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General Chemistry Name:__________________ Unit 6 Note Pocket Stoichiometry Date: _________ Per: ____ Empirical and Molecular Formulas and Percentage Composition The formula of a compound indicates: 1. 2. We can use a formula and atomic masses to determine the percentage (by mass) of each atom in a compound. Consider carbon dioxide (C02). Assume you have one mole of carbon dioxide. One mole of carbon dioxide has a molar mass of g. Because we also know the molar mass of carbon is g, it easy to calculate what percentage of the carbon dioxide is carbon. Complete the following calculation: 12.01 g C 44.01 g CO2 x 100 = What percentage of sulfur in S03? What is the percentage of potassium in potassium sulfate? What is the percentage of water in the mineral gypsum (CaS0 4•2H2O)? If you know the percentage composition of each element in a compound, you can determine the formula of the compound by using (more or less) the same process as above but in reverse. The formula you determine in this way is called the empirical formula. The empirical formula is Let’s say we have a sample of an unknown hydrocarbon and we want to figure out the empirical formula. Through experimentation, we determine that the compound is: 17.31% hydrogen 83.67% carbon Steps to determine the empirical formula 1. Assume 100g of the substance. 2. Break down the 100 g by percentage (in this case l7.31g H and 83.67g C). 3. Use the molar mass conversion factor (factor label) of each element to find the moles of each atom. 4. Find the smallest whole number ratio of the elements by dividing each mole value by the smallest of the values found in step three above; remember that you may have to round off to get a whole number because you are dealing with experimental data. Unit 6 combined student notes 1 Once you have the empirical formula, you might not be done. The empirical formula tells you the lowest ratio of each type of atom, but it might not tell you the actual number of atoms in a compound. The molecular formula gives you more information. For example, if the empirical formula of a compound is C 3H8 , its molecular formula may be C3H8 , C6H16 , etc. The molecular formula is Suppose we know that the molar mass of a compound is 180 g/mol. Let's determine the empirical formula for this compound with the following elemental composition: 40.00% carbon 6.72% hydrogen 53.29% oxygen Steps to determine the molecular formula from the empirical formula 1. Assume 100g of the substance. 2. Break down the 100 g by percentage (in this case 40.00 g C, 6.72 g H and 53.29 g O). 3. Use the molar mass conversion factor (factor label) of each element to find the moles of each atom. 4. Find the smallest whole number ratio of the elements by dividing each mole value by the smallest of the values found in step three above; remember that you may have to round off to get a whole number because you are dealing with experimental data. 5. Write the empirical formula of the compound in the example. 6. Find the formula mass of the empirical formula 7. Divide the given molecular mass by the calculated empirical formula mass to find a multiple. 8. Multiply the empirical formula by this ratio to find the molecular formula. Unit 6 combined student notes 2 Measurements in Chemistry - understanding and using the mole Chemical measurements In chemistry we make some measurements by counting and other measurements by weighing, determining volume, etc. Which technique we employ is determined, in large part, by our purpose. It is also necessary, when determining which technique to use, to consider what type of measurement is easiest to make or even feasible. Consider the chemical reaction below: C12(g) + 2KI(s) —> 2KCl(s) + I2 (l) You may at first think about the fact that one molecule of chlorine reacts with two formula units of potassium iodide to produce two formula units of potassium chloride and one molecule of iodine. While this is true, it’s impossible to actually study this reaction at the atomic level in a chemistry laboratory; we cannot count out just one or two atoms or molecules of anything! We have to use another way to measure out the necessary quantities of chlorine and potassium iodide. The mole A mole is a quantity of something just like a dozen (12), gross (144), or ream (500). We use these more familiar quantities to count out large quantities because it’s more convenient. We can convert from the individual item to the quantity units by using simple conversion factors. Try the problems below using factor labeling: 5 dozen eggs = ______________ eggs 1152 pencils = _______________ gross of pencils 5000 pieces of paper = ______________reams of paper A mole relates the number of atoms or molecules of a substance