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10/5/2013 Today: ◦ Mass Percent Composition: Empirical Formulas Molecular Formulas Understand the differences between these two types of formulas Course Survey 2: ◦ Please provide constructive feedback on this course and the instruction Using experimental data to calculate: UPDATE: ◦ Suggested homework problems are include for each reading ◦ Introduction to Chemical Reactions: Balancing Equations Next Meeting ◦ Concept Check: Law of Conservation of Matter General Types of Chemical Reactions Covers today’s material SCORE DISTRIBUTIONS FOR MIDTERM EXAM 1 60 Percentage of Class 50 40 30 20 10 0 < 60 % 60-70 % 70-80 % 80-90 % 90-100 % Exam Score 1 10/5/2013 Graded Concept Check: Gram to Mole Conversions How many moles of hydrogen atoms are present in 39.0 g of C6H6? A. ½ mole of H atoms B. 1.5 mole of H atoms Information that MIGHT be important: C = 12.01 amu H = 1.00794 amu Density of C6H6 = 0.876 g/mL C. 3 moles of H atoms D. 6 moles of H atoms E. 44.5 moles of H atoms How is the formula of a compound determined? We cannot count individual atoms to determine the formula. BUT: We can measure the composition of a substance in MASS PERCENT for each element in a compound. For Example: 100. grams of Natural Gas consists of: 74.8 grams of Carbon 25.2 grams of Hydrogen THEN: The elemental composition by mass can be related to the composition in moles of each element. ◦ Knowing the mole to mole ratio of each element in a compound gives us the Empirical formula for that substance. ◦ The empirical formula provides the LOWEST whole number ratio of the atoms in the substance. 2 10/5/2013 Calculating the Empirical Formula from Mass % 100. grams of Natural Gas consists of: 74.8 grams of Carbon & 25.2 grams of Hydrogen 1. Convert the masses of each element to the number of moles of each element (using the atomic mass): 2. THEN: Divide the number of moles of each element by the smallest value present. This gives a mole to mole ratio (or atom to atom ratio) for each element. THIS IS THE EMPIRICAL FORMULA. Elemental Composition in Mass Percent • The mass percent composition of a substance can be calculated from the chemical formula The MASS PERCENT tells us the portion of each element by MASS: % Mass Element in compound = (atomic mass of element )·(# atoms) x 100% molar mass of the compound 3 10/5/2013 100 g of Benzene 100 g of Acetylene Liquid Solvent Gas Used in welding Carbon Hydrogen Carbon Hydrogen 92.26 g 7.74 g 92.26 g 7.74 g 100 g of Benzene (C6H6) Carbon C1H1 Hydrogen 100 g of Acetylene (C2H2) Carbon C1H1 Hydrogen • Benzene & Acetylene have the same composition of Carbon and Hydrogen BY MASS. They have the same EMPRICAL FORMULA. BUT: How are they different? They have different… MOLECULAR FORMULAS: • The molecular formula indicates the TOTAL number of atoms in each individual molecule 4 10/5/2013 iClicker Question: Converting between units of VOLUME Glucose is known to have a molar mass of approximately 180.16 g/mol. Which of the following is a possible EMPIRICAL FORMULA for glucose? A. C12H24O12 B. C6H12O6 C. CHO D. CH2O E. None of the above Empirical and Molecular Formulas Empirical Formula - the smallest whole number ratio of atoms (or moles of atoms) in a compound. Calculated from Percent Composition by MASS. Molecular Formula - the actual number of atoms in a molecule (some multiple of the empirical formula). Calculated from two pieces of information: 1) the Empirical Formula, AND 2) the Molecular Weight (the molar mass of the molecule) Example: Octane 5 10/5/2013 Sample Problem: Empirical & Molecular Formulas What is the molecular formula of the sugar ribose if it is 40.0% C, 6.7% H, and 53.3% O and it has a molecular mass of 150.1 g/mol? Sample Problem Solution First, determine the empirical formula: Step 1. Assume 100.0 g of the compound. If you have 100 g of the material, you will have the following masses of C, H, & O: 40.0 g C 6.7 g H 53.3 g O 6 10/5/2013 Step 2. Convert grams to moles: Mass Ratio 40.0 g C 1 mol C 12.01 g C Particle Ratio = 3.33 mol C 6.7 g H 1 mol H 1.01 g H = 6.6 mol H 53.3 g O 1 mol O 16.00 g O = 3.33 mol O Step 3. Find the smallest ratio of atoms. C H O 3.33 / 3.33 = 1 6.6 / 3.33 3.33 / 3.33 = 1 = 1.98 2 Therefore, the empirical formula of ribose is CH2O 7 10/5/2013 Next, determine the molecular formula → This requires information on the mass of an individual molecule. Step 4. Calculate the empirical formula mass. C = H = O = 12.01 x 1 = 1.01 x 2 = 16.00 x 1 = 12.01 g/mol 2.02 g/mol 16.00 g/mol 30.03 g/mol Step 5. Calculate the number of empirical formula units that can fit within the molecular formula. 150.1 g/mol = 4.998 5 30.03 g/mol THEN: Multiply the subscripts of the empirical formula by this number to get the molecular formula. C1x5H2x5O1x5 = C5H10O5 8 10/5/2013 Example: Empirical Formulas Boron carbide, containing only boron and carbon, is one of the hardest materials known. An analysis of 5.000 grams of boron carbide was found to contain 1.0868 grams of carbon. Determine the empirical formula of boron carbide from this information. Example: Empirical & Molecular Formulas A 30.5-g sample of acrylic acid, used in the manufacture of acrylic plastics, is found to contain 15.25 g C, 1.71 g H, and 13.54 g O. In a separate mass spectrometer experiment, the acrylic acid is found to have a molar mass of approximately 72 g/mol. What are the empirical and molecular formulas of acrylic acid? 9 10/5/2013 Chemical Reactions “The water a cow drinks turns to milk; the water a snake drinks turns to poison.” —Basho, Poet of Japan (1694) "It’s amazing that the body feeds the brain sugar and amino acids and what comes out is poetry and pirouettes.“ --Neurologist Robert Collins 10 10/5/2013 Interpreting a Chemical Equation: A chemical equation describes a chemical reaction much like a sentence describes some action. • Element Symbols Letters • Formulas Words • Equations Sentences (s) = Solid (l) = Liquid (g) = Gas (aq) = Aqueous (dissolved in water) + = "and" = "reacts to form" or "yields" Balancing Chemical Equations Balancing chemical equations is an application of both the Modern Atomic Theory and the Law of Conservation of Matter. BALANCING EQUATIONS: The same number of each type of element must occur on the left (BEFORE the reaction) and on the right (AFTER the reaction) 11 10/5/2013 Decomposition Reactions Decomposition: Starts with one substance and breaks it into many simpler products. Consider the following chemical reaction: 2 H2O2 → O2(g) + 2 H2O Sentence form: Hydrogen peroxide breaks apart (decomposes) into oxygen gas and water. Combustion Reactions: Hydrocarbons & Carbohydrates • Combustion (aka “burning”): Reacting with OXYGEN (O2) • A hydrocarbon (something containing Hydrogen and Carbon) burns to produce carbon dioxide (CO2) and water (H2O) • Special balancing rule for combustion: Balance Carbon first, Hydrogen second, and Oxygen last. DOUBLE CHECK 12