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Math Review: Study Guide #2 AN INTRODUCTION TO THE METRIC SYSTEM, DIMENSIONAL ANALYSIS AND DENSITY I.) Standard Units of Measurement: Our system of measurement finds its roots in the English System which came to America with the founding of the first British Colony. Most of the world along with Britain has since switched to the metric system. English System of Weights and Measures Mass pound Volume fluid ounce, quart and gallon Length Inch, foot and mile Temperature Fahrenheit Energy calorie Starting in 1960, the international Bureau of Weights and Measures, in a series of international meetings proposed a standard set of units that they believed all scientists should use. This Systeme International d^Unites, abbreviated as the SI system, has been largely adopted throughout the scientific community. Table 1: SI Units of Physics and Chemistry. Quantity Unit Length meter (m) Mass kilogram (kg) gram (g) Typical unit of mass (1000 g = 1 kg) (Non-SI unit) Temperature Kelvin (K) Defined by absolute zero and the increment of °C. Celsius (°C) Typical unit of temp. (Non-SI unit) Time second (s) Minutes, hours and days also used. Amount Mole (mol) # of particles present, will be of Substance introduced in chapter 8. Volume m3 Typical unit is the cubic centimeter (cm3) or milliliter (mL) 106 cm3 = 106 mL =1m3 L Common unit of solution volume is the liter (L). mL 1000 milliliters (mL) = 1 L = 1000 cm3 Study Guide 2 Math Review The SI system is closely related to the metric system, long used in Europe. However, the SI system represents a further simplification by expressing all scientific measurements in terms of just seven base units. Despite these good intentions some units are in such common usage that they are retained. As you can see on Table 1 the base unit for time is in seconds. But, time in minutes, hours and days is still accepted. These are commonly used non-SI units. The standard unit of mass is the kilogram (kg) (1 kg = 1000 g), however, most small weights are expressed in the non-SI unit of grams (g) as a matter of convenience. The same can be said for units of volume. The SI unit is meters cubed. The unit of Liters (L) is most often substituted for m3 (1 m3= 1000 L). The SI system permits specification of large units or small units by choosing the appropriate prefix. Example 1: 0.010m (meter) = 1.0 cm (centimeter) = 10. mm (millimeter) 10,000 m = 10 km (kilometer) = 1,000,000 cm SI Prefixes: Prefix Symbol Power of 10 Prefix Symbol Power of 10 tera T 10 12 milli m 10 -3 giga G 10 9 micro µ 10 -6 mega M 10 6 nano n 10 -9 kilo k 10 3 pico p 10 -12 deci d 10 -1 fempto f 10 -15 centi c 10 -2 atto a 10 -18 The standard unit of temperature is Kelvin (K). However, most temperatures are collected in the laboratory using the metric unit of degrees Celsius (°C). Fortunately conversion between degrees Celsius and Kelvin is an easy endeavor. K = °C + 273 and °C = K - 273 Example 2: Convert 212°C to Kelvin. K = °C + 273 = 212 + 273 = 485 K On rare occasions we may want to convert between Celsius and Fahrenheit or vice versa. The equations below can be used for that purpose. 5 9 °C = 9 (#°F -32) and °F = 5 #°C + 32 Units of density, force, velocity, energy and pressure are examples of algebraic combinations of the S.I. base units. We will visit these individually during the semester, but density is a quantity we will calculate or apply through most of this course. So perhaps it will be helpful to define it now and get it out of the way. Density has units of mass per volume. According to the SI format the (2) Study Guide 2 Math Review units would be kg/m3. However, in practical use the unit is typically g/cm3. The quantity of density is a physical property of matter and is often used to describe substances. A given volume of lead for instance weighs more than the same volume of aluminum due to the difference in density of the two metals. Example 3: What is the density of 35.3 mL of a solution that has a mass of 38.2 g? mass 38.2 g density = volume = 35.3 mL = 1.08 g/mL 1 mL = 1 cm3 by definition mass density = volume = 1.08 g/mL = 1.08 g/cm3 Example 4: What is the density of a metal cube that measures 1.1 cm on each side and weighs 12.31 g. mass g g density = volume = length x width x height = cm3 12.31 g g density = = 9.2 1.1 x 1.1 x 1.1 cm3 cm3 The equation for density involves three variables; density, volume and mass. If any two variables are known the one which is unknown can be calculated. Example 5: What is the mass of 45.32 mL of a salt water solution which has a density of 1.12 g/cm3? Solve mass density = for mass = density x volume volume mass 45.32 mL = 45.32 cm3 1.12 g mass = x 45.32 cm3 = 50.8 g cm3 (3) Study Guide 2 Math Review Example 6: The density of a particular gas sample is 0.0893 g/L. What is the volume of 200. g of the gas? Solve mass mass density = for volume = density volume volume 200. g volume = 0.0893 g/L = 2240 L II.) Dimensional Analysis: Dimensional analysis is a large term for a type of mathematics that you have probably done at some level since you were a child. For example, you