Moles and Stoichiometry

Moles and Stoichiometry - Moles and Stoichiometry I Moles A...

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1 Moles and Stoichiometry I. Moles A. Definition. 1. A mole (mol) = 6.022 x 10 23 units. 6.022 x 10 23 = Avogadro's constant, with a dimension of particles mol –1 or mol –1 b. A mass in grams equal to the formula mass of a substance contains 6.02 x 10 23 formula units. c. The practical unit of formula mass or atomic mass = grams/mol. 2. The following should be familiar to you. Type of Formula Composition of 10.0 g of substance Formula substance mass in moles _ in no. of units Na element 23.0 g/mol 10.0 /23.0= (0.435)(6.02 x10 23 ) = 0.435 mol 2.62 x 10 23 atoms H 2 molecular 2.0 g/mol 5.0 moles H 2 3.01 x 10 24 molecules of H 2 10.0 moles H 6.02 x 10 24 atoms of H C 2 H 4 molecular 28.0 g/mol 0.357 mol C 2 H 4 2.15 x 10 23 molecules C 2 H 4 0.714 mol C 4.30 x 10 23 atoms C 1.428 mol H 8.60 x 10 23 atoms H Al 2 O 3 ionic 102 g/mol 0.098 mol Al 2 O 3 5.90 x 10 22 formula units 0.196 mol Al 3+ 1.18 x 10 23 Al 3+ ions 0.294 mol O 2- 1.77 x 10 23 O 2- ions
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2 3. Additional Example. Consider a 48.60 g sample of the compound C 4 H 10 O 2 a. How many moles of C 4 H 10 O 2 are present? MM of C 4 H 10 O 2 = 4(12.0) + 10(1.0) + 2(16.0) = 90.0g/mol Mol C 4 H 10 O 2 = 48.60g 90.0g/mol = 0.540 mol b. How many moles and how many grams of C are present in the sample? The formula shows 4 C atoms for every molecule there are 4 moles of C for every mole of C 4 H 10 O 2 . Moles C = 0.540 mole C 4 H 10 O 2 x 4 mol C 1 mol C 4 H 10 O 2 = 2.16 mol C grams C = (2.16 mol C)(12.0 g/mol C) = 25.92 g or grams C = 48.60g 48.0 g C 90.0 g compound = 25.92g C moles C = 25.92 g 12.0 g/mol = 2.16 mol C c. How many O atoms are present in the sample? Mol O = 0.540 mole C 4 H 10 O 2 x 2 mol O 1 mol C 4 H 10 O 2 = 1.08 mol O Atoms O = 1.08 mol O x 6.02x10 23 atoms/mol = 6.50x10 23 d. How many atoms of H and how many grams of H are present in the sample? Moles of H = 0.540 mole C 4 H 10 O 2 x 10 mol H 1 mol C 4 H 10 O 2 = 5.40 mol H grams H = 5.40 molx1.0g/mol = 5.40g Note: mol H could also be obtained by using the moles of C: Mol H = 2.16 mol Cx 10mol H 4 mol C = 5.40 mol
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3 B. Empirical formulas from analysis data. 1. A compound was analyzed and found to contain 29.1% Na, 40.6% S, and 30.3% O by mass. Calculate the empirical formula. moles of Na = 29.1g / 23.0 g /mol= 1.265 mol.----------> 1.000----------> 2 moles of S = 40.6g / 32.1g /mol = 1.265 mol -----------> 1.000 ---------> 2 moles of O = 30.3g / 16.0g/ mol = 1.894 mol ------------>1.500--------- > 3 formula : Na 2 S 2 O 3 Sodium thiosulfate 2. A 0.401 g sample of a compound was analyzed and found to contain 0.320 g of C and 0.081 g of H. In another experiment the molar mass was estimated to be equal to 30. Calculate the empirical and molecular formulas of the compound. moles C = 0.320g / 12.0g /mol = 0.0267 -------------> 1.00 ---------> 1 moles H = 0.081g / 1.0 g /mol = 0.081---------------> 3.04 ----------> 3 empirical formula = CH 3 empirical formula mass (EFM) = 12.0 + 3(1.0) = 15 30 15 = 2 = empirical formula units / molecule C 2 H 6 = molecular formula. 3. Analysis by combustion. A 1.000 g sample of a compound containing only C, H, and O was burned to give 2.197 g of CO 2 and 1.199 g of H 2 O. Calculate the empirical formula. + O 2 (C x H y O z )---------------> CO 2 + H 2 O 1.000 g 2.197 g 1.199 g all the C goes to form CO 2 mol C = mol CO 2 = 2.197 g 44.0 g/mol = 0.04992 mol grams C in the sample = (.04922 mol )(12.0 g/mol ) = 0.600 g all the H ends up in H 2 O mol H = 2xmol H 2 O = 2(1.199 g) 18.0 g/mol = 0.1331 mol grams H in sample = (0.1331 mol )(1.0 g/mo l) = 0.133 g total mass of sample = 1.000 g = g of C + g of H + g of O therefore g of O = 1.000 g - 0.600 g - 0.133 g = 0.266 g mol O = 0.266 g 16.0 g/mol = 0.0166 mol mol C = .04992--------------> 3.00
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4 mol H = 0.1331 -------------> 8.01 empirical formula = C 3 H 8 O mol O = 0.0166 -------------> 1.00 II. Stoichiometry: Calculations from balanced chemical equations.
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