Observations in Science
1.5 Problem Solving in Chemistry and Engineering
The rules above apply to any numbers that result from most of the measurements that we might make. But in the special case of countable objects, we must also consider one additional rule.
Rule 3: When we count discrete objects, the result has no ambiguity. Such measure- ments use exact numbers, so effectively they have infi nite signifi cant fi gures.
Thus if we need to use information such as four quarts in a gallon or two hydrogen atoms in a water molecule, there is no limitation on significant fi gures. Note that this rule also applies when we work with the various prefi xes in the SI system. There are exactly 100 centimeters in a meter, so the factor of 100 would never limit the number of signifi cant fi gures in a calculation.
1.5 Problem Solving in Chemistry
and
B/A (tells how much B is in one unit of A)
Example Problem 1.4 shows how this type of manipulation fi ts into a problem solving strategy with non-chemistry examples before we look at it in a chemistry context.
E X A M P L E P RO B L E M 1. 4
Shrimp are usually labeled with a “count” that indicates the average number of shrimp per pound. The bigger the shrimp, the smaller the count. Suppose that your supermarket is offering 20-count shrimp for $5.99 per pound. How much should you expect to pay for one dozen shrimp?
Strategy We can determine the average cost per shrimp and then multiply by the number of shrimp needed to fi nd the total price.
Solution
$5.99
————— 20 shrimp × 12 shrimp = $3.59
Analyze Your Answer When we calculate a numerical result, it is generally a good idea to pause and see if the answer makes sense. Such a check can often catch er- rors that might arise from pushing the wrong key on a calculator or even errors in the way we set up the calculation itself. In this case, we found that the cost of 12 shrimp was $3.59, which is a little more than half of the $5.99 price for 20 shrimp. Because 12 is a bit more than half of 20, this seems sensible.
Check Your Understanding A farmer purchases pesticide in 5-gallon drums that cost $23.00 each. If 65 gallons of the pesticide are applied to a fi eld, what was the total cost? How might the situation be more complicated if the amount needed were not a multiple of 5 gallons?
For some problems in chemistry, the units of measurement can be used to determine how to write the appropriate ratio. In the previous example, by considering the units in an algebraic sense, the “shrimp” (with the 20) in the denominator are “canceled” by the same unit (with the 12) in the numerator.
$5.99
—————
20 shrimp × 12 shrimp = $3.59
This type of reasoning is called dimensional analysis, or the factor-label method for calculation. In some cases, the dimensions can provide clues about what ratio to defi ne. We will point out this type of reasoning in problems where it may be used.
Ratios in Chemistry Calculations
We use ratios in a variety of common calculations in chemistry. One that you will undoubtedly also encounter in engineering is the need to convert between units of different sizes. Example Problem 1.5 introduces this type of manipulation.
E X A M P L E P RO B L E M 1. 5
Visible light is commonly described in terms of its wavelength, which is usually given in units of nanometers. In subsequent calculations, this measurement often needs to be expressed in units of meters. If we are considering orange light of wavelength 615 nm, what is its wavelength in meters?
Strategy We need to establish a ratio that relates nanometers and meters. Then we can use that ratio to determine the answer.
Solution
1 m = 1 × 109 nm
We can write this as a ratio. Because we want to convert from nm to m, we’ll need m in the numerator and nm in the denominator.
1 m
————
109 nm Then we just complete the calculation:
615 nm × ———— 1 m
109 nm = 6.15 × 10–7 m
Analyze Your Answer Writing units on all quantities involved helps ensure that we use the correct ratio. Even if we aren’t familiar with the typical wavelengths of light, we can still check that the answer makes sense. Because nanometers are much smaller than meters, the fact that we got a much smaller number for our answer seems correct. Note that conversions between SI units are between exact numbers when considering signifi cant fi gures.
Check Your Understanding Large amounts of electrical power, measured in watts (W), are used to produce aluminum from ore. If a production plant uses 4.3 GW in a certain period of time, how many W are used?
Another common type of problem that uses ratios occurs when two quantities are related to each other either by some chemical or physical property. One such property is mass density. Mass density is defi ned as the mass of a substance per unit volume.
This defi nition is itself a ratio that allows us to convert between mass and volume, as shown in Example Problem 1.6.
E X A M P L E P RO B L E M 1. 6
The mass density of water at 25°C is 0.997 g per mL. A child’s swimming pool holds 346 L of water at this temperature. What mass of water is in the pool?
