Stoichiometry mass-mole conversions

This is a critical chapter in your study of chemistry. Our goal is to help you master the mole concept. You will learn about balancing equations and the mole/mass relationships (stoichiometry) inherent in these balanced equations. You will learn, given amounts of reactants, how to determine which one limits the amount of product formed. You will also learn how to determine the empirical and molecular formulas of compounds. All of these will depend on the mole concept. Make sure that you can use your calculator correctly. If you are unsure about setting up problems, refer back to Chapter 1 of this book and go through Section 1-4, on using the Unit Conversion Method. Review how to find atomic masses on the periodic table. Practice, Practice, Practice. 

In a stoichiometry problem, (a) if the mass of a reactant is given, what conversions (if any) should be made (b) If a number of molecules is given, what conversions (if any) should be made (c) If a number of moles is given, what conversions (if any) should be made ... 

This balanced equation can be read as 4 iron atoms react with 3 oxygen molecules to produce 2 iron(III) oxide units. However, the coefficients can stand not only for the number of atoms or molecules (microscopic level) but they can also stand for the number of moles of reactants or products. So the equation can also be read as 4 mol of iron react with 3 mol of oxygen to produce 2 mol ofiron(III) oxide. In addition, if we know the number of moles, the number of grams or molecules may be calculated. This is stoichiometry, the calculation of the amount (mass, moles, particles) of one substance in the chemical equation from another. The coefficients in the balanced chemical equation define the mathematical relationship between the reactants and products and allow the conversion from moles of one chemical species in the reaction to another. 

We need to calculate the amount of methyl tert-bu tyl ether that could theoretically be produced from 26.3 g of isobutylene and compare that theoretical amount to the actual amount (32.8 g). As always, stoichiometry problems begin by calculating the molar masses of reactants and products. Coefficients of the balanced equation then tell mole ratios, and molar masses act as conversion factors between moles and masses. 

For work in the laboratory, it s necessary to weigh reactants rather than just know numbers of moles. Thus, it s necessary to convert between numbers of moles and numbers of grams by using molar mass as the conversion factor. The molar mass of any substance is the amount in grams numerically equal to the substance s molecular or formula mass. Carrying out chemical calculations using these relationships is called stoichiometry. 

Figure 10.2 Mass and Mole Conversions for Stoichiometry Problems...

The conversion factor for converting between mass and amount in moles is the molar mass of the substance. The molar mass is the sum of atomic masses from the periodic table for the atoms in a substance. Skills Toolkit 3 shows how to use the molar mass of each substance involved in a stoichiometry problem. Notice that the problem is a three-step process. The mass in grams of the given substance is converted into moles. Next, the mole ratio is used to convert into moles of the desired substance. Finally, this amount in moles is converted into grams. 

Recall that stoichiometry is the study of quantitative relationships between the amounts of reactants used and the amounts of products formed by a chemical reaction. What are the tools needed for stoichiometric calculations All stoichiometric calculations begin with a balanced chemical equation, which indicates relative amounts of the substances that react and the products that form. Mole ratios based on the balanced chemical equation are also needed. You learned to write mole ratios in Section 12.1. Finally, mass-to-mole conversions similar to those you learned about in Chapter 11 are required. 

Most chemical reactions that occur on the earth s surface, whether in living organisms or among inorganic substances, take place in aqueous solution. Chemical reactions carried out between substances in solution obey the requirements of stoichiometry discussed in Chapter 2, in the sense that the conservation laws embodied in balanced chemical equations are always in force. But here we must apply these requirements in a slightly different way. Instead of a conversion between masses and number of moles, using the molar mass as a conversion factor, the conversion is now between solution volumes and number of moles, with the concentration as the conversion factor. 

Now you are ready to try your first complete stoichiometry problems, where you quantitatively analyze chemical reactions. The mass-mass problem is where you either know the mass of the product that you want to produce and calculate the mass of the reactant(s) you start with, or you know the mass of the reactant(s) you start with and calculate the mass of the product you will end up with. As with mole conversion problems, there are a variety of these types of problems, with a varying range of difficulty. We will start off with some of the easier types and work our way up to harder problems. 

Between this chapter and Chapter 10, we have now seen three different ways to convert between a measurable property and moles in equation stoichiometry problems. The different paths are summarized in Figure 13.10 in the sample study sheet on the next two pages. For pure liquids and solids, we can convert between mass and moles, using the molar mass as a conversion factor. For gases, we can convert between volume of gas and moles using the methods described above. For solutions, molarity provides a conversion factor that enables us to convert between moles of solute and volume of solution. Equation stoichiometry problems can contain any combination of two of these conversions, such as we see in Example 13.8. 

