﻿ Mass Calculations - Chemical Quantities

# Chemical QuantitiesMass Calculations

Objective

· To learn to relate masses of reactants and products in a chemical reaction.

In the last section we saw how to use the balanced equation for a reaction to calculate the numbers of moles of reactants and products for a particular case. However, moles represent numbers of molecules, and we cannot count molecules directly. In chemistry we count by weighing. Therefore, in this section we will review the procedures for converting between moles and masses and will see how these procedures are applied to chemical calculations.

To develop these procedures we will consider the reaction between powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is

Suppose we have g of aluminum. What mass of should we weigh out to react exactly with this amount of aluminum?

To answer this question, let’s use the problem-solving strategy discussed in Chapter 8.

Where Are We Going?

We want to find the mass of iodine that will react with g of aluminum . We know from the balanced equation that

This can be written as the mole ratio

We can use this ratio to calculate moles of needed from the moles of present. However, this leads us to two questions:

1. How many moles of are present?

2. How do we convert moles of to mass of as required by the problem?

We need to be able to convert from grams to moles and from moles to grams.

How Do We Get There?

The problem states that we have g of aluminum, so we must convert from grams to moles of aluminum. This is something we already know how to do. Using the table of average atomic masses inside the front cover of this book, we find the atomic mass of aluminum to be . This means that mole of aluminum has a mass of g. We can use the equivalence statement

to find the moles of in g.

Now that we have moles of , we can find the moles of required.

We now know the moles of required to react with the moles of ( g). The next step is to convert moles of to grams so that we will know how much to weigh out. We do this by using the molar mass of . The atomic mass of iodine is g (for mole of atoms), so the molar mass of is

Now we convert the moles of to grams of .

We have solved the problem. We need to weigh out g of iodine (contains molecules) to react exactly with the g of aluminum. We will further develop procedures for dealing with masses of reactants and products in Example 9.4. Aluminum (top left) and iodine (top right) react vigorously to form aluminum iodide. The purple cloud results from excess iodine vaporized by the heat of the reaction (bottom).

Reality Check We have determined that g of is required to react with g . Does this answer make sense? We know from the molar masses of and ( g/mol and g/mol) that the mass of mole of is almost ten times as great as that of mole of . We also know that we need a greater number of moles of compared with (by a ratio). So, we should expect to get a mass of that is well over ten times as great as g, and we did.

Interactive Example 9.4. Using Mass—Mole Conversions with Mole Ratios

Propane, , when used as a fuel, reacts with oxygen to produce carbon dioxide and water according to the following unbalanced equation:

What mass of oxygen will be required to react exactly with g of propane?

Solution

Where Are We Going?

We want to determine the mass of required to react exactly with g .

What Do We Know?

· The unbalanced equation for the reaction is

· We know the atomic masses of carbon, hydrogen, and oxygen from the periodic table.

What Do We Need to Know?

· We need to know the balanced equation.

· We need the molar masses of and .

How Do We Get There?

To deal with the amounts of reactants and products, we first need the balanced equation for this reaction:

Our problem, in schematic form, is Using the ideas we developed when we discussed the aluminum—iodine reaction, we will proceed as follows:

1. We are given the number of grams of propane, so we must convert to moles of propane .

2. Then we can use the coefficients in the balanced equation to determine the moles of oxygen required.

3. Finally, we will use the molar mass of to calculate grams of oxygen.

We can sketch this strategy as follows: Thus the first question we must answer is, How many moles of propane are present in g of propane? The molar mass of propane is . The moles of propane present can be calculated as follows:

Next we recognize that each mole of propane reacts with moles of oxygen. This gives us the equivalence statement

from which we construct the mole ratio

that we need to convert from moles of propane molecules to moles of oxygen molecules.

Notice that the mole ratio is set up so that the moles of cancel and the resulting units are moles of .

Because the original question asked for the mass of oxygen needed to react with g of propane, we must convert the moles of to grams, using the molar mass of .

