Practice Work 28: Empirical and Molecular Formulas, Lecture notes of Chemistry

Download Practice Work 28: Empirical and Molecular Formulas and more Lecture notes Chemistry in PDF only on Docsity! CHEM 110 BEAMER Page 1 of 6 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 51) C6H6O2 Question 52) C4H10 Question 145) C13H18O2 (The empirical formula is identical to the molecular formula) Question 150) C3H8O3 CHEM 110 BEAMER Page 2 of 6 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? CHEM 110 BEAMER Page 5 of 6 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 Step 3 Step 4 Step 5 Step 6: Step 7: Step 8 ( 75.7 g C 1 ) ( 1 mol C 12.01 g C ) = 6.30 mol C ( 6.30 mol C 0.969 mol )  6.51 C Values are not close to whole numbers. Must use a coefficient. × 2  13 ( 8.80 g H 1 ) ( 1 mol H 1.01 g H ) = 8.71 mol H ( 8.71 mol H 0.969 mol )  8.99 H × 2  18 ( 15.5 g O 1 ) ( 1 mol O 16.00 g O ) = 0.969 mol O ( 0.969 mol O 0.969 mol )  1.000 O × 2  2 Step 9: C13H18O2 Step 3: Empirical Mass = (13 × 12.01 g/mol) + (18 × 1.01 g/mol) + (2 × 16.00 g/mol) = 206.31 g/mol Step 4: ( molecular mass empirical mass ) = ( 206 g/mol 206.31 g/mol ) = 0.998 ≈ 010 0.998 is very close to a whole number. Continue. Step 5: (C13H18O2)010  C13H18O2 answer Remember: The molecular formula can be the same as the empirical formula. Don’t let that throw you. CHEM 110 BEAMER Page 6 of 6 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 Step 3 Step 4 Step 5 Step 6: Step 7: Step 8 ( 39.12 g C 1 ) ( 1 mol C 12.01 g C ) = 3.257 mol C ( 3.257 mol C 3.257 mol )  1.000 C Values are not close to whole numbers. Must use a coefficient. × 3  3 ( 8.75 g H 1 ) ( 1 mol H 1.01 g H ) = 8.66 mol H ( 8.66 mol H 3.257 mol )  2.66 H × 3  8 ( 52.12 g O 1 ) ( 1 mol O 16.00 g O ) = 3.257 mol O ( 3.257 mol O 3.257 mol )  1.000 O × 3  3 Step 9: C3H8O3 Step 3: Empirical Mass = (3 × 12.01 g/mol) + (8 × 1.01 g/mol) + (3 × 16.00 g/mol) = 92.11 g/mol Step 4: ( molecular mass empirical mass ) = ( 92.11 g/mol 92.11 g/mol ) = 010 0.998 is very close to a whole number. Continue. Step 5: (C3H8O3)010  C3H8O3 answer