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 69.58g 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 3:3: 4: the simplest formula is BaCO 3 mol (reduced) mol 1.520/ 0.50666 = 3.000 0.50708/ 0.50666 = 1.001 0.50666/ 0.50666 = 1 1.5200.507080.50666 OCBa

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 AB 3 A4B3A4B3 A7B3A7B3 A2B3A2B3 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.A compound consists of 28.9 % K, 23.7 % S, and 47.7 % O. Determine the simplest formula.

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 3:3: 4: the simplest formula is Na 2 S 2 O 3 mol (reduced) mol 1.90/ 1.263 = 1.50 1.263/ 1.263 = 1 1.266/ 1.263 = 1.00 1.901.2631.266 OSNa 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 3:3: 4: the simplest formula is CH 2 O mol (reduced) mol 0.60/ 0.5995 = 1.0 1.188/ 0.5995 = 1.98 0.5995/ 0.5995 = 1 0.601.1880.5995 OHC

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 3:3: 4: the simplest formula is CH 2 O mol (reduced) mol 0.60/ 0.5995 = 1.0 1.188/ 0.5995 = 1.98 0.5995/ 0.5995 = 1 0.601.1880.5995 OHC

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 CH 2 O 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: C 5 H 10 O 5 is the molecular formula. (models) Q- For OF, give the molecular formula if the compound is 70 g/mol O 2 F 2 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 3:3: 2.2232.184.444 2.223/2.18 = 1.02 2.18/ 2.18 = 1 4.444/2.18 = 2.04 4: the simplest formula is CHO 2 5: factor = 90/45=2. Molecular formula: C 2 H 2 O 4

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 C 8 H 8

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 3:3: 4.43011.092.225 4.43/2.225 = 1.99 11.09/2.225 = 4.98 2.225/2.225 = 1 4: the simplest formula is C 2 H 5 O 5: factor = 90/45=2. Molecular formula: C 4 H 10 O 2

13. 1.Calculate the percentage composition of each substance: a) SiH 4, b) FeSO 4 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 CH 2 O 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. substancesimplest formula molar mass(g/mol) a) octaneC 4 H 9 114 b)ethanolC 2 H 6 O46 c)naphthaleneC 5 H 4 128 d)melamineCH 2 N 2 126 5.The percentage composition and approximate molar masses of some compounds are listed below. Calculate the molecular formula of each percentage compositionmolar mass(g/mol) 64.9% C, 13.5% H, 21.6% O74 39.9% C, 6.7% H, 53.4 % O60 40.3% B, 52.2% N, 7.5% H80