Tutoring Services: Conversions, Formulas, Calculations, Lecture notes of Chemistry

Download Tutoring Services: Conversions, Formulas, Calculations and more Lecture notes Chemistry in PDF only on Docsity! Provided by Tutoring Services 1 Handout Chemical Conversions and Problems Contents: Temperature Conversions: pg. 1 Basic Unit Conversions: pg. 1-3 Chemical Quantity Conversions: pg. 3-5 Density: pg. 5-6 Percent Composition & Empirical & Molecular Formula Determination: pg. 6-8 Stoichiometric Calculations: pg. 8-9 Limiting Reagents: pg. 9-11 Percent Yield: pg. 11 Temperature Conversions To convert between temperature in Kelvins (TK) and degrees Celsius (T°C): TK = T°C + 273.15 or T°C = TK – 273.15 To convert between temperature in degrees Celsius (T°C) and degrees Fahrenheit (T°F): T°C = T°F – 32 or T°F = 1.80(T°C) + 32 1.80 Example 1: What is the boiling point of water, 100°C, in degrees Fahrenheit? T°F = 1.80(T°C) + 32 & T°C = 100 T°F = 1.80(100) + 32 T°F =180 + 32 T°F = 212 Basic Unit Conversions To do a basic conversion from one unit to another: 1) Start with the original number you are given. In example 2, the original number given is 34 minutes. 2) Multiply/divide that original number by a known relationship between that original unit and the unit you want to end up with. In example 2, the known relationship is 60 seconds equals 1 minute. How you determine whether to put the 60 s on the top and the 1 min on the bottom or put the 1 min on the top and the 60 s on the bottom depends on which unit you are starting with and which unit you need to end up with. You want to be able to cancel out the original unit leaving the ending unit on top; put the unit you began with, min, on bottom and the unit you want to end up with, s, on top. In Example 2, you want to be able to cancel out the min, so you put min on bottom, and you want to end up with s, so put s on top. If you were to do a problem converting from seconds to minutes, however, you would put min on top and s on the bottom as in the Example 3. Provided by Tutoring Services 2 Handout Example 2: How many seconds are in 34 minutes? Example 3: How many minutes are in 25 seconds? Often in conversion problems, you will have to use more than one conversion factor (known relationship) as in Example 4. Just take these multi-step conversions one step at a time, canceling out all units except the one you want to end up with. Example 4: How many centimeters are in 10 miles? Known relationships: 1 mi = 5280 ft, 1 ft = 12 in, and 1 in = 2.54 cm (If you need to review how to put numbers in scientific notation, see our handout Exponents, Radicals, and Scientific Notation.) Another type of common conversion problem deals with conversions between some unit and a prefix of that unit such as a conversion from meters to millimeters. The following table provides a list of some widely used prefixes. For example, 1 gigameter (Gm) = 1,000,000,000 meters (m) or 109 m, and 1 microinch (μm) = 0.000001 in or 10-6 in. Prefix Value Name Symbol Numerical Exponential giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 103 hector h 100 102 deka da 10 101 deci d 0.1 10-1 centi c 0.01 10-2 milli m 0.001 10-3 micro μ 0.000001 10-6 nano n 0.000000001 10-9 pico p 0.000000000001 10-12 femto f 0.000000000000001 10-15 Provided by Tutoring Services 5 Handout Density The density of a substance is the ratio of that substance’s mass to its volume. The higher the density of a substance, the less space it takes to fill that space with some amount of the substance. For example, imagine you have 100 pounds of lead and 100 pounds of cotton balls. It would take a space with a much larger volume to accommodate 100 pounds of cotton balls than lead. This is because lead is much denser than cotton balls. To determine the density of a substance, divide the amount of that substance by the volume it takes to accommodate that amount. If you know the density, you can solve for the volume a certain mass will occupy or you can solve for the mass a certain volume will accommodate. Density = Mass or Volume = Mass or Mass = Volume x Density Volume Density Example 12: What is the density of 908 g of cadmium metal that occupies 105 mL? Density = Mass = 908 g = 8.65 g/mL Volume 105 mL The important thing is to first make sure you are dealing with the correct units before solving for density. Mass should be in grams, and volume should either be in mL or cm3 since 1 mL = 1 cm3. This means density should always be in g/mL or g/cm3. You may have to do some basic unit converting first. Example 13: What mass of Ne gas with a density of 0.00090 g/mL occupies a volume of 70 ounces? Known relationships: 32 oz = 1 qt (quart), 1.0567 qt = 1 L, and 1 mL = 10-3 L (or 0.001 L) Sometimes density can be incorporated into other types of problems, for example, chemical quantity conversions. In Example 14, you must find the mass from the density and volume before the number of moles can be found. Example 14: How many moles are there in a sample of titanium metal, density 4.54 g/mL, that can occupy a volume of 25.0 mL? Known relationships: 1 mol = molar mass of Ti = 48 g Provided by Tutoring Services 6 Handout Percent Composition & Empirical & Molecular Formula Determination The molecular formula of a compound is the chemical formula that specifies the number of atoms of each element in one molecule of that compound and specifies the number of moles of each element in one mole of that compound. The molecular formula for sugar glucose is C6H12O6. This means that for every 1 molecule of glucose, there are 6 atoms of C, 12 atoms of H, and 6 atoms of O. This also means that for every 1 mole of glucose, there are 6 moles of C, 12 moles of H, and 6 moles of O. The empirical formula of a compound is the smallest whole number ratio of the molecular formula and gives the relative number of atoms of each element in the compound. The empirical formula for glucose is CH2O. This means that the ratio of C, H, and O in glucose is 1 : 2 : 1. The percent composition of a compound gives the percentage of each element in that compound. The percent composition of K2CO3, for example, is 56.52% K, 8.70% C, and 34.78% O. Knowing percent composition can help you determine empirical formula, and knowing empirical formula can help you determine molecular formula, so we will go over how to determine each of these. To determine the percent composition of a compound: 1) Convert moles of each element into grams. 