CCL 1110 1st Edition Lecture 3Outline of Last Lecture I. Chemical EquationsA. Types of Reactions II. Formula MassIII. MolesB. Avogadro’s NumberIV. Stoichiometry and Balanced EquationsOutline of Current Lecture I. Empirical Formulaa. Converting to empirical formulaII. CombustionCurrent LectureI. Empirical Formulaa. Empirical formula- smallest integer ration in a chemical formulab. Molecular- a formula giving the number of atoms of each of the elements present in one molecule of a specific compoundi. Finding empirical formula:1. Divide each # by smallest subscript value2. Multiply each result by same integer, if necessaryThese 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.a. Empirical formulas may not always be unique; different compounds will have different molecular formulas, but may have same empirical formulab. Empirical formula C4H5N2O is empirical formula and molar mass is 194 g. Find moleculari. 4(12.01 g C)+5(1.008 g H)+2(14.01 g N)+16 g O = 97.100 g Empirical molar massii. 194 g/ 97.1 g = 1.998 or about 21. C4H5N2O x 2 = C8H10N4O2II. Combustiona. CxHyOz + nO2 —> xCO2 + y/2H2Oi. All C converted to CO2ii. All H converted to H2O1. % C + % H + % O =
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