Conversions of moles of one substance to moles of any other in the balanced chemical equation is straightforward just remember that it is moles not mass that is related to the coefficients in the equation, ffowever, stoichiometry problems often give students more trouble than they should because the problems are often asked in terms of masses or other quantities that can be related to moles of reactant or product. These problems are multistep problems, but should not present too much difficulty because each individual step is straightforward. 

This is "simple" because the stoichiometry is one mole of reactant goes to one mole of product, and because the conversion of A to B follows first-order kinetics, as does the conversion of B back to A. Thus, when we assemble the two-component mass balance equations in a constant volume batch reactor, we find ... 

Chemical stoichiometry is the area of study that considers the quantities of materials in chemical formulas and equations. Quite simply, it is chemical arithmetic. The word itself is derived from stoicheion, the Greek word for element and metron, the Greek word for measure. When based on chemical formulas, stoichiometry is used to convert between mass and moles, to calculate the number of atoms, to calculate percent composition, and to interpret the mole ratios expressed in a chemical formula. Most topics in chemical arithmetic depend on the interpretation of balanced chemical equations. Mass/mole conversions, calculation of limiting reagent and percent yield, and various relationships among reactants and products are commonly included in this topic area. 

Molar flow rates are mostly useful because using moles instead of mass allows you to write material balances in terms of reaction conversion and stoichiometry. In other words, there are a lot less unknowns when you use a mole balance, since the stoichiometry allows you to consolidate all of the changes in the reactant and product concentrations in terms of one variable. This will be discussed more in a later chapter. 

Recall that the coefficients in a balanced equation give the relative number of moles of reactants and products, aexs (Section 3.6) To use this information, we must convert the masses of substances involved in a reaction into moles. When dealing with pure substances, as we did in Chapter 3, we use molar mass to convert between grams and moles of the substances. This conversion is not valid when working with a solution because both solute and solvent contribute to its mass. However, if we know the solute concentration, we can use molarity and volume to determine the number of moles (moles solute = M X F). T Figure 4.17 summarizes this approach to using stoichiometry for the reaction between a pure substance and a solution. 

The heart of any stoichiometry problem is the balanced chemical equation that provides the mole ratio we need. We must convert between masses and the number of moles in order to use this ratio, first for the reactant, 5, and at the end of the problem for the product, P4S3. Molar mass is used to provide the needed conversion factors. The phrase, excess of phosphorus, tells us that we have more than enough P4 to consume 153 g of 8 completely. 

Corollary In warm-up examples 4 and 5 (chemical reactors), we had information about stoichiometry and conversion, and the proposed procedure was to construct a table to take into account the moles entering the reactor, moles reacting, and the moles leaving the reactors (reagents and products). This is a convenient procedure and facilitates the material balance in the reactor. On the other hand, in the bioreactor problems, we had information about the disappearance of the substrate (kinetics), and in that case it was easier just to formulate the mass balance like (8.6). 

Before you can solve any stoichiometry problem, you must have the reaction equation and the conversion factors between moles and quantities of Given and Wanted substances. For convenience, we will use the expression starting steps to describe these items. Thus, complete the starting steps means to write and balance the equation, if it is not given, and determine the conversion factors. Molar mass is the conversion factor in this section others will appear later. 

First, we complete the starting steps for all stoichiometry problems. We write the reaction equation and calculate the molar mass or other conversion relationships between moles and the measured quantities of the substances in the problem. In this example the equation is given. We must calculate the molar masses of the given and wanted substances. They are C2Hg 2(12.01 g/mol) + 6(1.008 g/mol) = 30.07 g/mol O2. 2(16.00 g/mol) = 32.00 g/mol. The Plan for the problem is... 

The given quantity is moles of heptane. In other words, Step 1 of the stoichiometry path is completed. The wanted quantity of Oj is to be expressed in grams. Therefore, you need its molar mass for the mol g conversion. 

Recall that the molar mass is the mass, in grams, of one mole of a substance. The molar mass is the conversion factor that relates the mass of a substance to the amount in moles of that substance. To solve reaction stoichiometry problems, you will need to determine molar masses using the periodic table. 

We can use the heat of vaporization of a liquid to calculate the amount of energy required to vaporize a given mass of the liquid (or the amount of heat given off by the condensation of a given mass of liquid), using concepts similar to those covered in Section 6.6 (stoichiometry of AH). The heat of vaporization is like a conversion factor between number of moles of a liquid and the amount of heat required to vaporize it (or the amount of heat emitted when it condenses), as demonstrated in Example 11.3. 

Such problems as this one involve many steps or conversions. Try to break the problem into simpler ones involving fewer steps or conversions. It may also help to remember that solving a stoichiometry problem involves three steps (1) converting to moles, (2) converting between moles, and (3) converting from moles. Use molarities and molar masses to carry out volume-mole conversions and gram-mole conversions, respectively, and stoichiometric factors to carry out mole-mole conversions. The stoichiometric factors are constructed from a balanced chemical equation. 

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