3.10 Determining a Chemical Formula from Experimental Data

Analysis of a Compounds Constituent Elements The mass of elements that make up a compound can be used to determine the compounds chemical formula. When determining a compounds empirical formula from mass data: 1st convert mass to moles. (If % comp. is given, convert to mass.) 2nd divide all subscripts (number of moles) by the smallest one to determine the number of atoms of each element. 3rd we must make the number of atoms of each element a simple whole number.

Let’s Try a Sample Problem A sample of a compound is decomposed in a laboratory and produces 165 g carbon, 27.8 g hydrogen, and 220.2 g oxygen. Calculate the empirical formula of the compound. Step 1: 165 g C 12.01 g C ------------ = --------------- X = 13.7 mol C X 1.00 mol 27.8 g H 1.01 g H ------------ = --------------- X = 27.5 mol H X 1.00 mol 220.2 g O 16.00 g O ------------- = ----------------- X = 13.7 mol O X 1.00 mol O

Sample Problem 1 (Continued) A sample of a compound is decomposed in a laboratory and produces 165 g carbon, 27.8 g hydrogen, and 220.2 g oxygen. Calculate the empirical formula of the compound. 13.7 mol C --------------- = 1 mol C 13.7 27.5 mol H --------------- = 2 mol H 13.7 mol O --------------- = 1 mol O This compounds empirical formula is: CH2O

Calculating Molecular Formulas for Compounds If we know the empirical formula of a compound and its molar mass, we can calculate the molecular formula. Since a molecular formula is a whole number multiple of its empirical formula, we can calculate the simple whole number multiple (n) by: molar mass n = ------------------------------------------ empirical formula molar mass Next, all we have to do is multiple the empirical formula by (n) to determine the molecular formula of the compound. Molecular formula = empirical formula X n

Let’s Try a Sample Problem A compound with the percent composition shown next has a molar mass of 60.10 g/mol. Determine its molecular formula. C = 39.97% , H= 13.41%, and N= 46.62% Step 1: Convert % to mass 39.97% C = 39.97 g 13.41% H = 13.41 g 46.62% N = 46.62 g Step 2: Convert mass to moles (I will use dimensional analysis this time) 1.00 mol C 39.97 g C X --------------- = 3.33 mol C 12.01 g C 1.00 mol H 13.41 g H X --------------- = 13.3 mol H 1.01 g H 1.00 mol N 46.62 g N X --------------- = 3.33 mol N 14.01 g N

Sample Problem 2 (Continued) A compound with the percent composition shown next has a molar mass of 60.10 g/mol. Determine its molecular formula. Step 3: Convert mol to atoms 3.33 mol C -------------- = 1 3.33 mol 13.3 mol H -------------- = 4 3.33 mol N The empirical formula of this compound is: CH4N (Now we must determine its molecular formula)

Sample Problem 2 (Continued) A compound with the percent composition shown next has a molar mass of 60.10 g/mol. Determine its molecular formula. Step 4: Determine molecular formula molar mass 60.10 g/mol n = ----------------------------------------- = ----------------- = 2 empirical formula molar mass 30.06 g/mol Molecular formula = empirical formula X n = (CH4N)2 = C2H8N2

Chapter 3 pg. 133 #’s 86, 87, 91 and 94 (only do the a’s) Read 3.11 and 4.2 pgs. 119-122 and pgs. 140-145