Empirical, molecular, percent composition

Percent Composition Percent by mass of each element in a compound %mass of element = mass element/mass compound x 100%

Percent from mass data When a 13.60 g sample of a compound containing only magnesium and oxygen is decomposed, 5.40 g of oxygen is obtained. What is the percent composition? Total sample: 13.60 + 5.40 = 19.0 g Percent Mg: 13.60/19.0 =71.6% Percent O: 5.40/19.0=28.4% Double check: 71.6 + 28.4 = 100! Got it!

Percent from formula Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Mass carbon: 3(12.01) = 36.03 g/mol Mass hydrogen: 8(1.01) = 8.08 g/mol Total molar mass: 36.03 + 8.08 =44.11 g/mol Percent carbon: 36.03/44.11 = 81.7% Percent hydrogen: 8.08/44.11 = 18.3% Double check: 81.7 + 18.8 = 100!! Got it!

Empirical Formula Lowest whole number ratio of atoms in a compound May or may not be the same as the molecular formula Can be experimentally determined Example: HO is the lowest ratio of H to O

Determining empirical formula A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Percent to grams: 25.9% is 25.9 g, 74.1% is 74.1 g Grams to moles Nitrogen: 25.9/14.01 = 1.85 mol Oxygen: 74.1/16 = 4.63 mol Divide by least, then multiply to whole numbers Nitrogen: 1.85/1.85 = 1 x 2 = 2 Oxygen: 4.63/1.85 = 2.5 x 2 = 5 Use as subscripts: N2O5

Molecular Formula Same as empirical formula or a simple whole-number multiple of the empirical formula Use the molar mass to determine Example: HO can be paired as H2O2. Twice the empirical formula

Finding the molecular formula Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N. Calculate molar mass of empirical formula: 12.01 + 4.04 + 14.01 = 30.06 g/mol Determine the multiplication factor: 60.0/30 = 2 Molecular formula is 2 times empirical CH4N x 2 = C2H8N2

Comparison Formula Classification Molar mass CH Empirical 13 C2H2 (ethyne) Molecular 26 (2 x 13) C6H6 (benzene) 78 (6 x 13) CH2O (methanal) Empirical and molecular 30 C2H4O2 (ethanoic acid) 60 (2 x 30) C6H12O6 (glucose) 180 (6 x 30)

More Let Mr. Bozeman take it away! https://www.youtube.com/watch?v=QcC4OsSxWYU

Now you try: 1. A compound is formed when 9.03 g Mg combines completely with 3.48 g N. What is the percent composition of this compound? 2. Calculate the percent composition of these compounds: A. ethane (C2H6) B. sodium hydrogen sulfate (NaHSO4). 3. Calculate the empirical formula of each compound: A. 94.1% O, 5.9% H B. 67.6% Hg, 10.8% S, 21.6% O 4. Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62 g/mol and the empirical formula is CH3O.