Therefore, g of oxygen is required to burn g of propane. We can summarize this problem by writing out a “conversion string” that shows how the problem was done. This is a convenient way to make sure the final units are correct. The procedure we have followed is summarized below. Reality Check According to the balanced equation, more is required (by moles) than by a factor of . Because the molar mass of is not much greater than that of , we should expect that a greater mass of oxygen is required, and our answer confirms this.

Self-Check: Exercise 9.2

· What mass of carbon dioxide is produced when g of propane reacts with sufficient oxygen?

See Problems 9.23, 9.24, 9.25, and 9.26.

Self-Check: Exercise 9.3

· Calculate the mass of water formed by the complete reaction of g of propane with oxygen.

See Problems 9.23, 9.24, 9.25, and 9.26.

So far in this chapter, we have spent considerable time “thinking through” the procedures for calculating the masses of reactants and products in chemical reactions. We can summarize these procedures in the following steps:

Steps for Calculating the Masses of Reactants and Products in Chemical Reactions

Step 1.

Balance the equation for the reaction.

Step 2.

Convert the masses of reactants or products to moles.

Step 3.

Use the balanced equation to set up the appropriate mole ratio(s).

Step 4.

Use the mole ratio(s) to calculate the number of moles of the desired reactant or product.

Step 5.

Convert from moles back to masses.

The process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction is called stoichiometry (pronounced stoiˊkē-ŏmˊ i-trē). Chemists say that the balanced equation for a chemical reaction describes the stoichiometry of the reaction.

We will now consider a few more examples that involve chemical stoichiometry. Because real-world examples often involve very large or very small masses of chemicals that are most conveniently expressed by using scientific notation, we will deal with such a case in Example 9.5.

Critical Thinking

· Your lab partner has made the observation that we always measure the mass of chemicals in lab, but then use mole ratios to balance equations. What if your lab partner decided to balance equations by using masses as coefficients? Is this possible? Why or why not?

Interactive Example 9.5. Stoichiometric Calculations: Using Scientific Notation

Solid lithium hydroxide has been used in space vehicles to remove exhaled carbon dioxide from the living environment. The products are solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can g of lithium hydroxide absorb?

Solution

Where Are We Going?

We want to determine the mass of carbon dioxide absorbed by g of lithium hydroxide.

What Do We Know?

· The names of the reactants and products.

· We can obtain the atomic masses from the periodic table.

What Do We Need to Know?

· We need to know the balanced equation for the reaction, but we first have to write the formulas for the reactants and products.

· We need the molar masses of lithium hydroxide and carbon dioxide. NASA

Astronaut Sidney M. Gutierrez changes the lithium hydroxide canisters on the Space Shuttle Columbia.

How Do We Get There?

Step 1

Using the description of the reaction, we can write the unbalanced equation

The balanced equation is

Check this for yourself.

Step 2

We convert the given mass of to moles, using the molar mass of , which is .

Step 3

The appropriate mole ratio is

Step 4

Using this mole ratio, we calculate the moles of needed to react with the given mass of .

Step 5

We calculate the mass of by using its molar mass ( g).

Thus g of can absorb g of .

We can summarize this problem as follows: The conversion string is

Reality Check According to the balanced equation, there is a mole ratio of to . There is about a molar mass ratio of . We should expect about the same mass of as , and our answer confirms this ( g compared to g).

Self-Check: Exercise 9.4

· Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with the silica, , in the glass to produce gaseous silicon tetrafluoride and liquid water. The unbalanced equation is

a. Calculate the mass of hydrogen fluoride needed to react with g of silica. Hint: Think carefully about this problem. What is the balanced equation for the reaction? What is given? What do you need to calculate? Sketch a map of the problem before you do the calculations.

b. Calculate the mass of water produced in the reaction described in part a.

See Problems 9.23, 9.24, 9.25, and 9.26.

Interactive Example 9.6. Stoichiometric Calculations: Comparing Two Reactions

Baking soda, , is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach. The balanced equation for the reaction is

Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, , is also used as an antacid. The balanced equation for the reaction is

Which antacid can consume the most stomach acid, g of or g of ?

