Chemical Quantities Chapter 7 (10)

How many dozen objects are present when you have: a) 60 objects: 5 dozen b) 3 objects: 1/4 dozen or 0.25 dozen c) 41 objects: 3.4 dozen d) 2.5 objects: 0.21 dozen What did you use as the basis for the conversions?

1 dozen objects = 12 objects 12 objects = 1 dozen 1 dozen 60 objects x = 5 dozen 12 objects 1 dozen 2.5 objects x = .21 dozen 12 objects

How many objects are present when you have: a) 6.0 dozen: 72 objects b) 3.2 dozen: 38.4 objects c) 0.025 dozen: 0.30 objects d) 3.25 x 10-3 dozen: 0.0390 objects What did you use as the basis for the conversions?

1 dozen objects = 12 objects 12 objects = 1 dozen 6.0 dozen x = 72 objects 1 dozen 12 objects = 0.30 objects 0.025 dozen x 1 dozen

Solve these problems: a) 1.3 x 102 dozen: 1.56 x 103 objects b) 5.5 objects: 0.458 dozen c) 2.5 x 10-4 objects: 2.08 x 10-5 dozen d) 7.4 x 10-1 dozen: 8.88 objects

1 object 90 g x = 9 objects 10 g 12 objects 10 g 3.0 dozen = 360 g If each object weighs 10 grams, how many objects do you have when you have 90 grams worth? 1 object 90 g x = 9 objects 10 g If each object weighs 10 grams, how many grams of objects do you have when you have 3.0 dozen objects? 12 objects 10 g 3.0 dozen = 360 g 1 dozen 1 object

Solve these problems: a) How many dozen objects are present when you have 100. g of objects? Each dozen weighs 32 g. 1 dozen 100. g = 3.13 dozen 32 g

Solve these problems: b) How many objects are present when you have 100. g of objects? Each dozen weighs 32 g. 1 dozen 100. g 12 objects = 37.5 objects 32 g 1 dozen

Solve these problems: c) How much do 1.25 dozen objects weigh if each dozen weighs 32 g? 1.25 doz 32 g = 40.0 grams 1 dozen

I. The Mole Concept 1 mole = 6.02 x 1023 Avogadro’s Number A. Just as “1 dozen = 12”, chemistry uses this ratio: 1 mole = 6.02 x 1023 B. This number is frequently referred to as: Avogadro’s Number

1) How many atoms of carbon are present in 0.250 moles of carbon? 6.02 x 1023 atoms 1 mole = 1.51 x 1023 atoms

2) How many moles of carbon atoms are present in 5 2) How many moles of carbon atoms are present in 5.00 x 1010 atoms of carbon? 1 mole 5.00 x 1010 atoms 6.02 x 1023 atoms = 8.31 x 10-14 mol

a) What is the mass of 0.330 mol of carbon? 3) The weight of 1 mole of an element (6.02 x 1023 atoms of that element) = it’s atomic weight a) What is the mass of 0.330 mol of carbon? 0.330 mole C 12.011 g = 3.96 g 1 mole 14

b) How many moles are present in 1.00 x 102 grams of carbon? c) How many atoms of carbon are present in 2.00 grams of carbon? 2.00 g 1 mole 6.02 x 1023 atoms 12.011 g 1 mole = 1.00 e23 15

d) How much do 3.0 x 109 atoms of carbon weigh? 1 mole 6.02 x 1023 atoms 1 mole = 6.0 x 10-14 g 16

e) How many grams do 0.125 moles of H2O weigh? 0.125 mol H2O 18 g = 2.25 g 1 mole H2O f) How many molecules of water are present in 90. grams of water? 90. g H2O 6.02 x 1023 molec 1 mole H2O 18 g 1 mole H2O = 3.0 x 1024 molecules 17

II. Percent Composition A. Percent = Part x 100 whole Ex: What is the percent boys in a class made up of 5 boys and 15 girls? % boys = 5 boys x 100 = 25% boys 20 total 18

Ex: What is the percent oxygen in H2O, by weight? 16 g oxygen x 100 18 g total = 88.9% oxygen 19

III. Empirical and Molecular Formulas A. The subscripts in a chemical formula are the mole ratios of atoms Ex: H2O has 2 moles of hydrogen : 1 mole of oxygen in every 1 mole of water Ex: C6H12O6 has 6 moles of carbon : 12 mole of hydrogen: 6 moles of oxygen in every 1 mole of C6H12O6 20

B. If the percent composition is known, the molar ratio of each element can be calculated: Ex: A sample is known to consist of 88.9% O and 11.1% H. What is the formula of this compound? Div by a.w. Div by smallest H2O 88.9 O 11.1 H 1 5.56 11.1 2 21

(Smallest whole number ratio) Ex: A sample is known to consist of 40.0% C and 6.70% H and the rest is oxygen. What is the empirical formula of this compound? If the molecular weight is 180 g, what is the molecular formula? Div by a.w. Div by smallest CH2O Empirical Formula 40.0 C 6.70 H 53.3 O 1 3.30 (Smallest whole number ratio) 6.70 2 3.30 1 Molecular Formula = C6H12O6 22