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CHEM 110 BEAMER Last Name Date First Name Practice Work 28: Empirical and Molecular Formulas • You should do this on your own paper. • I suggest a sideways format to the paper. This assignment is not due. • Appendix 11 (available on the website) will be helpful. Remember that you will not have the appendix with the steps for the Final. Also remember that the first step of finding the molecular formula is to find the empirical formula. I apologize about this blank page. It was the only way to get all of the solutions to fit nicely on pages of their own. Also, the solution sets are highly color coded. I hope that this is helpful, instead of distracting. I put the answer key on the first page to save space. -- beamz Answers (Solutions start on page 3) Question Question Question Question 51) 52) 145) 150) C6H6O2 C4H10 C13H18O2 (The empirical formula is identical to the molecular formula) C3H8O3 Page 1 of 6 CHEM 110 BEAMER PAGE 335, QUESTION 51 51) Analysis of a chemical compound used in photographic developing fluid indicates a chemical composition of 65.45% C, 5.48% H, and 29.08% O. The molecular mass is found to be 110.0 g/mol. Determine the molecular formula. 10 min maximum PAGE 335, QUESTION 52 52) A compound was found to contain 49.98 g carbon and 10.47 g hydrogen. The molecular mass of the compound is 58.12 g/mol. Determine the molecular formula. 8 min max PAGE 349, QUESTION 145 145) Determine the molecular formula for ibuprofen, a common headache remedy. Analysis of ibuprofen yields a molar mass of 206 g/mol and a percent composition of 75.7% C, 8.80% H, and 15.5% O. Determine the molecular formula. 10 minutes maximum. PAGE 349, QUESTION 150 150) Glycerol is a thick, sweet liquid obtained as a byproduct of the manufacture of soap. Its percent composition is 39.12% C, 8.75% hydrogen, and 52.12% oxygen. The molecular mass is 92.11 g/mol. What is the molecular formula for glycerol? Page 2 of 6 CHEM 110 BEAMER Part 1: Molecular Formulas PAGE 335, QUESTION 51 51) Analysis of a chemical compound used in photographic developing fluid indicates a chemical composition of 65.45% C, 5.48% H, and 29.08% O. The molecular mass is found to be 110.0 g/mol. Determine the molecular formula. 10 min maximum DETERMINING MOLECULAR FORMULA Step 1: The molecular mass was given. (110.0 g/mol) Step 2: Determine the empirical formula DETERMINING EMPIRICAL FORMULA Step 1: Mass wasn’t given. Go to Step 2. Step 2 ( Step 3 65.45 g C ) 1 ( 5.48 g H ) 1 ( ( 29.08 g O ( ) 1 Step 9: Step 4 1 mol C ) 12.01 g C = 1 mol H ) 1.01 g H = 1 mol O ( ) 16.00 g O = 2.998 mol C 2.99 mol H 1.000 mol O Step 5 ( Step 6: 2.998 mol C ) 1.000 mol 2.99 mol H ) 1.000 mol ( 1.000 mol O ( ) 1.000 mol Step 8 All values 2.998 C 3 3 1 are very close to 2.99 H whole numbers. 1.000 O Skip Step 7 C3H3O1 Step 3: Empirical Mass = (3 × 12.01 g/mol) + (3 × 1.01 g/mol) + (1 × 16.00 g/mol) Step 4: ( molecular mass 110.0 g/mol ) = ( ) = 1.998 ≈ empirical mass 55.06 g/mol 0 20 1.998 is very close to a whole number. Continue. = 55.06 g/mol Step 5: (C3H3O1) 2 0 0 C6H6O2 answer Page 3 of 6 CHEM 110 BEAMER PAGE 335, QUESTION 52 52) A compound was found to contain 49.98 g carbon and 10.47 g hydrogen. The molecular mass of the compound is 58.12 g/mol. Determine the molecular formula. 