molecular-formula.ppt
- 2. Simplest formula calculations Q- a compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula? Step 1: imagine that you have 100 g of the substance. Thus, % will become mass in grams. E.g. 69.58 % Ba becomes 69.58 g Ba. (Some questions will give grams right off, instead of %) Step 2: calculate the # of moles (mol= g g/mol) Step 3: express moles as the simplest ratio by dividing through by the lowest number. Step 4: write the simplest formula from mol ratios.
- 3. Simplest formula: sample problem Q- 69.58% Ba, 6.090% C, 24.32% O. What is the empirical (a.k.a. simplest) formula? 1: 69.58 g Ba, 6.090 g C, 24.32 g O 2: Ba: 69.58 g 137.33 g/mol = 0.50666 mol Ba C: 6.090 g 12.01 g/mol = 0.50708 mol C O: 24.32 g 16.00 g/mol = 1.520 mol O 3: 4: the simplest formula is BaCO3 mol (reduced) mol 1.520/ 0.50666 = 3.000 0.50708/ 0.50666 = 1.001 0.50666/ 0.50666 = 1 1.520 0.50708 0.50666 O C Ba
- 4. Mole ratios and simplest formula Given the following mole ratios for the hypothetical compound AxBy, what would x and y be if the mol ratio of A and B were: A = 1 mol, B = 2.98 mol A = 1.337 mol, B = 1 mol A = 2.34 mol, B = 1 mol A = 1 mol, B = 1.48 mol AB3 A4B3 A7B3 A2B3 1. A compound consists of 29.1% Na, 40.5 % S, and 30.4 % O. Determine the simplest formula. 2. A compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen. Find the empirical formula for this compound 3. - 6. Try questions 3 - 6 on page 189.
- 5. Question 1 1: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O 2: Na: 29.1 g 22.99 g/mol = 1.266 mol Na S: 40.5 g 32.06 g/mol = 1.263 mol S O: 30.4 g 16.00 g/mol = 1.90 mol O 3: 4: the simplest formula is Na2S2O3 mol (reduced) mol 1.90/ 1.263 = 1.50 1.263/ 1.263 = 1 1.266/ 1.263 = 1.00 1.90 1.263 1.266 O S Na For instructor: prepare molecular models
- 6. Question 2 1: 7.20 g C, 1.20 g H, 9.60 g O 2: C: 7.20 g 12.01 g/mol = 0.5995 mol C H: 1.20 g 1.01 g/mol = 1.188 mol H O: 9.6 g 16.00 g/mol = 0.60 mol O 3: 4: the simplest formula is CH2O mol (reduced) mol 0.60/ 0.5995 = 1.0 1.188/ 0.5995 = 1.98 0.5995/ 0.5995 = 1 0.60 1.188 0.5995 O H C
- 7. Question 3 1: Assume 100 g: 28.9 g K, 23.7 g S, 47.7 g O 2: C: 7.20 g 12.01 g/mol = 0.5995 mol C H: 1.20 g 1.01 g/mol = 1.188 mol H O: 9.6 g 16.00 g/mol = 0.60 mol O 3: 4: the simplest formula is CH2O mol (reduced) mol 0.60/ 0.5995 = 1.0 1.188/ 0.5995 = 1.98 0.5995/ 0.5995 = 1 0.60 1.188 0.5995 O H C
- 8. Molecular formula calculations • There is one additional step to solving for a molecular formula. First you need the molar mass of the compound. E.g. in Q2, the molecular formula can be determined if we know that the molar mass of the compound is 150 g/mol. • First, determine molar mass of the simplest formula. For CH2O it is 30 g/mol (12+2+16). • Divide the molar mass of the compound by this to get a factor: 150 g/mol 30 g/mol = 5 • Multiply each subscript in the formula by this factor: C5H10O5 is the molecular formula. (models) Q- For OF, give the molecular formula if the compound is 70 g/mol O2F2 70 35 = 2
- 9. 7. Combustion analysis gives the following: 26.7% C, 2.2% hydrogen, 71.1% oxygen. If the molecular mass of the compound is 90 g/mol, determine its molecular formula. 8. What information must be known to determine a) the empirical formula of a substance? b) the molecular formula of a substance? 9. A compound’s empirical formula is CH, and it weighs 104 g/mol. Give the molecular formula. 10. A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by mass. Calculate the molecular formula of the unknown if its molar mass is 90 g/mol.