have four friends, two dollars and want to buy candy bars for everyone at 45 cents each. You intuitively do the math in your head by dividing 50 cents into two dollars and know that you have enough. Dimensional analysis is a more formal way of doing the same thing while providing an internal proof regarding the correctness of your equation. Lets look at this simple example to get a solid idea of how dimensional analysis is applied. We'll find the number of candy bars that can be purchased with 2 dollars by dimensional analysis. The first step is to pull the pertinent information out of the text and organize it as below. a.) One candy bar requires 45 cents to purchase. b.) There are 100 cents in one dollar. c.) You have 2 dollars to purchase the candy. The ultimate goal of our exercise is to convert dollars to candy bars both figuratively through calculation and in reality through purchase. Statements a and b are true relationships. For every 45 cents we can purchase one candy bar. This is called a conversion factor. It expresses a true relationship between two different units (candy bars and cents). At the time of purchase these two quantities are equivalent. one candy bar = 45 cents The equality can be expressed as 45 cents 1 = one candy bar if we divide both sides by one candy bar. Or if we divide both sides by 45 cents. one candy bar 1= 45 cents (4) Study Guide 2 Math Review Notice that both relations equal one. Therefore, to multiply any term by these is to multiply it by one. This type of relationship is called a conversion factor. It is used to convert units from those of the conversion factor's denominator to that of the numerator. 1 candy bar Example 7: 200 cents = 4 candy bars 45cents Conversion factors permit conversion from a given number (200) and units (cents) to another unit (candy bars) based upon a true relationship(s) (the conversion factor(s). The statement in b.) is a definition indicating one dollar = 100 cents. 100 cents 1 dollar It can also be stated as: 1 = 1 dollar = 100 cents . Similar to the above, multiplying any term by either of these is to multiply it by one. Definitions are also true relationships and can be used for unit conversion. 100 cents Example 8a: 2 dollars = 200 cents 1dollar Notice in both examples how the units cancel to permit you to obtain the desired units. The chief utility of this technique is that if a problem is correctly set up, the equation will obtain the correct units. 1dollar 2 Example 8b: 2 dollars = 0.02 dollar /cents 100 cents Oops, the units don't work out and the error is apparent. So to obtain the correct equation for conversion of 2 dollars to candy bars we assemble the equation so the units correctly cancel. The following is a map of the conversions we'll use. dollars --->cents ---> candy bars The equation then is: 100 cents onecandybar 2 dollars = 4 candy bars 1dollar 45 cents (5) Study Guide 2 Math Review Well you are now saying to yourself why bother with all this hassle. From the above you can see that you obtained the correct answer based on the fact that all, the units correctly cancel. This probably is not enough for you. So lets prove the point by doing a problem that is just a little more challenging. Could a stack of 1 dollar bills representing the national debt of 5 trillion dollars reach the moon. Assume each 20 dollar is 0.3 mm thick and the average distance to the moon is 250, 000 miles. 1.) Lets write down what we have. a.) One bill = 1 dollar. b.) One dollar bill is 0.3 mm thick. c.) One stack represents 5 x 1012 dollars (5 trillion dollars). d.) The distance to the moon is approximately 250,000 miles. 2.) What we want to determine: Is the stack of dollar bills higher than 250,000 miles? 3.) Since the thickness of an individual bill is in mm we will need to convert at some point from millimeters (1 mm = 0.001 meters) to miles. Some additional definitions of length will be required. a.) 1 meter (m) = 1000 millimeters (mm) b.) 1000 meters (m) = 1 kilometer (km) c.) 1 kilometer (km) = 0.621 miles (mi) 4.) One possible solution map for this problem is: 5.0 x 1012 dollars ---> bills ---> mm ---> m ---> km ---> mi 1bill 0.3mm 1m 1km 0.621 mi 5.0 x 1012 dollars = 931,500 mi 1dollars 1bill 1000 mm 1000 m km Inverting any of these conversion factors would lead to an error in the final calculation. In the absence of some expert to check your work the error might go unnoticed. Using dimensional analysis permits calculation at an advanced level by mere amateurs. Like a cross word puzzle you can continue to peck away at the problem until the correct result is obtained and confirmed. You can prove your result and easily communicate how and why you obtained it. This is the style of mathematics that will be applied to all calculations during this course. Resist the temptation to use alternative methods. The wrong answers on the multiple choice exams are often the answers obtained by inverting conversion factors. (6) Study Guide 2 Math Review III.) Applied Unit Conversions: Mathematical operations in chemistry involve unit conversions. For instance, a volume of water can be expressed in fluid ounces, cups, quarts, gallons, milliliters or liters or in units of weight such as ounces, pounds, tons, grams, kilograms to name but a few. Inevitably this leads to the necessity of converting from one unit to another. Dimensional analysis provides the mathematical where-with-all for the unit conversion and a proof that we set up the mathematics correctly. Before using this technique it is important to remember that units are treated mathematically in the same way we treat numbers. Example 9: When like units appear in the numerator and denominator of a fraction they cancel. 