CHEM 111 1st Edition Lecture 4Calculate the empirical formula given percent composition:Given:C= 26.68%H=2.22%O=71.09%Step 1: Assume a weight of 100g C=26.68gH=2.22gO=71.09gStep 2: Convert to molesC: 26.68g x (1 mol/12.011g)=2.221 molH: 2.22g x (1 mol/1.008g)=2.202 molO: 71.09g x (1 mol/15.999g)=4.443molStep 3: Express the molar ratio using integersC: 2.221mol/2.202 mol=1.009H: 2.202 mol/2.202 mol=1O: 4.443 mol/2.202 mol=2.018Empirical Formula: CHO2Empirical Formula of CHO2. If the molar mass is 90.022g/mol, what is the molecular formula?1. Calculate the formula weight of CHO2C+H+2(O)=12.011+1.01+2(16)=45.02 g/mol2. Multiply bythe appropriate integer until it equals the molar mass90.022/45.02=2CH O2x 2=C2H2O4Waters of Hydration MgSO4× 7 H2OCaS O4×2 H2ON a2S2O3× 5 H2ON a2S O4×10 H2OCaS O4×.5 H2ONaC O3×10 H2ONames with out the water “anhydrous CaSO4”The molar mass includes hydrated waterAssume 5 g CuS O4× n H2O is heated to dryness, with final weight of 3.196g. What is n?Final moles of CuSO4: 3.196g x (1mol/159.607g)=.0200 molesMass of H2O drive off 5.000g-3.196g=1.804gMoles of H2O driven off: 1.804g x (1 mol/18.015g)=.1001 molesMolar ratio of H2O to CuSO4: .1001/.0200=5N=5If 400.34g of CaCO3 decomposes, how many grams of CO2 are produced?1. Write a balanced equationCaC O3−→CaO +C O2100.086g/mol 56.077g/mol 44.009g/mol2. Convert pounds to moles400.34 g x (1mol/100.086g)=4.0000moles3. Multiply by appropriate stoichiometric factor4 moles of CaCO3 leads to 4 moles of CO24. Convert back to grams4.1 moles Co2 x 44.009 g/mol=176.04gLimiting reactants/ingredients Whatever element is in short supply limits how many compounds can you make?Example: You have 36 slices of meat, 15 slices of cheese, and 40 slices of bread. Which is the Limiting? Meat limits because it can only make 12 sandwichesIf you have 100g each of P, I2, and H2O, which is the limiting reagent?2 P+3 I2+6 H2O−→ 2 HPO(O H)2+6 HI1. Balance Equation (already done)2. Convert to moles# of moles of P: 100g/31 g mole=3.23 moles P2.23 moles P x 6 moles HI/ 2 moles P=9.69 moles HI# of moles of I2: 100g/254 g mole=.394 moles I26 moles HI/3 moles I2=.788 moles HI# moles H2O=100g/18 g mole=5.56 moles H2O6 moles HI/6 moles H2O=5.56 moles of HIHI is the limiting reactant!Carbon monoxide reacts with O2 to form carbon dioxide. If you have 200.0 grams each of carbon monoxide and O2, how many grams of carbon dioxide are formed? What is the limiting reagent?1. Write a balanced equation2CO +O2−→ 2C O22. Convert to molesCO: 200.0 g x (1 mol/28.01g)=7.140 molO2: 200.0 g x (1 mol/31.998g)=6.250 mol3. Calculate CO2 produced assuming CO is limiting CO: 7.140 mol x (1 molCO2/1 mol CO)=7.14 mol CO2O2: 6.250 mol x (2 mol CO2/1 mole O2)=12.50 CO2Limiting Reagant: CO4. Convert moles of CO2 to grams7.140 mol CO2 x