to the mass in grams. The conversion factor used is based on the mass of each element in atomic mass units and how that relates to the mass of the element in grams. • mole = the number of atoms of an element that would be found in, say, 12.01 grams of carbon-12 or 14.01 grams of nitrogen-14 (the molar mass) • mole = ______________________ atoms (this number is called “________________ number”) • mole = __________ L of any gas at STP (standard temperature and pressure: 273 K and 1.0 atm) This means that we can use the atomic masses on the periodic table to convert instantly to grams. The mass of one mole of atoms is called the molar mass. Use the periodic table to determine the following: one atom of Ar _____ amu one atom of Pb _____ amu one mole of Ar atoms ____ g = molar mass of Ar one mole of Pb atoms ______ g molar mass of Pb one atom of Na = __________ amu one atom of B ______ amu one mole of Na atoms ____ g = molar mass of Na one mole of B atoms ______ g = molar mass of B Unit 6 combined student notes 3 Simple conversions using mole equivalencies We use mole equivalencies to convert: • moles to ________ 1 mole of element = __________________________ • moles to the number of ________ 1 mole of element = __________________________ • moles of gas to _______________ 1 mole of gas = _____________________________ For example: 1.50 mol Na = ______ Na 1.50 mol Na x 23.0 g Na = 34.50 g Na = 34.5 g Na (sig figs!) 1 mol Na Do the following conversions using the mole equivalencies above. Use factor labeling and show your work. 2.0 mol Ni = _______ g Ni _________mol C = 24.02 g C 5 mol He = _______ atoms He ___________mol Ca = 4.75 x 1026 atoms Ca 0.115 mol Xe = __________L of Xe _____________ mol Ne = 67.2 L of Ne Atomic masses of elements formula mass of a compound molar mass of a compound We can work easily with the mass of a compound in the same way we do with elements. Determine the mass of a unit of sodium chloride or calcium fluoride by adding up __________________of the component elements in their proper _______________. The sum is called the ________________ one atom of Na = ___________ amu one atom of Ca = ____________ amu one atom of Cl = ____________ amu two atoms of F = ____________ amu formula mass of NaCl = _______amu formula mass of CaF2 = ________ amu Unit 6 combined student notes 4 The molar mass of a substance is determined the same way, but you are finding the mass of _________________________ instead of the mass of just one _______________ (for a covalent compound) or one _____________________ (for an ionic compound). molar mass of NaCl = ________ g molar mass of CaF2 = ________ g molar mass of H2O _________ 9 molar mass of K2SO4 = _______ g molar mass of SO3 _________ g molar mass of Cl2 = __________ g More simple conversions using mole equivalencies 2.0 mol S02 = ___________g SO2 _____________ mol F2 = 1.4 L F2 2.44 x 1023 molecules H2O = ______________ mol H2O 0.98 mol Fe = ______________ atoms Fe ___________g 02 = 4.00 L 02 5.643 g CO2 = ______________ molecules CO2 0.500 L NO = ____________ moles NO Multi-step conversions, using mole equivalencies Conversions involving moles, like other conversions, can involve more than one step. When this is the case, it is convenient to use the following “mole map” to help you plan the conversion. Unit 6 combined student notes 5 Example problem: 2.3 x 1026 molecules H2 = _____ L of H2 2.3 x 1026 molecules H2 x 1 mol H2 x 23 6.02 x 10 molecules H2 22.4 L H2 1 mol H2 = Practice Problems: 1. 34 g BeCl2 = __________ formula units BeCl2 2. 44 g H2O = __________ atoms H2O 3. 3.7 x 1023 molecules I2 = ____________ moles I2 4. 6.90 L CH4 = _____________ g CH4 5. 0.034 g NaF = ____________ formula units NaF 6. 5.66 x 1028 atoms Ar = ____________ L Ar 7. 9.0 x 1022 molecules Br2 = __________________ g Br2 8. 0.05 L CO = ______________ molecules CO 9. How many atoms of uranium are in 5.00 g of uranium? 10. What is the volume of 2.6 g of nitrogen gas? 11. What is the mass of 1.0 x 1024 molecules of sulfur dioxide? 12. What is the mass of Neon in a sign when the gas tubes have a total volume of 166 mL? Unit 6 combined student notes 6 STOICHIOMETRY Stoichiometry : In many ways, stoichiometry is the backbone of the most practical part of chemistry; it helps us relate actual quantities (measured by mass or volume) of reactants to products in a chemical reaction. It’s a lot like using a chocolate chip cookie recipe to produce a certain number of cookies of a given size. You have to put the ingredients together in the proper proportions to make a cookie that tastes good! Consider the following balanced equation: 2Na + Cl2 2NaCl We can look at this equation in two different ways: 1. at the atomic scale: 2. at the real-world, measurable scale: Dimensional analysis