Solution

Where Are We Going?

We want to compare the neutralizing power of two antacids, and . In other words, how many moles of will react with g of each antacid?

What Do We Know?

· The balanced equations for the reactions.

· We can obtain atomic masses from the periodic table.

What Do We Need to Know?

· We need the molar masses of and .

How Do We Get There?

The antacid that reacts with the larger number of moles of is more effective because it will neutralize more moles of acid. A schematic for this procedure is Notice that in this case we do not need to calculate how many grams of react; we can answer the question with moles of . We will now solve this problem for each antacid. Both of the equations are balanced, so we can proceed with the calculations.

Using the molar mass of , which is , we determine the moles of in g of .

Next we determine the moles of , using the mole ratio .

Thus g of neutralizes mole of . We need to compare this to the number of moles of that g of neutralizes.

Using the molar mass of , which is , we determine the moles of in g of .

To determine the moles of that react with this amount of , we use the mole ratio .

Therefore, g of neutralizes mole of . We have already calculated that g of neutralizes only mole of . Therefore, is a more effective antacid than on a mass basis.

Self-Check: Exercise 9.5

· In Example 9.6 we answered one of the questions we posed in the introduction to this chapter. Now let’s see if you can answer the other question posed there. Determine what mass of carbon monoxide and what mass of hydrogen are required to form kg of methanol by the reaction

See Problem 9.39.

Chemistry in Focus Cars of the Future

There is a great deal of concern about how we are going to sustain our personal transportation system in the face of looming petroleum shortages (and the resultant high costs) and the challenges of global warming. The era of large gasoline-powered cars as the primary means of transportation in the United States seems to be drawing to a close. The fact that discoveries of petroleum are not keeping up with the rapidly increasing global demand for oil has caused skyrocketing prices. In addition, the combustion of gasoline produces carbon dioxide (about lb of per mile for many cars), which has been implicated in global warming.

So what will the car of the future in the United States be like? It seems that we are moving rapidly toward cars that have an electrical component as part of the power train. Hybrid cars, which use a small gasoline motor in conjunction with a powerful battery, have been quite successful. By supplementing the small gasoline engine, which would be inadequate by itself, with power from the battery, the typical hybrid gets to miles per gallon of gasoline. In this type of hybrid car, both the battery and the engine are used to power the wheels of the car as needed.

Another type of system that involves both a gasoline engine and a battery is the so-called “plug-in hybrid.” In this car, the battery is the sole source of power to the car’s wheels. The gasoline engine is only used to charge the battery as needed. One example of this type of car is the Chevrolet Volt, which is designed to run about miles on each battery charge. The car can be plugged into a normal household electric outlet overnight to recharge the battery. For trips longer than miles, the gasoline engine turns on to charge the battery.

Another type of “electrical car” being tested is one powered by a hydrogen—oxygen fuel cell. An example of such a car is the Honda FCX Clarity. The Clarity stores hydrogen at a pressure of lb per square inch. The is sent to a fuel cell, where it reacts with oxygen from the air supplied by an air compressor. About of these cars were delivered in Southern California in early 2017, leased to people who live near one of the three -hour public hydrogen stations. The Clarity has a range of miles per tank of hydrogen. One obvious advantage of a car powered by an fuel cell is that the combustion product is only . However, there is a catch (it seems there is always a catch). Currently, of hydrogen produced is obtained from natural gas , and is a by-product of this process. Intense research is now being conducted to find economically feasible ways to produce from water. Horizon Fuel Cell Technologies Pte. Ltd.

Even model cars are becoming “green.” The H-racer from Horizon Fuel Cell Technologies uses a hydrogen—oxygen fuel cell.

It appears that our cars of the future will have an electrical drive component. Whether it will involve a conventional battery or a fuel cell will depend on technological developments and costs.

See Problem 8.40 Courtesy of American Honda Motor Co., Inc.

Honda FCX Clarity at a hydrogen refueling station.

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