Download Determining Empirical and Molecular Formulas of Compounds from Mass Analysis and more Study notes Chemistry in PDF only on Docsity! Lecture #7 Molecular Formula Chemistry 142 Younan Xia, Instructor Autumn Quarter, 2005 Information in the Molecular Formula Example: caffeine (C8H10N4O2) Chemical composition of this molecule: Ratio between the elements: Each caffeine molecule is made of: We can also count the number of molecules/atoms in terms of mole: How do we consider this problem in terms of mass, which can be directly and easily weighed using various methods. Calculating Mass Percent and Masses of Elements in a Sample of a Compound Problem: Sucrose (C12H22O11) is a common table sugar. (a) What is the mass% of each element in sucrose? (b) How many grams of carbon are in 24.35 g of sucrose? Determining the mass percent of each element: mass of C per mol sucrose = mass of H / mol = mass of O / mol = total mass per mole = Finding the mass fraction of C in Sucrose & % C : mass of C per mole mass of 1 mole sucrose To find mass% of C = Mass Fraction of C = = Molecular vs. Empirical Formula Molecular Formula: The formula of the compound as it exists. It may be a multiple of the Empirical formula. Empirical Formula: The simplest formula for a compound that agrees with the elemental analysis. The smallest set of whole numbers of atoms. Example: molecular formula empirical formula benzene C6H6 CH glucose C6H12O6 CH2O Some Examples of Compounds with the Same Elemental Ratios Empirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4, C3H6, C4H8 OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6 Determining Empirical Formula From the Masses of Elements - II Constructing the preliminary formula: Converting to integer subscripts: (dividing all by the smallest subscript) Rounding off to whole numbers: CnHm + (n+ ) O2(g) n CO2(g) + H2O(g) m 2 m 2 Elemental Analysis by Combustion CxHyNz + O2(g) CO2(g) + H2O(g) + N2(g) Determining a Chemical Formula from Combustion Analysis - I Problem: Erythrose (M = 120 g/mol) is an important chemical compound used often as a starting material in chemical synthesis, and contains carbon, hydrogen, and oxygen. Combustion analysis of a 700.0 mg sample yielded: 1.027 g CO2 and 0.4194 g H2O. From this data calculate the molecular formula. Plan: We find the masses of hydrogen and carbon using the mass fractions of H in H2O, and C in CO2. The mass of carbon and hydrogen are subtracted from the sample mass to get the mass of oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.