can be used to relate the moles of reactants and products in a chemical reaction to one another. It is essentially a matter of determining the ratio of reactants and products by looking at the coefficients in the balanced chemical equation. In the reaction above, the ratio of sodium to chlorine is 2:1. the ratio of sodium to sodium chloride is the ratio of chlorine to sodium chloride is If you’re starting out with 0.650 moles of sodium and you want to know how many moles of chlorine you need to fully react with the sodium, you can use dimensional analysis: 0.650 mol Na x 1 mol Cl2 = 2 mol Na If you want to produce 6.4 moles of sodium chloride, how many moles of chlorine do you need to use? Because we can literally measure out the moles of sodium by weighing it (one mole would have a mass of 22.99g) and we can measure out the moles of chlorine by volume (each mole of chlorine gas would have a volume of 22.4 L), we can literally put the reactants together in their proper proportions. This makes stoichiometry an incredibly powerful but very simple tool - one that you must master in order to be successful in chemistry! When solving a stoichiometry problem, you’ll always be given a quantity of either a product or reactant that you have to relate to a quantity of another product or reactant. To do this, you’ll use the balanced chemical equation as well as the mole equivalencies you’ve been using in mole problems. These equivalencies are: 1 mole = 1 mole = 1 mole = Unit 6 combined student notes 7 The equivalencies you’ll use depend on the specific problem, but the logic is always the same: quantity of given moles of given moles of unknown quantity of unknown Use the “new and improved” mole map shown below to help you map out a problem-solving plan for any stoichiometry problem. liters of substance A grams of substance A liters of substance B moles of substance A moles of substance B particles of substance A grams of substance B particles of substance B Given the following equation: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) What mass of oxygen will react with 36.0 g of methane (CH 4)? 36.0 g CH4 x 1 mol CH4 16.04 g CH4 x 2 mol O2 1 mol CH4 x 32.00 g O2 = 1 mol O2 What mass of water would be produced when oxygen reacts with the 36.0 g of methane? How many liters of carbon dioxide would be produced from the reaction if 44.0 g of oxygen are consumed? What volume of methane is needed to produce 26.8 g of water? How many molecules of methane are needed to produce 2.4 moles of carbon dioxide? Unit 6 combined student notes 8 Limiting Reactant/Reagent Problems: a stoichiometry problem where quantities of reactants are not available in the EXACT molar ratios one reactant is “in excess” and the other is the “limiting reactant” given mass/volume/moles of more than one reactant you find the actual amount of product(s) produced by determining which reactant LIMITS the reaction Steps for Identifying Limiting Reactants: 1. Write the balanced chemical equation. 2. Solve two (or more) stoichiometry problems – one for each reactant given. 3. Compare the results from step #2 above. The reactant that would produce the smaller amount of product is the limiting reactant; the other reactant is in excess. or 1. 2. 3. Write the balanced chemical equation. Find the number of moles of each reactant given. Compare the ratio of moles from step #2 above to the ratio of moles in the balanced chemical equation from step #1. Example Problem: What is the limiting reactant when 1.5 moles of hydrogen react with 0.5 moles of oxygen to produce water? Example Problem: When 2.60 g of hydrogen react with 10.20 g of oxygen, what mass of water will be produced? What is the limiting reactant? Unit 6 combined student notes 9 Percent Yield Problems the amount of product produced from a reaction based on a calculation is the expected yield the amount of product that is really obtained during the reaction (in a lab or industrial setting) is the actual yield reasons for differences between actual and expected yield: 1. some reactants didn’t fully react 2. there were side reactions that differ from the reaction on which the expected yield was based 3. some product was lost during recovery or transfer it is useful to determine what percent of the expected yield was actually obtained percent yield = actual yield expected yield x 100% Example Problem: A reaction occurs between 500.0 g of carbon disulfide and an excess amount of oxygen gas. The mass of sulfur dioxide produced is 768 g. What is the percent yield for this reaction? Step 1. Write the balanced chemical equation. Step 2. Determine the expected yield using stoichiometry Step 3. Compare the actual yield to the expected yield using the percent yield equation. Unit 6 combined student notes 10