SCORE DISTRIBUTIONS FOR MIDTERM EXAM
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1 Percentage of Class 10/5/2013 Today: Mass Percent Composition: Using experimental data to calculate: Empirical Formulas Molecular Formulas Understand the differences between these two types of formulas Introduction to Chemical Reactions: Balancing Equations Law of Conservation of Matter General Types of Chemical Reactions Course Survey 2: Please provide constructive feedback on this course and the instruction UPDATE: Suggested homework problems are include for each reading Next Meeting Concept Check: Covers today s material SCORE DISTRIBUTIONS FOR MIDTERM EXAM < 60 % % % Exam Score % % 1
2 Graded Concept Check: Gram to Mole Conversions How many moles of hydrogen atoms are present in 39.0 g of C 6 H 6? A. ½ mole of H atoms B. 1.5 mole of H atoms C. 3 moles of H atoms Information that MIGHT be important: C = amu H = amu Density of C 6 H 6 = g/ml D. 6 moles of H atoms E 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
3 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 = (atomic mass of element ) (# atoms) x 100% in compound molar mass of the compound 3
4 Liquid Solvent 100 g of Benzene 100 g of Acetylene Gas Used in welding Carbon Hydrogen Carbon Hydrogen g 7.74 g g 7.74 g (C 6 H 6 ) (C 2 H 2 ) 100 g of Benzene 100 g of Acetylene Carbon Hydrogen Carbon Hydrogen C 1 H 1 C 1 H 1 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
5 iclicker Question: Converting between units of VOLUME Glucose is known to have a molar mass of approximately g/mol. Which of the following is a possible EMPIRICAL FORMULA for glucose? A. C 12 H 24 O 12 B. C 6 H 12 O 6 C. CHO D. CH 2 O 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
6 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 g/mol? Sample Problem Solution First, determine the empirical formula: Step 1. Assume 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
7 Step 2. Convert grams to moles: Mass Ratio Particle Ratio 40.0 g C 1 mol C = 3.33 mol C g C 6.7 g H 1 mol H = 6.6 mol H 1.01 g H 53.3 g O 1 mol O = 3.33 mol O g O Step 3. Find the smallest ratio of atoms. C 3.33 / 3.33 = 1 H 6.6 / 3.33 = O 3.33 / 3.33 = 1 Therefore, the empirical formula of ribose is CH 2 O 7
8 Next, determine the molecular formula This requires information on the mass of an individual molecule. Step 4. Calculate the empirical formula mass. C = x 1 = g/mol H = 1.01 x 2 = 2.02 g/mol O = x 1 = g/mol g/mol Step 5. Calculate the number of empirical formula units that can fit within the molecular formula g/mol = g/mol THEN: Multiply the subscripts of the empirical formula by this number to get the molecular formula. C 1x5 H 2x5 O 1x5 = C 5 H 10 O 5 8
9 Example: Empirical Formulas Boron carbide, containing only boron and carbon, is one of the hardest materials known. An analysis of grams of boron carbide was found to contain 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 g C, 1.71 g H, and 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 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
11 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
12 Decomposition Reactions Decomposition: Starts with one substance and breaks it into many simpler products. Consider the following chemical reaction: 2 H 2 O 2 O 2(g) + 2 H 2 O Sentence form: Hydrogen peroxide breaks apart (decomposes) into oxygen gas and water. Combustion Reactions: Hydrocarbons & Carbohydrates Combustion (aka burning ): Reacting with OXYGEN (O 2 ) A hydrocarbon (something containing Hydrogen and Carbon) burns to produce carbon dioxide (CO 2 ) and water (H 2 O) Special balancing rule for combustion: Balance Carbon first, Hydrogen second, and Oxygen last. DOUBLE CHECK 12
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