11inchapter3andthusthemolecularformulaisidentical

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Unformatted text preview: ; 5.10; 9th ed. p. 187. Equation (5.12). Example 5.9. From PM = dRT where P is in unit of atm and T is in unit of Kelvin M = dRT/P = [(7.10/5.40)x0.082x(40+273.15)]/(741/760) = 35.0 g/mol 2 Application: What is the density of HBr gas in grams per liter at 733 mmHg and 46 oC? (a) 0.54 (b) 1.36 (c) 2.24 (d) 2.97 (e) 3.57 Hint: 11th ed.: p. 190 and Example 5.8; 10th ed. p.p. 190‐191 Example 5.8; 9th ed. p. 186. Equation (5.11). Example 5.8. From PM= dRT d = PM/RT = (733/760)x(1.008+79.90)/[0.082x(46+273.15)] = 2.97 g/L 14. A compound has the empirical formula as SF4. At 20 oC, 0.100 gram of the gaseous compound occupies a volume of 22.1 mL and exerts a pressure of 1.02 atm. What is the molecular formula of the gas? (b) SF6 (c) S2F10 (d) S4F16 (e) S5F20 (a) SF4 Hint: 11th ed. p.p. 191‐193 Example 5.10; 10th ed. p.p. 193‐194 Example 5.10; 9th ed. p. 211 (5.50) or p. 187. Equation (5.12). Example 5.9. To find molecular formula, we need to find the molar mass from PM = dRT and convert 22.1 mL to 0.0221 L M = dRT/P = (0.100/0.0221)x0.082x(20+273.15)/1.02 = 106.64 g/mol. Ratio = molar mass/empirical molar mass = 106.64/(32.07+19x4) = 0.99 round to whole number as 1. So the molecular formula contains one empirical form...
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