CHEM161: Chapter 3 – Independent Review page 1 of 6 Chapter 3: CHEMICAL REACTIONS AND EARTH’S COMPOSITION (Topics to Review) 3.2 The Mole Avogadro’s Number (NA) = 6.022×1023 (to 4 sig figs) How big is this? • If 6.022×1023 hydrogen atoms were laid side by side, the total length would encircle the earth about a million times. • The mass of 6.022×1023 Olympic shotput balls is about equal to the mass of the Earth. • The volume of 6.022×1023 softballs is about equal to the volume of the Earth. 1 mole (abbreviated mol) = 6.022×1023 entities Similar to: 1 dozen = 12 entities: 1 dozen doughnuts = 12 doughnuts 1 mole of doughnuts = 6.022×1023 doughnuts Ex. 1 How many eggs are in 3 dozen eggs? 3 x 12 eggs = 36 eggs Ex. 2 How many eggs are in 3 moles of eggs? 3 x 6.022x1023 eggs = 1.807x1024 eggs Ex. 3 How many moles of carbon contain 7.25×1024 carbon atoms? 7.25x1024 C atoms =××atoms C 106.022C mol 12312.0 mol C Atomic weights and molar masses: – The mass of 1 C atom (on average) is 12.01 amu – The mass of 1 mole of C atoms is 12.01 g (or 12.01 g/mol) → 1 mole (6.022×1023) is the amount of atoms of any element that has a mass in grams equal to the mass of ONE atom in amu. → The atomic masses reported in the Periodic Table give the atomic weight (or molecular/formula weight for compounds) in amu and the molar mass in g/mol.CHEM161: Chapter 3 – Independent Review page 2 of 6 MOLAR MASS (MM): Mass in grams of 1 mole of any element/compound – To obtain, multiply the molar mass of each element by the number of each present, then add up all the constituent parts. Example: Determine the molar mass of each of the following compounds: a. O2: 2 (molar mass of O) = 2 (16.00 g/mol) = 32.00 g/mol b. H3PO4: H: 3(1.008) c. Al2(SO4)3: Al: 2(26.98) P: 30.97 S: 3(32.07) O: 6(16.00) O: 12(16.00) 97.994 g/mol 310.18 g/mol Note: Don’t worry about rounding molar masses since they rarely limit the number of sig figs for your final calculated answer. Mole Calculations Ex. 1 How many moles of Ne are in 0.500 g Ne? 0.500 g Ne =×Ne g 20.18Ne mol 10.0248 mol Ne Ex. 2 How many Ne atoms are in 0.500 g of Ne? 0.500 g Ne Ne g 20.18Ne mol 1×=××Ne mol 1atoms Ne 106.022231.49x1022 Ne atoms Ex. 3 How many moles of CO2 are in 0.500 g of CO2? 0.500 g CO2 =×22CO g 44.01CO mol 10.0114 mol CO2 Ex. 4 How many CO2 molecules are in 0.500 g of CO2? 0.500 g CO2 22CO g 44.01CO mol 1×=××2223CO mol 1molecules CO 106.0226.84x1021 CO2 molecules Ex. 5 How many oxygen atoms are in 0.500 g of CO2? 0.500 g CO2 22CO g 44.01CO mol 1×2223CO mol 1molecules CO 106.022 ××molecule COatoms O 22× = 1.37x1022 O