Strategy We want to use the density as a ratio, but fi rst we must express the volume in appropriate units. We can convert the volume of the pool from liters to milliliters and then use the density as given.
Solution
1 L = 1000 mL, so 1000 mL/1 L provides a conversion from L to mL.
The given density is 0.997 g/1 mL, and this provides the connection between volume and mass.
346 L × ————— 1000 mL 1 L × 0.997 g
———— 1 mL = 3.45 × 105 g
Analyze Your Answer It may be diffi cult to have an intuitive feel for the size of an answer. But 346 L would be nearly 100 gallons of water, which would be fairly heavy.
A pound is equivalent to 454 g, so the answer we found (3.45 × 105 g) is more than 700 pounds, and this is at least plausible. If we had inverted the ratio used to convert from mL to L, for example, we would have gotten an answer of just 345 g, or less than a pound. Clearly, that would be too small for even a very tiny pool, so we would have had a chance to fi nd and correct the error.
Discussion Here we have written our solution as a single two-step process in which we fi rst converted the volume from liters to milliliters and then used the density to fi nd the mass. Some students feel more comfortable breaking these steps down into separate calculations; we could have found the volume explicitly as 346,000 mL and then multiplied that by the density. As long as the operations are carried out correctly, the result will be the same. If you do choose to break calculations down into smaller pieces, though, be careful not to round your answers until you reach the fi nal result.
Check Your Understanding A liquid pesticide has a density of 1.67 g per mL.
What is the mass of liquid in a full 20.0-L container?
Calculations in chemistry use ratios very often. Example Problem 1.6 also shows how they are used in combinations. Learning how to carry out these sequential ma- nipulations will be a key part of solving many of the problems we encounter in chem- istry. One guide that often helps to determine how to construct the appropriate ratio is to note the units used. For example, the equality between 1000 mL and 1 L leads to two ways to express the ratio, 1000 mL/1 L or 1 L/1000 mL. Which ratio did we need? Because we started with liters, we required the former ratio, with 1 L in the denominator. We will often include discussions about the ratios we form in the “strategy” part of the example problems.
Mass density is more than a convenient way to defi ne a ratio for calculations. It is also an important characteristic of materials. Among the reasons that aluminum is a popular structural material is that its relatively low density is often important. When a design calls for a strong but lightweight material, aluminum is a good candidate. We’ll look at this aspect further in the closing insight in Section 1.6 when we look at various materials used in bicycle frames.
Conceptual Chemistry Problems
Although numerical calculations will always be an important component of chemistry, they are only part of the fi eld. Sometimes, to ensure that the concepts involved in chemistry are understood, we’ll also work problems that focus on the particulate representation and other concepts. Often, the strategies employed in this type of question are different from those we have just described.
When we try to visualize a chemical concept using drawings, we will start from the conventions that were introduced for the particulate perspective of chemistry in Section 1.2. In this type of problem, we could be asked, for example, to draw a dia- gram that depicts what happens to the molecules when steam condenses into liquid water. Example Problem 1.7 illustrates this type of thinking.
E X A M P L E P RO B L E M 1. 7
Dry ice is solid carbon dioxide; it is called “dry” in large part because it goes directly from a solid to a gas without becoming a liquid under ordinary conditions. Draw a picture that shows what the carbon dioxide molecules might look like as a solid and as a gas.
Strategy To answer conceptual questions such as this, we must depict the differ- ences between the two phases. Our drawing can be schematic, but it must imply the differences clearly. In this case, we know that a solid will have closely packed mol- ecules, whereas a gas will exhibit signifi cant space between particles.
Solution Our picture shows several things. First, we have included some informa- tion about the chemical composition of the molecules by using two different colors for the spheres that represent the atoms. Second, the solid is indicated at the bottom and is distinguishable because the molecules are shown closely packed together in
Mass density also is important in determining buoyancy because less dense objects fl oat on more dense ones.
Mass density also is important in determining buoyancy because less dense objects fl oat on more dense ones.
an orderly array. Finally, the gas is depicted by only a few molecules, and they are widely spaced.
Discussion Obviously, the solutions to conceptual questions like this are less exact than numerical answers. So your drawing might look somewhat different, yet still be
“correct.” Neither the gas nor the solid needs to look exactly like the picture we have drawn, but the drawing itself should impart the essential concepts we need to under- stand when thinking about the particulate perspective of this process.
