Determining the Molecular Formula

Determining the Molecular Formula1. Determine the atoms in the material usingelemental analysis2. Use various chemical techniques likecombustion to determine the relativenumber of atoms in the substance.This is the empirical formula.3. Determine the molecular mass of thesubstance using mass spectrometry, gasdensity measurements, etc.4. Deduce the molecular formula

Empirical / Molecular FormulasDetermine the empirical formula of thecompounds with the followingcompositions:a. 0.0130 mol C, 0.0390 mol H,and 0.0065 mol Ob. 11.66 g iron and 5.01 g oxygenc. 40 % C, 6.7 % Hand 53.3 % O by massWhat is the molecular formula of each of thefollowing compounds?a. empirical formula CH 2 ,molar mass = 84 g/molb. empirical formula NH 2 Cl,molar mass = 51.5 g/mol

Molecular formula from combustionGiven that the material contains C, H & OBurn 0.225 g of C x H y O zC H O + O(g) → CO(g) + H O(g)x y z2 2 2form 0.209 g H 2 O & 0.512 g CO 21. The number of moles of CO 2 is thenumber of moles of C in C x H y O z2. The number of moles of H 2 O is twice thenumber of moles of H in C x H y O z3. The number of moles of O in C x H y O zmust be determined by difference.

0.512gmoles of CO = 00116 . mol244.01 g/mol =0.209gmole of HO2= 0. 0116 mol18.00g/mol =grams of C = 0.0116(12.00) = 0.1392grams of H = 2(0.0116)(1.00) = 0.0232grams of O = 0.225 – 0.1392 – 0.0232 = 0.061

moles O =0.061g16.00g/mol =0.0038 molatoms of C relative to O = 0.0116/0.0038 = 3.04atoms of H relative to O = 0.0232/0.0038 = 6.08C 3 H 6 OFormula Mass is 58.09Molecular mass as 116 g/molC 6 H 12 O 2

Limiting ReactantsReaction will stop when one of the reactantsis gone3H (g) + N (g) → 2NH (g)2 2 3How many moles of product will result from3.0 mol N 2 and 6.0 mole H 23.0 mol N 2 can use 9.0 mol H 2 , have 6.06.0 mol H 2 can use 2 mol N 2 , have 3.0All H 2 will be consumed and so it is thelimiting reactant

Given the reaction2Na PO ( aq ) + 3Ba( NO ) ( aq ) → Ba ( PO ) ( s ) + 6NaNO ( aq )3 4 3 2 3 4 2 3350 . g Na3PO 4and 6.40g Ba(NO3) 2How many grams of Ba3(PO 4)2will be formed?Which species is the limiting reactant?Compound3 43 2MolesNa PO 0.0213Ba(NO ) 0.0245Ba(NO3) 2is limiting reactantMoles of Ba (PO ) formed3 4 21 moles of Ba(NO3 )2 reacted3which is 0.00816

Concentration of Solutionsmolarity=moles of soluteliters of solutionM=nLMolarity of solution prepared by mixing23.4g of Na 2 SO 4 with enough water to makea solution whose volume is 125ml?Formula weight of Na2SO 4is 142.0 g/mol23. 4 / 142.0M = = 132 .0.125

Dilution ProblemGiven a solution with a known molarity oneis to dilute it to form a solution of a lessermolarity.Given a 1.00M solution of CuSO 4 , make250ml of a 0.1M solution.How many moles of CuSO 4 do we need?n = ( 0250 . )x( 01 . ) = 0025 . molWhat volume of the 1.0M solution containsthis amount ?0025 . molV = = 0.025L100 . mol / LAdd 25ml of 1.00M solution to 225ml of water

TitrationThe process of reacting a solution of anunknown concentration with one of a knownconcentrationOne can determine the Cl - concentration in awater sample by titrating with AgNO 3 .+ −Ag ( aq ) + Cl ( aq ) → AgCl( s )How many grams of Cl - are present in asample if 20.2 ml of 0.100M Ag + is neededto react with all of the Cl - ?moles Ag + = moles Cl -moles Ag + = (0.100mol/L)(0.020L) =0.0020 molg of Cl - = (0.0020 mol)(35.5g/mol) = 7.17 x 10 -2