2) Add the mass in grams of each element to get a total mass. 3) Divide each element’s mass in grams by the total mass in grams and multiply by 100%. 4) Check your answer by making sure that the sum of the percentages equals about 100%. The sum may not equal exactly 100% depending on how much you rounded your answers. Example 15: What is the percent composition of K2CO3? Provided by Tutoring Services 7 Handout You can find the empirical formula of a compound if you know the percent composition of that compound. To do so: 1) Designate the total amount of sample as 100 g so that the percentage of each element will equal the amount in grams of each element. 2) Convert grams of each element into moles. 3) Divide each number of moles by the smallest number of moles. In example 16, the smallest number of moles is Cr, so moles of K, moles of Cr, and moles of O will each be divided by moles of Cr. 4) Put the numbers obtained from dividing by the smallest number into a ratio. 5) Multiply that ratio by the smallest whole number that will give a ratio of close to whole numbers. Example 16: What is the empirical formula of a compound containing 26.57% K, 35.36% Cr, and 38.07% O? The molecular formula of a compound can be determined if you know the empirical formula and the molar mass of the compound. To do so: 1) Convert moles of each element into grams. 2) Add the grams of each element together to get a total empirical formula weight. 3) Divide the molar mass by the empirical formula weight. 4) Multiply the subscripts of each element by the number obtained from step 3. Example 17: What is the molecular formula of ethylene glycol if its empirical formula is CH3O and its molar mass is 62.1 g? Provided by Tutoring Services 10 Handout Example 20: A solution containing 2.00 g of C5H11OH was added to a solution containing 2.00 g of O2. Which reagent is the limiting reagent? How much of the reagent that is not limiting will be left over? Since the amount of C5H11OH that will react with all the O2, 1.67 x 10-2 mol is less than the amount of C5H11OH actually used, 2.27 x 10-2 mol, there will be leftover C5H11OH, and O2 will be the limiting reagent. This could also be determined if you determined how much O2 would be needed to react with all the C5H11OH. That amount would be more than the available amount of O2. Since O2 is the limiting reagent, all of it will be consumed, and there will be leftover C5H11OH. Example 21: How many grams of Ca3(PO4)2 and grams of KCl can be produced by mixing 5.00 g of CaCl2 with 8.00 g of K3PO4? 3CaCl2 + 2 K3PO4 → Ca3(PO4)2 + 6KCl Provided by Tutoring Services 11 Handout CaCl2 is the limiting reagent since there is not enough of it for all the K3PO4 being used to be consumed. 0.0566 mol CaCl2 would be needed for all the K3PO4 to be consumed, but we only have 0.0455 mol CaCl2. Percent Yield The theoretical yield of a reaction is the amount of product that was calculated to be produced. Example 21 shows how to calculate the theoretical yield of a product or products. Percent yield is the percentage of the theoretical yield that was actually obtained when the chemical reaction was performed. To calculate the percent yield, use the following formula: Percent Yield = Actual Yield x 100% Theoretical Yield Example 22: The theoretical yield of H2O in the combustion of CH4 was calculated to be 55 g, but when the reaction was actually performed, the amount of H2O yielded was 51.7 g. What is the percent yield? Percent Yield = Actual Yield x 100% Theoretical Yield Provided by Tutoring Services 12 Handout *If you found this handout helpful, try the Tutoring Center’s other chemistry handouts! Pick one up outside either of our campus’ offices or print them out from http://www.gcc.vccs.edu/tutor/helpful_handouts.asp.* • Periodic Table – Gives a traditional periodic table with atomic masses and atomic numbers with additional trends and values such as electronegativity and ionization energy in one convenient handout. This handout is available in color and in black/white, but the color table is suggested since it is easier to read. • Balancing Chemical Equations – Gives step-by-step instructions on how to balance chemical equations and contains many practice problems to help you perfect your balancing. • Naming Compounds – Gives step-by-step instructions on how to name ionic compounds, binary molecular compounds, and acids with practice problems. • Aqueous Reactions – Explains the following types of aqueous reactions: precipitation, acid/base (neutralization) , and oxidation/reduction. Also gives helpful examples and practice problems. • Significant Figure Rules – Explains the rules of determining significant figures and how to use these rules in addition, subtraction, multiplication, division, and converting. Also has practice problems. • Exponents, Radicals, and Scientific Notation – Under the math handout section, this handout includes how to turn a number into scientific notation and vice versa with practice problems. * Information for this handout was obtained from the following sources: • Brown, LeMay, & Bursten. Chemistry: The Central Science. 9th Ed. Pearson Education, Inc. 2003. • Gilbert, Kriss, & Davies. Chemistry: The Science in Context. 1st Ed. W. W. Norton & Company. 2004. • Rosenberg & Epstein. Schaum’s Outlines: College Chemistry. 8th Ed. McGraw-Hill. 19