8 min max DETERMINING MOLECULAR FORMULA Step 1: The molecular mass was given. (58.12 g/mol) Step 2: Determine the empirical formula DETERMINING EMPIRICAL FORMULA Step 1: Mass was given. Skip to Step 2. Mass Values ( 49.98 g C ) 1 Step 3 ( 1 mol C ) 12.01 g C Step 4 = 4.162 mol C Step 5 ( 4.162 mol C ) 4.162 mol Step 6: 1.000 C Step 7: Values are not close to Step 8 × 2 2 × 2 5 whole ( 10.47 g H ) 1 ( 1 mol H ) 1.01 g H = 10.37 mol H ( 10.37 mol H ) 4.162 mol 2.492 H numbers. Must use a coefficient. Step 9: C2H5 Step 3: Empirical Mass = (2 × 12.01 g/mol) + (5 × 1.01 g/mol) = 29.07 g/mol Step 4: ( molecular mass 58.12 g/mol ) = ( ) = 1.999 ≈ empirical mass 29.07 g/mol 0 20 1.999 is very close to a whole number. Continue. Step 5: (C2H5) 2 0 0 C4H10 answer Page 4 of 6 CHEM 110 BEAMER PAGE 349, QUESTION 145 145) Determine the molecular formula for ibuprofen, a common headache remedy. Analysis of ibuprofen yields a molar mass of 206 g/mol and a percent composition of 75.7% C, 8.80% H, and 15.5% O. Determine the molecular formula. 10 minutes maximum. DETERMINING MOLECULAR FORMULA Step 1: The molecular mass was given. (206 g/mol) Step 2: Determine the empirical formula DETERMINING EMPIRICAL FORMULA Step 1: Mass wasn’t given. Go to Step 2. Step 2 75.7 g C ) 1 ( 8.80 g H ) 1 ( ( ( ( Step 9: Step 3 15.5 g O ) 1 ( Step 4 1 mol C ) 12.01 g C = 1 mol H ) 1.01 g H = 8.71 mol H 1 mol O ) 16.00 g O = 0.969 mol O 6.30 mol C Step 5 6.30 mol C ) 0.969 mol 8.71 mol H ) 0.969 mol 0.969 mol O ) 0.969 mol ( ( ( Step 6: 6.51 C Step 7: Values are × 2 13 × 2 18 × 2 2 not close to 8.99 H whole numbers. Must use a 1.000 O coefficient. Step 8 C13H18O2 Step 3: Empirical Mass = (13 × 12.01 g/mol) + (18 × 1.01 g/mol) + (2 × 16.00 g/mol) Step 4: ( molecular mass 206 g/mol ) = ( ) = 0.998 ≈ empirical mass 206.31 g/mol 0 10 0.998 is very close to a whole number. Continue. = 206.31 g/mol Step 5: (C13H18O2) 1 0 0 C13H18O2 answer Remember: The molecular formula can be the same as the empirical formula. Don’t let that throw you. Page 5 of 6 CHEM 110 BEAMER PAGE 349, QUESTION 150 150) Glycerol is a thick, sweet liquid obtained as a byproduct of the manufacture of soap. Its percent composition is 39.12% C, 8.75% hydrogen, and 52.12% oxygen. The molecular mass is 92.11 g/mol. What is the molecular formula for glycerol? DETERMINING MOLECULAR FORMULA Step 1: The molecular mass was given. (92.11 g/mol) Step 2: Determine the empirical formula DETERMINING EMPIRICAL FORMULA Step 1: Mass wasn’t given. Go to Step 2. Step 2 ( 39.12 g C ) 1 ( 8.75 g H ) 1 ( ( ( Step 9: Step 3 52.12 g O ) 1 ( Step 4 1 mol C ) 12.01 g C = 1 mol H ) 1.01 g H = 8.66 mol H 1 mol O ) 16.00 g O = 3.257 mol O 3.257 mol C Step 5 3.257 mol C ) 3.257 mol 8.66 mol H ) 3.257 mol 3.257 mol O ) 3.257 mol ( ( ( Step 6: 1.000 C Step 7: Values are Step 8 × 3 3 × 3 8 × 3 3 not close to 2.66 H whole numbers. Must use a 1.000 O coefficient. C3H8O3 Step 3: Empirical Mass = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) + (3 × 16.00 g/mol) Step 4: ( molecular mass 92.11 g/mol ) = ( ) = empirical mass 92.11 g/mol 0 10 0.998 is very close to a whole number. Continue. = 92.11 g/mol Step 5: (C3H8O3) 1 0 0 C3H8O3 answer Page 6 of 6