Check Your Understanding Draw a picture that shows a molecular scale view of steam condensing into liquid water.
We will continue to use conceptual problems throughout the text to help provide practice in thinking about the behavior of molecules and atoms.
Visualization in Chemistry
One area where conceptual understanding in chemistry is different from that in most engineering pursuits is the way we visualize systems. Chemistry provides multiple simultaneous ways to view problems, including the completely abstract perspective of atoms and molecules that will never be observed directly. Methods to visualize this level of chemistry provide an important tool in the way that chemistry is taught and learned. We can explore this idea by thinking about the refi ning of aluminum.
The most common aluminum ore is bauxite, which consists of aluminum oxides and rock. The fi rst step in producing aluminum metal is separating the aluminum oxides from the remainder of the rock. As shown in Figure 1.10, this actually entails The chemical name for caustic soda is
sodium hydroxide, and its formula is NaOH. Lime is calcium oxide, CaO.
The chemical name for caustic soda is sodium hydroxide, and its formula is NaOH. Lime is calcium oxide, CaO.
Mining
Kiln Mud settler Mud lake
Precipitated aluminum hydroxide
Liquor for recycling
Precipitation
Calcination and shipping
Conveyor
Ball mill Steam
To Red-Mud Lake Clarification
Slurry mixing Digestion
Digestion vessel Filter press
Pregnant liquor
Caustic soda and lime
Bauxite storage
Blow-off tank
Flash tank
Figure 1.10 ❚ Several steps in the processing of bauxite are represented in this diagram.
Based on a drawing by Robert J. Lancashire, University of West Indies
several steps; the fi rst is digestion of the ore. Process engineers design “digesters” in which crushed ore, caustic soda, and lime are mixed at high temperatures to create a slurry. If we look at Figure 1.11, we can visualize this process at the microscopic level.
The molecular level pictures show alumina as atoms of aluminum and oxygen, and the rock is shown as silicon and oxygen (or silica). This is a simplifi ed depiction based on the fact that much of the earth’s crust contains a large percentage of these elements.
(Bauxite ore also contains signifi cant amounts of iron oxide and other minerals, but these behave much like silica in the digestion process.) Using this level of visualization to consider the reactions in the digester, we see that alumina reacts with the caustic soda and lime, but silica does not. In this way, we can look at a large-scale industrial process while thinking in the microscopic perspective.
The next step in aluminum refi ning is smelting. At this stage, a chemical reaction is induced where the aluminates, now dissolved in a material called cryolite (Na3AlF6), lose oxygen atoms to a carbon rod, forming relatively pure aluminum and carbon dioxide. This portion of the process is shown in Figure 1.12. Although this molecular
Note that the conceptual drawings in Figures 1.11 and 1.12 are schematic. They do not attempt to impart molecular detail but rather aim to provide a sense of the particles involved in the chemical processes.
Note that the conceptual drawings in Figures 1.11 and 1.12 are schematic. They do not attempt to impart molecular detail but rather aim to provide a sense of the particles involved in the chemical processes.
Treat with lime or caustic soda
Aluminates, dissolved in liquid phase Alumina
(Al2O3)
Unreacted silica Silica
(SiO2)
Figure 1.11 ❚ Bauxite is treated with caustic soda and lime in a pro- cess called digestion. The alumina in the ore reacts and dissolves. This step separates the aluminum from the rest of the ore, which contains a variety of other minerals.
Carbon rod, immersed in cryolite
Carbon dioxide gas carries away oxygen atoms.
Molten cryolite, containing dissolved aluminum oxides
Aluminum falls to bottom of tank and is drawn off for further purification.
Figure 1.12 ❚ These particulate level illustrations provide a simpli- fi ed view of the atomic-scale process involved in smelting aluminum.
level visualization again shows how the chemical reaction rearranges atoms at the par- ticulate level, it is worth remembering that this process is carried out on a massive scale: the U.S. aluminum industry produces about 2.6 million metric tons of alumi- num per year.
The visualization techniques introduced in this section will help us to develop a particulate level understanding of many concepts we encounter throughout the text.
At this point, let’s take some time to look at one application where the properties of aluminum metal make it a popular material choice for engineering a consumer product—a bicycle.
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