Chapter 3 Mass Relationships in Chemical Reactions Chemistry: McMurry and Fay, 6th Edition Chapter 3: Mass Relationships in Chemical Reactions 2/23/2019 1:34:39 AM Chapter 3 Mass Relationships in Chemical Reactions Copyright © 2011 Pearson Prentice Hall, Inc.

3.8 Mass Percent Composition Percentage of each element in a compound Express by identifying the element presents and give percent by mass of each Can be determined from the formula of the compound the experimental mass analysis of the compound The percentages may not always total to 100% due to rounding

Example - Mass Percent as a Conversion Factor Calculate the mass percent of Na in NaCl Benzaldehyde is 79.2% carbon. What mass of benzaldehyde contains 19.8 g of C?

Empirical formula and Molecular Formulas Empirical formula is the whole number ratio of atoms in a compound Molecular formula is the true number of atoms of each element in a formula of a compound To determine the molecular formula you need to know the empirical formula and the molar mass of the compound E.g Empirical formula: CH Molecular formula: C6H6 Empirical mass: 13.02 g/mol Molecular mass: 78.54 g/mol molecular mass empirical formula mass Multiple (n) = Molecular formula = empirical formula x n where n = 1, 2, 3, 4

Steps in determine the Empirical formula Example: Determine the empirical formula for an unknown sample AxBy Step 1: Obtain the mass of each element (in grams) Step 2: Determine the number moles of each atom present Step 3: Divide the smallest moles by numbers of each atom to obtain the closet integer as possible. Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible.

Example A compound containing nitrogen and oxygen is decomposed in the laboratory and produces 24.5 g nitrogen and 70.0 g oxygen. Calculate the empirical formula of the compound.

Example An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula?

Example Laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula and molecular formula C = 60.00% H = 4.48% O = 35.53%

Determining Empirical Formulas: Elemental Analysis Chapter 3: Formulas, Equations, and Moles Determining Empirical Formulas: Elemental Analysis 2/23/2019 Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO2 and H2O, which are separated and weighed by an automated instrument called a gas chromatograph. hydrocarbon + O2(g) xCO2(g) + yH2O(g) carbon hydrogen If the unknown compound also contains oxygen, then the masses of carbon and hydrogen are determined first. The mass of oxygen is determined by subtracting those masses from that of the unknown compound. Copyright © 2008 Pearson Prentice Hall, Inc.

Combustion Analysis Unknown formula: CxHyOx (Oxygen can be replaced with other nonmetal) gCO2  moles CO2  moles C  gC gH2O  moles H2O  moles H  gH g O = g sample – (g H + g C) gO  moles O Follow steps in determine the empirical formula and molecular formula

Percent Composition and Empirical Formulas Chapter 3: Formulas, Equations, and Moles 2/23/2019 Percent Composition and Empirical Formulas Upon combustion, a compound containing only carbon and hydrogen produces 1.83 g CO2 and 0.901 g H2O. Find the empirical formula of the compound Copyright © 2008 Pearson Prentice Hall, Inc.

Example When a 1.0000 g sample of a vitamin C (an organic molecule contains H, C and O) was combusted, 1.4991 g of CO2 and 0.4092 g of H2O were isolated. Calculate the empirical formula of vitamin C. In another analysis the molecular mass of vitamin C was found to be 176.14 g/mol. What is the molecular formula of vitamin C? Calculate the percent composition