- 10. Question 7 1: Assume 100 g total. Thus: 26.7 g C, 2.2 g H, and 71.1 g O 2: C: 26.7 g 12.01 g/mol = 2.223 mol C H: 2.2 g 1.01 g/mol = 2.18 mol H O: 71.1 g 16.00 g/mol = 4.444 mol O 3: C H O mol mol (reduced) 2.223 2.18 4.444 2.223/2.18 = 1.02 2.18/ 2.18 = 1 4.444/2.18 = 2.04 4: the simplest formula is CHO2 5: factor = 90/45=2. Molecular formula: C2H2O4
- 11. Question 8, 9 • For the empirical formula we need to know the moles of each element in the compound (which can be derived from grams or %). For the molecular formula we need the above information & the molar mass of the compound • Molar mass of CH = 13 g/mol Factor = 104 g/mol 13 g/mol = 8 Molecular formula is C8H8
- 12. Question 10 1: Assume 100 g total. Thus: 53.2 g C, 11.2 g H, and 35.6 g O 2: C: 53.2 g 12.01 g/mol = 4.430 mol C H: 11.2 g 1.01 g/mol = 11.09 mol H O: 35.6 g 16.00 g/mol = 2.225 mol O 3: C H O mol mol (reduced) 4.430 11.09 2.225 4.43/2.225 = 1.99 11.09/2.225 = 4.98 2.225/2.225 = 1 4: the simplest formula is C2H5O 5: factor = 90/45=2. Molecular formula: C4H10O2
- 13. 1. Calculate the percentage composition of each substance: a) SiH4, b) FeSO4 2. Calculate the simplest formulas for the compounds whose compositions are listed: a) carbon, 15.8%; sulfur, 84.2% b) silver,70.1%; nitrogen,9.1%; oxygen,20.8% c) K, 26.6%; Cr, 35.4%, O, 38.0% 3. The simplest formula for glucose is CH2O and its molar mass is 180 g/mol. What is its molecular formula? Assignment
- 14. 4. Determine the molecular formula for each compound below from the information listed. substance simplest formula molar mass(g/mol) a) octane C4H9 114 b) ethanol C2H6O 46 c) naphthalene C5H4 128 d) melamine CH2N2 126 5. The percentage composition and approximate molar masses of some compounds are listed below. Calculate the molecular formula of each percentage composition molar mass(g/mol) 64.9% C, 13.5% H, 21.6% O 74 39.9% C, 6.7% H, 53.4 % O 60 40.3% B, 52.2% N, 7.5% H 80
- 15. 1 a) Si= 87.43% (28.09/32.13 x 100), H= 12.57% b) Fe= 36.77% (55.85/151.91 x 100), S= 21.10% (32.06/151.91 x 100), O= 42.13% 2 a) Assume 100 g. Thus: 15.8 g C, 84.2 g S. C: 15.8 g 12.01 g/mol = 1.315 mol C S: 84.2 g 32.06 g/mol = 2.626 mol S the simplest formula is CS2 2.626/1.315 = 2.00 1.315/1.315 = 1 Mol reduced 2.626 1.315 Mol S C
- 16. 2 b) Ag: 70.1 g 107.87 g/mol = 0.6499 mol Ag N: 9.1 g 14.01 g/mol = 0.6495 mol N O: 20.8 g 16.00 g/mol = 1.30 mol O .6499/.6495 = 1.0 0.6499 Ag 1.30/.6495 = 2.00 .6495/.6495 = 1 Mol reduced 1.30 0.6495 Mol O N AgNO2 2 c) K: 26.6 g 39.10 g/mol = 0.6803 mol K Cr: 35.4 g 52.00 g/mol = 0.6808 mol Cr O: 38.0 g 16.00 g/mol = 2.375 mol O .6803/.6803 = 1 0.6803 K 2.375/.6495 = 3.49 .6808/.6803 = 1.00 Mol reduced 2.375 0.6808 Mol O Cr K2Cr2O7
- 17. 3 C6H12O6 (CH2O = 30 g/mol, 180/30 = 6) 4 a) C8H18 (C4H9 = 57 g/mol, 114/57 = 2) b) C2H6O (C2H6O = 46 g/mol, 46/46 = 1) c) C10H8 (C5H4 = 64 g/mol, 128/64 = 2) d) C3H6N6(CH2N2 = 54 g/mol, 126/42 = 3) 5 a) C: 64.9 g 12.01 g/mol = 5.404 mol C H: 13.5 g 1.01 g/mol = 13.37 mol H O: 21.6 g 16.00 g/mol = 1.35 mol O 5.404/1.35 = 4.00 5.404 C 1.35/1.35 = 1 13.37/1.35 = 9.90 Mol reduced 1.35 13.37 Mol O H C4H10O C4H10O (C4H10O = 74 g/mol, 74/74 = 1)
- 18. 5 b)C: 39.9 g 12.01 g/mol = 3.322 mol C H: 6.7 g 1.01 g/mol = 6.63 mol H O: 53.4 g 16.00 g/mol = 3.338 mol O 3.322/3.322 = 1 3.322 C 3.338/3.322 = 1.00 6.63/3.322 = 2.0 Mol reduced 3.338 6.63 Mol O H CH2O C2H4O2 (CH2O = 30 g/mol, 60/30 = 2) 5 c) 3.728/3.726 = 1.00 3.728 B 7.43/3.726 = 2.0 3.726/3.726 = 1 Mol reduced 7.43 3.726 Mol H N B3N3H6 (BNH2 = 26.84 g/mol, 80/26.84= 2.98) For more lessons, visit www.chalkbored.com