Download NOTES: 10.3 – Empirical and Molecular Formulas and more Lecture notes Chemistry in PDF only on Docsity! NOTES: 10.3 – Empirical and Molecular Formulas What Could It Be? Calculating the Empirical Formula Step 1: Convert the masses to moles. Cumoles Cug Cumole Cug 03460.0 5.63 1 199.2 Copper: Oxygen: Omoles Og Omole Og 01731.0 0.16 1 277.0 Calculating the Empirical Formula Step 2: Divide all the moles by the smallest value. This gives the “mole ratio” O Om u 000.1 mol 0173.0 ol 0173.0 C1.999 mol 0.0173 Cumol 0.03460 Calculating the Empirical Formula Step 3: Round off these numbers, they become the subscripts for the elements. Cu2O 1) A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. H Hm 0.3 mol 65.6 ol 2.20 C1.0 mol 6.65 Cmol 6.65 1) A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. EMPIRICAL FORMULA = CH3 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? Nmoles Ng Nmole Ng 85.1 0.14 1 9.25 Omoles Og Omole Og 63.4 0.16 1 1.74 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? EMPIRICAL FORMULA = NO2.5 Is this formula acceptable? NO! (must be WHOLE NUMBERS) So, multiply by 2. 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? EMPIRICAL FORMULA: 2 x (NO2.5) = N2O5 Molecular Formulas • Indicates the actual formula for a compound; the true formula • Will be a multiple of the empirical formula • EXAMPLE: glucose -empirical: CH2O -molecular: C6H12O6 Molecular Formula Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH4N. 60.0 / 30.0 = 2 So, multiply the empirical form. by 2 2 x (CH4N) = C2H8N2 Example: A material is found to be composed of 38.7% Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formula for the compound. If you assume a sample weight of 100 grams, then the percents are really grams. Example: A material is found to be composed of 38.7% Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it is known that the molecular weight is 62.0 Calculate the empirical and molecular formula for the compound. Carbon: mol grams mole grams 23.3 0.12 1 7.38 Oxygen: mol grams mole grams 23.3 0.16 1 6.51 Hydrogen: mol grams mole grams 7.9 0.1 1 7.9 Remember, the empirical formula is not necessarily the molecular formula! MW of the empirical formula = 31 MW of the molecular formula = 62 Empirical Molecular FactorgMultiplyin 2 31 62 CH3O2 x ( ) = C2H6O2 Empirical Formula Molecular Formula Remember, the molecular formula represents the actual formula.