19 m 19 1m 19 3.5 kg x m = 3.5 kg x 1m = 3.5 kg x 1 = 5.4 kg -1 Units may be treated like numbers. They can be raised to a power, multiplied or divided. Fundamental to unit conversions is the use of conversion factors. As mentioned earlier these are obtained from any true relationship that can be expressed as an equality. Example 10: 1 cup contains 8 fluid ounces: which implies 1 cup 1 cup = 8 fl. ounces or = 1. 8 fl.ounces We could just as easily have divided both sides by 1 cup and obtained 8 fl. ounces 1 cup = 1. As we observed before, multiplying a number by a conversion factor is essentially the same as multiplying by 1. To convert from cups to fluid ounces we need a conversion factor that contains both the units we want to eliminate and the desired units. Example 11: Convert 33.65 cups to fluid ounces. By correctly setting up the conversion 8 fl. oz. factor the units of cups cancel, leaving 33.65 cups 269.2 fl. oz. the desired units of fluid ounces. 1 cup If the conversion factor had been set up incorrectly the result would have had units of cups2/ounces and our error would be apparent. (see below) (7) Study Guide 2 Math Review Example 12: Convert 33.65 cups to fluid ounces. The error is readily apparent 1 cup 2 when the conversion factor is 33.65 cups 4.206 cups inverted. 8 fl. oz. fl. oz. On many occasions a single conversion factor for the desired operation is not available, so we must choose a series of conversions that will take us ultimately to our goal. A solution map is helpful in that it allows us to try out ideas that might lead to a solution without writing out the entire solution and checking units. Example 13: The distance from Fresno to San Francisco is 180 miles what is that distance in meters? A proposed solution map: 180 mile --->ft --->in --->m (desired) The solution map always starts with the unit of measurement that is given in the problem and ends in the units desired. The individual steps are conversion factors or definitions. Units of ft result when a conversion factor is selected that has the undesired unit(miles) in the denominator and the desired unit of feet (ft) in the numerator. The concept is continued for conversion to inches (in) and then meters (m), which was the desired unit. The result based upon the solution map is: 5280 ft 12 in 1 m 5 180 mi = 2.9 x 10 m 1 mi 1 ft 39.37 in Conversions involving area or volume have units such as m2, m3 or cm3, that require some special handling. SINCE THE UNITS HAVE A POWER THE CONVERSION FACTOR MUST ALSO BE RAISED TO THE APPROPRIATE POWER FOR THE CONVERSION TO WORK OUT. (8) Study Guide 2 Math Review Example 14: How many cubic meters are present in 123,500 cm3? The definition necessary for relating meters to centimeters is 1 m = 100 cm. However the conversion won't work unless the conversion factor is also cubed as below. Solution map: cm3 ---> m3 1 m 1 m 1 m 123,500 cm3 100 cm100 cm100 cm = 3 1m 123,500 cm3 100 cm = 1m 3 1 m3 123,500 cm3 3 3 = 123500 cm3 6 3 = 0.1235 m3 100 cm 10 cm ______________________________________________________________________________ Math Review: Study Guide #2 Problem set 1: Do the temperature conversions indicated. a.) Convert 33.5 °C to Kelvin. b.) Convert 45.8 K to degrees Celsius. c.) Convert 75°F to degrees Celsius. d.) Convert 295 K to degrees Fahrenheit. Problem set 2: Do the density calculations as follows. a.) What is the density of an unknown metal if 45.36 g of metal occupies a volume of 3.62 cm3? b.) A 55.3 mL sample of a salt solution has a mass of 56.213 g. What is the density of the solution? (9) Study Guide 2 Math Review c.) What is the mass of 35.2 mL of a solution that has a density of 1.12 g/cm3? d.) A rock sample has a density of 10.3 g/cm3 and weighs 243.12 g. What is the volume of the rock? e.) A particular gas sample has a density of 1.12 g/cm3. What is the mass in kg of 4000 L of gas? Problem set 3: Use your conversion factor handout to make the necessary unit conversions and report the answer with correct sig. figs and units. Use the form presented in the study guide and show all work. a.) 746 miles --> ft b.) 15.3 yr. --> day c.) 8.31 x 105 in --> ft d.) 9.6 x 104 yd --> m e.) 4.671 x 106 L --> Gal f.) 2.37 x 104 g --> lb g.) 895 miles --> km (10) Study Guide 2 Math Review h.) 15.3 yr --> hr i.) 0.4837 km --> mm k.) 56 cm2 -->m2 l.) 94.72 km to cm m.) 7.1 x 103 L --> fl oz n.) 5.76 kg --> oz o.) 8.7 Gal --> mL p.) 184.3 in2 --> m2 q.) 7.71 x 104 µm --> km r.) 2.6 tons -->kg s.) 5.22 qt --> L (11)

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