Welcome Back CW 1. Convert the following numbers to scientific notation: a. 124 1.24 x 102 b. 0.004852 4.852 x 10-3 c. 54.21 5.421 x 101 d. 0.000084 8.4 x 10-5 2. Write the following numbers in standard notation a. 1.33 x 10-3 0.00133 b. 5.77 x 105 577,000 3. Round the following numbers to 3 sig figs a. 19.9994 20.0 b. 8,439 8,440 8 c. 5.3861 x 10 5.39 x 108 4. How many atoms are in 3.4 mol Cu? 3.4 (6.022 x 1023) = 2.0 x 1024 atoms of Cu 5. Calculate the mass of 1.28 mol of Fe 1.28 x 55.845 = 71.5 g of Fe 6. How many moles of atoms are in 4.6 x 1024 atoms of Pb? 4.6 ´10 24 = 7.6 moles of Pb 6.022 ´10 23 7. 285.33 g of NaNO3 represent how many moles? 285.33 = 3.3571 moles of NaNO3 84.994 8. How many atoms are in 9.7 x 10-3 mol C? (9.7 x 10-3) (6.022 x 1023) = 5.8 x 1021 atoms of C 9. Determine the mass of 2.48 x 1022 atoms of HCl 2.48 ´10 22 = 0.0411823 mol of HCl 6.022 ´10 23 (0.0411823) (36.4609) = 1.50 g of HCl 10. Calculate the mass percent composition for each of the following compounds. a. C2H2 2(12.011) (100) = 92.258% C 2(12.011) + 2(1.0079) 2(1.0079) (100) 2(12.011) + 2(1.0079) = 7.7418% H b. C3H6 3(12.011) (100) = 85.629% C 3(12.011) + 6(1.0079) 6(1.0079) (100) 3(12.011) + 6(1.0079) = 15.230% H 11. Calculate the empirical formula for the compound given the following masses of their constituent elements. 1.245 g Ni, 5.381 g I 1.245 5.381 = 0.02121 mol Ni = 0.04240 mol I 58.693 126.90 0.02121 = 0.02121 1 NI2 0.04240 = 0.02121 1.999 12. Calculate the empirical formula for pineapple oil, Ethyl butyrate: C – 62.04%, H – 10.41%, O – 27.55% 62.04 10.41 27.55 = 5.1657 mol C = 10.328 mol H = 1.722 mol O 12.011 1.0079 15.999 5.16757 10.328 = 3 mol C = 6 mol H 1 mol O 1.722 1.722 C3H6O 13. The following are the molar masses and empirical formulas of several compounds. Find the molecular formula of each compound. a. 284.77 g/mol – CCl CCl = 12.011 + 35.453 = 47.464 g/mol 284.77 =6 47.464 so C6Cl6 b. 131.39 g/mol – C2HCl3 C2HCl3 = 2(12.011) + 1.0079 + 3(35.453) = 131.39 g/mol Molar mass is the same as empirical mass so the EF = MF C2HCl3