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Chem notes gen - October 8 2007 How many moles are in a...

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The molar mass of silicon is 28.09 g/mol 28.1g SI x 1mol = 1.0 28.09g The molar mass of calcium is 40.08 g/mol 10.0g Ca x 1mol = 0.245mol Ca 40.08g 0.250mol x 6.022x10 23 atoms/mol = 1.50x10 23 atoms 1.00mol H 2 O = 18.0g 10.0g x 1mol = 0.556mol 18.0g 0.556mol x 6.022x10 23 molecules 1mol =3.35x10 23 molecules The molar mass of HNO 3 is 63.01 g/mol 1mol HNO 3 = 63.01g 3.26mol x 63.01g = 205.4g 1 mol Molar mass = the mass (in grams) of 1 mol of a substance 1mol CO 2 contains: 1mol of carbon & 2mol of The formula of sodium carbonate is Na 2 CO 3 1mol Na 2 CO 3 = 2 mol Na, 1 mol C, 3 mol O 1 mol Na 2 CO 3 = 2(23.0g) + 12.0g + 3(16.0g) = 106g The molar mass of Na 2 CO 3 is 106g There are 2 moles of Na in Na 2 CO 3 Mass% Na = 2x23.0g x 100% = 43.4% Na 106g The molar mass of Na 2 CO 3 is 106g There are 3 moles of O in Na 2 CO 3 Mass% O = 3x16.0g x 100% = 45.3% 106g The molar mass of Na 2 CO 3 is 106g There is 1 mole of C in Na 2 CO 3 Mass% C = 1x12.0g x 100% = 11.3% 106g Chromium(III) oxalate is Cr 2 (C 2 O 4 ) 3 MM = 2(52.00) + 6(12.01) + 12(16.00) = 368.1 g/mol Ammonium chloride is NH 4 Cl MM = 12.01 + 4(1.008) + 35.45 = 53.49 g/mol 1mol NH 4 Cl contains 1mol N Mass% = 14.01g/mol x 100 = 26.2% 53.49 g/mol MM of Mg = 24.31 g/mol 0.450g x 1mol = 0.0185mol 24.31g Not elemental nitrogen (N 2 ) In a compound, N is bonded to other atoms, so we only consider the mass of the nitroge MM of N = 14.01 g/mol 0.173g x 1mol = 0.0123mol 12.01g MM = 4(12.01) + 5(1.008) + 2(14.01) + 16.00 = 97.10 g/mol Moles of Carbon 5.657g C x 1mol C = 0.4714mol C 12.01g C Moles of Hydrogen 0.3165g H x 1mol H = 0.3140mol H 1.008g H Moles of Chlorine 5.566g Cl x 1mol Cl = 0.1570mol Cl 35.45g Cl Divide through by smallest molar amount 0.4714mol C = 3.003mol C 0.1570 0.3140mol H = 2.000mol H 0.1570 0.1570mol Cl = 1.000mol Cl 1.570 The empirical formula, CH 2 , has molar mass 12 + 2(1) = 14 g/mol The molar mass of the compound = 82 g/mol The ratio of molar masses is: 82 = 5.9 6 14 The molecular formula is 6 times the empirical formula (CH 2 ) 6 = C 6 H 12 Step1: assume 100.0g sample to find amounts and convert to moles of each element in the sample 12.1g C x 1mol = 1.01mol C 12.01g 16.2g O x 1mol = 1.01mol O 16.00g 71.7g Cl x 1mol = 2.02mol Cl 35.34g The molar ratios are essentially integer values, so no further manipulation is needed %mass = mass solute x 100% mass solution mass solution = mass solvent + mass solute %mass = 2.50g x 100% 50.0g + 2.50g %mass = 4.76% MM = 30.97 g/mol + 5(35.34 g/mol) MM = 208.22 g/mol 10.5g x 1mol = 0.0504mol 208.22g MM = 208.22 g/mol 10.5g x 1mol = 0.0504mol 208.22g 150.0mL x 1L = 0.1500L 1000mL M = 0.0504mol = 0.336 M 0.1500L Molarity = mol V 15.6g KBr x 1mol KBr = 0.131mol 119.0g Molarity = 0.131mol = 0.105mol/L 1.25L Mass% = mass solute x 100 mass solution mass solution = mass solute + mass solvent mass solute = mass solution + mass solvent = 153.4g – 125.2g = 28.2g Mass% = 28.2g x 100 = 18.38% 153.4g Molarity ( M ) = mol L 10.0g NaCl x 1mol = 0.171mol NaCl 58.44g 750.0mL x 1L = 0.7500L 1000mL M = 0.171mol = 0.288 M NaCl 0.7500L Molarity( M ) = mol L Sodium sulfate is Na 2 SO 4 , MM = 142.04 7.31g Na 2 SO 4 x 1mol = 0.0515mol Na 2 SO 4 142.04g 225mL x 1L = 0.225L 1000mL M = 0.0515mol = 0.229 M NaCl 0.225L Molarity ( M ) = mol L Solve for moles then use molar mass to convert to grams. #mol = 0.500mol x 5.91L = 2.96mol CaCl 2 L 2.96mol CaCl 2 x 110.98g = 328g 1mol CaCO 3 K 3 PO 4 Ca(OH) 2 AgCl NaCl M 1 V 1 = M 2 V 2 M 1 = molarity of concentrated solution (4.00 M ) V 1 = volume of concentrated solution (250.0mL) M 2 = molarity of dilute solution (?) V 2 = volume of dilute solution = V 1 + added water V 2 = 250.0mL + 300.0mL = 550.0mL M 2 = M 1 V 1 = 4.00 M x 250.0mL = 1.82 M V 2 550.0mL M 1 V 1 = M 2 V 2 M 1 = molarity of concentrated solution (0.1000 M ) V 1 = volume of concentrated solution (25.00mL) M 2 = molarity of dilute solution (0.025 M ) V 2
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