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Calculating Empirical Formula Using percentage or mass to find the Empirical Formula Empirical Formula • Empirical Formula is the lowest whole number ratio of elements in a compound. • Molecular Formula the actual ratio of elements in a compound • The two can be the same: H2O. • Or Different: – CH2 empirical formula – C2H4 molecular formula – C3H6 molecular formula Calculating Empirical Formula • Just find the lowest whole number ratio – C6H12O6 becomes CH2O – CH4N is the lowest ratio • It is not just the ratio of atoms, it is also the ratio of moles of atoms – In 1 molecule of CO2 there is 1 atom of C and 2 atoms of O – In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen Combustion Analysis • Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this – C is determined from the mass of CO2 produced – H is determined from the mass of H2O produced – O is determined by difference after the C and H have been determined Calculating Empirical Formula • • • • • • • If you have mass, skip the next 2 steps If you have %, pretend that you have a 100 gram sample of the compound. That is, change the % to grams. Convert the grams to moles for each element. Write the number of moles as subscripts in a chemical formula. Divide each number by the least number. Multiply the results to get rid of any fractions. Example • Calculate the empirical formula of a compound composed of 38.7 % C, 16.2 % H, and 45.1 %N. • Assume 100 g so: • 38.7 g C x 1mol C = 3.22 mole C 12.0 gC • 16.2 g H x 1mol H = 16.2 mole H 1.0 gH • 45.1 g N x 1mol N = 3.22 mole N 14.0 gN Determining Mole Ratio • 3.22 mole C • 16.2 mole H • 3.22 mole N •C3.22H16.2N3.22 If we divide all of these by the smallest one It will give us the empirical formula Divide by smallest to get Empirical Formula • The ratio is 3.22 mol C = 1 mol C 3.22 molN 1 mol N • The ratio is 16.2 mol H = 5 mol H 3.22 molN = 1 mol N • C1H5N1 is the Empirical Formula A compound is 43.64 % P and 56.36 % O. What is the empirical formula? • 43.6 g P x 1mol P 31.0 gP • 56.4 g O x 1mol O 16 gO P1.4O3.5 = 1.4 mole P = 3.5 mole O Divide both by the lowest one P1.4O3.5 • The ratio is 3.5 mol O = 2.5 mol O 1.4 mol P 1 mol P P1O2.5 • Multiply the result to get rid of any fractions. 2X P1O2.5 = P2O5 • Caffeine is 49.5% C, 5.15% H, 28.9% N and 16.5% O. What is its empirical formula? • 49.5 g C 1mol 12 g • 5.15 g H 1mol 1g • = 4.1 mol = 5.15 mol 1mol 28.9 g N 14 g = 2.06 mol 1mol 16 g = 1.03 mol • 16.5 g O We divide by lowest (1mol O) and ratio doesn’t change Since they are close to whole numbers we will use this formula C4.12H5.15N2.06O1 OR C4H5N2O1 empirical mass = 97g Empirical to molecular • Since the empirical formula is the lowest ratio, the actual molecule could weigh more (by whole number multiples • Divide the actual molar mass by the mass of one mole of the empirical formula. • Caffeine has a molar mass of 194 g. what is its molecular formula? • molar mass Find x if x empirical formula mass 194 g 97 g 2X C4H5N2O1 C8H10N4O2. =2 Example • A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? Example- Data to moles 1mol • 71.65 Cl 35.5 g 24.27C 1mol = 2.0mol = 2.0mol 12 g 4.07 H. 1mol 1g = 4.0mol Moles to Empirical Formula • C2H4Cl2 We divide by lowest (2mol ) •C1H2Cl1 Lowest ratio = Empirical Formula would give an empirical wt of 48.5g/mol Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? •Empirical to Molecular Formula would give an empirical wt of 48.5g/mol Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? molar mass x empirical formula mass = 98.96 g =2 48.5 g Molecular Formula 2 X C1H2Cl1 = C2H4Cl2