Today’s focus. Mole Concept Avogadro’s Number is 6.02x1023 The mole unit is used to express: 1. A mass quantity 2. A counting quantity 1 water molecule 1 mole of water molecules Conversion Factors: Conversion Factors: 6.02x1023 amu = 1 gram (Scale up so that we can weigh in the lab) (Scale up so that we can weigh in the lab) 6.02x1023 = 1 mole water molecules of water molecules amu molecule g mole Information in Chemical Formulae Sodium carbonate N 2CO3 Na What are the parts p needed to make ONE formula unit of sodium carbonate? 22.990 12.011 15.999 Na C O 11 6 8 Information in Chemical Formulae N 2CO3 Na N Na : C : O C O 2 2 : 1 : 3 1 3 Information in Chemical Formulae N b Number of ratio of ions: (parts) f i fi ( ) N 2CO3 Na 2 : 1 Particles can be: # of atoms, Parts of a formula unit Parts of a formula unit # of ions, # of molecules, # of formula units Conversion Factor Conversion Conversion Factor * * * Conversion Conversion Factor * Particles can be: # of atoms, Parts of a formula unit Parts of a formula unit # of ions, # of molecules, # of formula units *In a problem, look for In a problem look for unit given and unit sought. Q: How many hydrogen atoms are in y y g 54.15 g of ammonium phosphate? (NH4)3PO4 Dimensional Analysis: Dimensional Analysis: Unit given x conversion = unit sought factor(s) Do you expect the answer to be a big number or a small number? Conversion Factor: Molar Mass = 149.05 Conversion Factor: Molar Mass = 149 05 g/mole 2 1 4 A: 2.62x1024 H atoms x H atoms ( NH ) PO 3 4 ( NH ) PO mole 4 x 3 4 . 3 x 2 0 1 4 3 4 . 2 0 6 5 0 1 9 4 1 54.15 g (NH4)3PO4 x mole g ( NH ) PO = ? H atoms This is an example of a theoretical calculation of Percent Composition of the elements in a the chemical formula. Percent Composition (NH4)3PO4 Element Formula Mass = 149.05 g/mole Mass Contribution to the formula mass Percent Composition p g mole g mole x g mole g mole l x 5 0 9 4 1 14.007 x 3 = 42.021 14 007 x 3 = 42 021 g/mole . 0 0 1 1 2 0 2 4 Nitrogen . 5 0 9 4 1 Hydrogen 1.008 x 12 = 12.10 g/ . mole 0 0 1 0 1 2 1 . = 28.193% 30.9738 x 1 = 30.9738 g/mole 5 0 9 4 1 Phosphorus g mole g mole . 0 0 1 8 3 7 9 0 3 . x 5 0 9 4 1 15.999 x 4 = 63.996 g/mole g mole g mole . = 20.781% 0 0 1 6 9 9 3 6 Oxygen . = 8.118% x The percent composition of nitrogen in (NH4)3PO4 is 28.193%. etc. = 42.936% 100% Percent Composition Difference in mass = mass of oxygen Experimentally: ll Start with KClO3 of a certain mass + After heating, solid remains should have less mass Oxygen escaped Empirical Formula The simplest formula for a compound that has the SIMPLEST whole Th i l f l f d h h h SIMPLEST h l number ratio of the atoms present. Example: Water Example: Water H2 O H2O is the chemical formula of water. 2 : 1 Since this is the SIMPLEST whole number ratio of H:O. Since this is the SIMPLEST whole number ratio of H:O Therefore, H2O is also the empirical formula for water. Empirical Formula The simplest formula for a compound that has the SIMPLEST whole p p number ratio of the atoms present. Example: Hydrogen peroxide Example: Hydrogen peroxide H2O2 H2O2 is the chemical formula of hydrogen peroxide. 2 : 2 The chemical formula is NOT the SIMPLEST whole number ratio of H:O. Therefore, the empirical formula for hydrogen peroxide is HO. Empirical Formula The simplest formula for a compound that has the SIMPLEST whole p p number ratio of the atoms present. Chemical formula of the compound of the compound H:O Ratio Empirical Formula Multiple H2O water 2:1 H2O 1x H2O2 hydrogen peroxide 22 2:2 HO 2 2x What do we know from this: 1. Sometimes the chemical formula IS the empirical formula, sometimes it is not. 2. The chemical formula IS a multiple of the empirical formula. When the chemical formula is the empirical formula, the multiple is one. 3. If we know the empirical formula and we know the molar mass of the compound, then we can determine the chemical formula of the compound! From Empirical Formula to Chemical Formula Question: Q i H Hexane has a molar mass of 86.05 h l f 86 05 g/mole. The Th empirical formula of hexane is C3H7. What is the chemical formula of hexane? Chemical formula of the compound C:H Ratio Empirical Formula Multiple Hexane ? 3:7 C3H7 2 A: C6H14 x2 Empirical formula has a mass of: (12.011 x 3) + (1.008x7) = 43.027 g/mole. . . g mole g mole 7 2 0 3 4 molar mass of compound Multiple empirical formula mass 5 0 6 8 To determine the multiple, =2 Percent Composition Empirical Formula Chemical Formula Road Map Question: A sample was sent to the lab and the analysis comes back showing us g p that the following mass percent of the elements was found. What is the empirical formula? 15.88% Carbon 2.22 % Hydrogen 63.42% Oxygen 18.50% Nitrogen 1 . . . . (Nitroglycerine or TNT) 53 . 3 Answer: C3H5O9N3 7 6 6 1 = 1.320 moles 1 mole . g . . x 3 = 3 00 22 33 11 g l mole = 3.964 moles . g Divide through by the smallest ll t number 40 62 93 31 . = 2.203 moles 7 0 0 4 1 18.50 g 0 5 8 1 N g mole . g . . 0 22 23 21 . = 1.322 moles 9 9 9 5 1 63.42 g 63.42 g 2 4 3 6 O g . mole . g 0 2 3 1 2.22 g g Mole Ratio 2 2 3 1 H . 8 0 0 1 15.88 g 1 1 0 2 1 C Moles of Elements 2 2 2 100 gg 8 8 5 1 Elements x 3 = 9 x 3 = 3 x 3 3 x 3 = 5 Do the Percent Composition, Empirical h l Formula, Chemical Formula Worksheet. Answers will be posted later in the week. p Try some practice problems in Maple TA. Use Dimensional Analysis in ALL your calculations!!