Covalent Calculations **Calculations mean the return of SIGNIFICANT FIGURES**
Representative Particles The form in which a substance exists is its “representative particle”. Covalent compounds: molecules Substance Representative Particle Element Atom Ion Ionic Compound Formula Unit Molecular Compound Molecule
Solving Problems with Avagadro’s Number 1 mole = 6.02 x 1023 representative particles Representative Particles: unit needs to be applicable to what is being calculated.
Packet #1. How many moles are 3.891 x 1024 molecules of CO2? 1 mol CO2 3.891 x 1024 molecules CO2 6.02 x 1023 molecules CO2 6.463 mol CO2
Packet #2. How many molecules are in 0.445 mol of sulfur hexafluoride? 0.445 mol SF6 6.02 x 1023 molecules SF6 1 mol SF6 2.68 x 1023 molecules SF6
Packet #3. How many atoms of Hydrogen are in 0.621 mol of ammonia? 0.621 mol NH3 6.02 x 1023 molecule NH3 3 atoms H 1 mole NH3 1 molecule NH3 = 1.12 x 1024 atoms H
Packet #4. How many atoms of nitrogen are in 2 Packet #4. How many atoms of nitrogen are in 2.03 mol of dinitrogen tetroxide? 2.03 mol N2O4 6.02 x 1023 molecule N2O4 2 atoms N 1 mole N2O4 1 molecule N2O4 = 2.44 x 1024 atoms N
Mole-Mass and Mole-Volume Relationships
#5. What is the molar mass of carbon tetrafluoride? C 1 x 12.01 = 12.01 F 4 x 19.00 = 76.00 88.01 g Finish b, c, and d for homework!
Terminology Summary Mass terminology Used For Example Gram atomic mass Atoms What is the mass of Zn? Gram molecular mass Molecules (2 or more non-metals together) What is the mass of CO2? Gram formula mass Formula units (metal and nonmetal combinations) What is the mass of MgF2?
Mole – Mass Conversions Conversion factor 1 mol = molar mass (g)
#6. What is the mass of 0.667 moles of silicon dioxide? 0.667 mol SiO2 60.09 g SiO2 1 mol SiO2 = 40.1 g SiO2
#7. How many moles is equivalent to 7.055 grams of dinitrogen monoxide? N2O = (2 x 14.01) + 16.00 = 44.02 g 7.055 g N2O 1 mol N2O 44.02 g N2O = 0.1603 mol N2O
Multi-step Mole Problems Using multiple conversion factors.
#8. What is the mass of 8.120 x 1023 molecules of tetranitrogen monoxide? 8.120 x 1023 mc N4O 1 mol N4O 72.04 g N4O 6.02x1023 mc N4O 1 mol N4O = 97.17 g N4O
#9. How many molecules are in 12.5 grams of chlorine gas? 12.5 g Cl2 1mol Cl2 6.02 x 1023 mc Cl2 70.90 g Cl2 1 mol Cl2 = 1.06 x 1023 mc Cl2
#10. How many atoms of phosphorus are present in 20 #10.How many atoms of phosphorus are present in 20.0 g of diphosphorus tetrahydride? 20.0 g P2H4 1 mol P2H4 6.02x1023 mc P2H4 2 atoms P 65.98 g P2H4 1 mol P2H4 1 mc P2H4 = 3.65 x 1023 atoms P
Calculating Empirical Formulas Empirical formula - lowest whole number ratio of the elements in a compound It may or may not be the same as the molecular formula. Ex. H2O is both the empirical & molecular formula H2O2 is a molecular formula while HO is the empirical formula for H2O2.
Steps to calculate empirical formula: Find moles of each element. Set up mole ratio for each element in the compound. Simplify mole ratio (divide by smallest). If your answers are not in whole numbers, you must multiply by 2,3,4,or 5 to get whole numbers. Use mole ratio as subscripts in the formula. If given % composition, assume 100 g of compound. Memory Trick: % to mass, mass to moles, divide by smallest and round ‘till whole!
#24. A compound is 79.8% C and 20.2% H. Find its empirical formula. 79.8gC 1mol C = 6.64 mol C /6.64 = 1 12.01 g C 20.2 g H 1 mol H = 20.0 mol H /6.64 = 3 1.01 g H CH3
#25. An oxide of aluminum is formed by the complete reaction of 4 #25. An oxide of aluminum is formed by the complete reaction of 4.151g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound. 4.151g Al 1 mol Al = 0.1539 mol Al 26.98 g Al 3.692g O 1 mol O = 0.2308 mol O 16.00 g O = 1 x 2 0.1539 = 2 = 1.5 x 2 0.1539 = 3 Al2O3
Calculating Molecular Formulas Molecular formulas are the actual formulas. They may be the same as the empirical formula or a multiple of it. To find the multiple (n), take the gram formula mass (gfm) and divide by the empirical formula mass (efm): Multiply each subscript in the empirical formula by n to get the molecular formula. n = gfm efm
Ex. A white powder is analyzed and found to have the empirical formula P2O5. The compound has a molar mass of 283.9 g. What is the compound’s molecular formula? P2O5 = 142.00 g/mol gfm/efm 283.9 g/mol 142.00 g/mol = 2 2(P2O5) = P4O10
27. A compound used as an additive for gasoline to help prevent engine knock shows the following percentage composition: 71.65% Cl 24.27% C 4.07% H The molar mass is known to be 98.96 g. Determine the empirical formula and the molecular formula for this compound. /2.021 = 1 71.65 g Cl 1 mol Cl = 2.021 mol Cl 35.45 g Cl 24.27g C 1 mol C = 2.021 mol C 12.01 g C 4.07g H 1 mol H = 4.030 mol H 1.01 g H ClCH2 or CH2Cl efm = 12.01+2.02+35.45=49.48 g/mol 98.96/49.48=2 C2H4Cl2 /2.021 = 1 /2.021 = 2