CHEM 1061 1st Edition Lecture 5Outline of Last Lecture I. Naming a. Covalent Compoundsb. Straight Chain AlkanesII. Massesa. Mass of Ionic Compoundsb. Mass of Covalent CompoundsIII. Stoichiometrya. Empirical Formulab. Mass % to Chemical Formula Outline of Current Lecture I. Continuation of Empirical FormulaII. Molar Mass of a CompoundIII. Writing and Balancing Chemical EquationsIV. Limiting ReactantsCurrent LectureI. Empirical and Molecular Formulaa. Percent by mass of elements in a formulaII. Mass Percentage to Chemical Formulaa. Compound is 40% C, 6.73% H, and 53.3% O By mass.b. Determine Empirical Formula (Smallest whole number ratio between the elements)c. Steps:i. Choose Sample Mass (Mass= 100g therefore C=40g H=6.73g and O=53.3g)These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.ii. Convert grams into moles using conversion factors of molar mass.1. C: 40.0gC x 1mol C/12g C = 3.33 mol CH: 6.73gH x 1mol H/1.008g H = 6.67 mol HO: 53.3gO x 1mol O/16g O = 3.33 mol O2. Empirical formula- ratio numbers become subscripts in formulaC1H2O1 = CH2Od. SAMPLE PROBLEM: If molar mass of the formula is 60g/mol, what is the molecular (actual) formula?Empirical mass is approximately equal to 30g/mol, which means it is NOT the actual formula. Multiply so empirical mass equals the actual mass.Molar Mass of a Compound/ Empirical Mass60 g/mol / 30 g/mol = 2Actual formula is C2H4O2III. Writing and Balancing Chemical Equationsa. Think about molesb. Must balance; make sure there is the same number of atom on each side.c. This is a way of keeping track of the reactants and products.Nitrogen Gas and Hydrogen Gas combined to yield Ammonia (NH3) Gas:N2(g) + 3H2(g) 2NH3(g)I. States of Mattera. Gas (g)b. Solid (s)c. Liquid (l)d. Aqueous (aq)II. Products: Things produced from a reactionIII. Reactants: Things combined to form a reactionBALANCE NUMBERS WITH COEFFICIENTS, DO NOT CHANGE SUBSCRIPTS.SAMPLE PROBLEM: Combustion of Methane (CH4) gas in the presence of Oxygen gas (02) yields carbon dioxide gas and water vapor.CH4 (g) + 2 02 (g) CO2 (g) + 2 H2O (l)SAMPLE PROBLEM: If 20g of CH4 reacts, how many molecules of H2O are produced?20g CH4 x (1 mol CH4/ 16g CH4) x (2 mol H2O/ 1 mol CH4) x (6.022 x 10^23 molecules of H20/1 molecule H2O) = 1.51 x 10^24 molecules of H2OI. Limiting Reactanta. Given quantities of two or more reactantsb. Limits amount of product formedc. Limiting reactant is that of which runs out first.SAMPLE PROBLEM: 2.5 mol of N2 reacts with 3.0 mol of H3. How many moles of NH3 are formed?N2(g) + 3H2(g) 3NH3(g)N2: 2.5 mol N2 x (2 mol NH3/ 1 mol N2) = 5 mol NH3H2: 3.0 mol H2 x (2 mol NH3/ 3 mol H2) = 2 mol NH3H2 is the limiting reactant and N2 is the reactant in
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