements involved, you can determine how much of a substance (in grams) you need or will produce. This process is essential in preparing for laboratory experiments and industrial processes to ensure that the correct amount of each reactant is used.

The exercise provided focuses on the first part of stoichiometry, converting moles to grams, which requires knowledge of the average atomic masses of the elements. Understanding this basic step paves the way for more complex stoichiometric calculations, like those involving multiple steps or limiting reactants.

Average Atomic Mass

The average atomic mass of an element takes into account the relative abundance of each naturally occurring isotope and its mass. It is essentially the weighted average of the isotopic masses, which is why it's not usually a whole number.

In moles to grams calculations, the average atomic mass serves as a conversion factor between the amount of a substance in moles and its mass in grams. One mole of any element always contains the same number of atoms—Avogadro's number, which is approximately $6.022\times {10}^{23}$ atoms. However, the mass of one mole of different elements varies according to their average atomic masses.

For the exercise, you find the average atomic mass of each element listed in the textbook or a periodic table and use it directly to calculate the mass. It's crucial to use the correct value to obtain an accurate mass, reflecting the mixture of isotopes found in nature.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we experience. A mole represents a specific number of particles, similar to how a dozen represents twelve items.

A mole of any substance contains Avogadro's number of entities, which can be atoms, molecules, electrons, or any other particles. For chemists, using moles is similar to counting by dozens or scores as it simplifies chemical calculations. Because individual atoms and molecules are exceedingly small and numerous, it's impractical to work with them in any other unit.

The exercises you're working on require you to use the mole concept to convert the given number of moles into grams, by multiplying the moles by the average atomic mass of each element. Remember, the number of moles reflects the number of atoms or molecules you're dealing with, and by multiplying by the average atomic mass, you're finding out the mass of that many particles.

Chemical Calculations

Chemical calculations are a set of methods chemists use to understand and predict the outcomes of chemical reactions. These include various types of calculations, such as determining the amount of reactants needed or the amount of products formed, based on stoichiometric ratios.

These calculations often involve the mole concept and the use of Avogadro's number. In the case of the exercise provided, which involves no actual reactions, the chemical calculations are straightforward: you are simply converting the number of moles of an element to its mass in grams. However, more complex chemical calculations may involve balancing chemical equations, computing empirical and molecular formulas, and working with concepts such as the limiting reagent and percent yield.

The key to mastering chemical calculations is to understand the relationships between different chemical quantities and to be meticulous with units and significant figures to ensure accurate and meaningful results.