Chapter 10 The Mole Objectives n Explain

Chapter 10 The Mole

Objectives n Explain how a mole is used to indirectly count the number of particles of matter n Relate the mole to common everyday counting unit n Calculate the molar mass of a compound n Convert between moles and number of representative particles n Convert between number of moles and the mass of an element n Convert between number of moles and number of atoms of an element

Chemical Quantities Mole – represents a number of items - is the SI base unit used to measure the amount of a substance n Just like a dozen = 12, gross = 144, ream =500 n Mole has 6. 02 x 1023 particles of any substance Avogadro’s Constant n 6. 02 x 1023 particles = number of particles in 1 mole of substance n We use Avogadro’s number as a grouping n Particles can be anything – Cars, people, atoms, molecules

How big is a mole? n n 100 average-sized peas n Roughly the volume of a cubic inch 1, 000 peas (106) n n Billion (109) peas n n Fill a refrigerator Fill a house basement to attic Trillion (1012) peas n Fill a 1000 houses

How big is a Mole? n Quadrillion (1015) peas n n Quintillion (1018) peas n n Blanket of peas 4 feet deep all over ND Sextillion (1021) peas n n Fill all the buildings in a larger city like Minneapolis Earth covered in 4 ft Septillion (1024) peas n 250, 000 planets covered 4 ft deep

Molecular Weight Molar mass – of any substance is the mass (in grams) of 1 mole of substance 3 different types Molar Masses n Each of the following represents 1 mole: 1) Gram Atomic mass (gam) n 1 mole of atoms (single element) n Can get from periodic table n write values to the tenth Examples: Oxygen 1 mole O = 16. 0 g Iron 1 mole Fe = 55. 9 g Cu+2 1 mole Cu+2 = 63. 6 g

Molecular Weight 2) Gram molecular mass (gmm) n 1 mole of molecules –> used with molecular compounds n Add the molar mass of each element multiplied by its subscript Example: What is the molar mass or Gram molecular mass of CO 2 ? C 12. 0 g x 1 = 12. 0 g O 16. 0 g x 2 = 32. 0 g 44. 0 g 1 mole of CO 2 = 44. 0 g

Molecular Weight 3) Gram formula mass (gfm) n 1 mole of formula units –>ionic compounds n Add the molar mass of each element multiplied by its subscript Examples: What is the molar mass or Gram Formula mass of Na. Cl ? Na Cl 23. 0 g x 1 = 23. 0 g 35. 5 g x 1 = 35. 5 g 58. 5 g 1 mole of Na. Cl = 58. 5 g

Molecular Mass What is the molar mass or Gram Formula mass of Na 2 SO 4 ? Na S O 23. 0 g x 2 = 46. 0 g 32. 1 g x 1 = 32. 1 g 16. 0 g x 4 = 64. 0 g 142. 1 g 1 mole of Na 2 SO 4 = 142. 1 g How many molecules of Na 2 SO 4 is there in 1 mole? 6. 02 x 1023 molecules of Na 2 SO 4 = 1 mole Na 2 SO 4

Moles to mass conversions 1 mole of substance is equal to its molar mass n n 1 mole of Na. Cl = 58. 5 g Na. Cl 2. 54 moles of Na. Cl contains how many grams? 1 ( ( 2. 54 moles Na. Cl ( 58. 5 g Na. Cl 1 mole Na. Cl ( n = 149 g Na. Cl 252 grams of Na. Cl is how many moles? ( 1 mole Na. Cl 58. 5 grams Na. Cl ( ( 252 grams Na. Cl = 4. 31 moles Na. Cl

Conversions – Mass/grams/particles 1 ( 252 grams Na. Cl ( 1 mole Na. Cl 58. 5 grams Na. Cl ( ( How many particles (Formula Units) of Na. Cl are there in 252 grams of Na. Cl ( 6. 02 x 1023 FU of Na. Cl 1 mole Na. Cl ( n = 2. 59 x 1024 FU Na. Cl 1 ( 3. 50 x 1024 molecules of NO 3 1 mole NO 3 ( 6. 02 x 10 23 ( ( 3. 50 x 1024 molecules of NO 3 has a mass of how many grams? molecules of NO 3 = ( 62. 0 grams NO 3 1 mole NO 3 360. grams NO 3 ( n

Flow chart for conversions Mass of Compound Number of Moles of compound Molar Mass Avagadro’s Number of particles n Atoms n Molecules n Formula Units

Volume of a Mole of a Gas n n n Volume of 1 mole of gas varies with n Δ in temperature n Δ in pressure Gases usually measured @ Standard temp and pressure (STP) n 0° and 101. 3 KPa or 1 ATM 1 Mole of any gas occupies a volume of 22. 4 L n 22. 4 L Know as Molar Volume of Gas

Mole to a gas Example: 25 g of Neon = ? L 20. 2 g Ne ( ( 1 mole Ne ( 22. 4 L Ne 1 mole Ne ( ( 25 g Ne = 28 L Ne

Percent Composition of a Compound Percent composition – amounts, by % mass, of each element in a compound 2 types of problems 1) Percent composition – relative amounts n you are provided amounts n % mass of element = grams of element x 100 grams of compound Example: Find the percent composition of the compound that is formed from 128. 0 g of sulfur and 192. 0 g of oxygen.

Example: Find the percent composition of the compound that is formed from 128. 0 g of sulfur and 192. 0 g of oxygen. 192. 0 g +128. 0 g 320. 0 g Total mass of compound % mass of S = 128. 0 g x 100 = 40. 00 % S 320. 0 g % mass of O = 192. 0 g x 100 = 320. 0 g 60. 00 % O

Percent Composition of a Compound 2) Percent composition of known compound You are given the composition Step 1 – use the chemical formula to find molar mass Step 2 – for each element n Find the mass n n %mass Divide it grams of element by molar mass of compound Multiply by 100 = mass of 1 mole of element x 100 molar mass of compound Example n Find the percent composition of Cu, SO 4 and H 2 O in Cu. SO 4 2 H 2 O

Empirical Formula – lowest whole-number ratio of elements in a compound Example n C 3 H 6 O n C 6 H 12 O 2 Empirical formula Molecular formula The percentage composition of diborane is 78. 1% B and 21. 9% H Determine the empirical formula.

Empirical Formula n n Step 1 n Change % to grams n 78. 1% B = 78. 1 g B 21. 9% H = 21. 9 g H Step 2 n Change the grams to moles for each element (78. 1 g B ) (1 mole B ) = 7. 23 mole B 1 (10. 8 g B) (21. 9 g H) (1 mole H ) = 22 mole H 1 (1. 0 g H )

Empirical Formula n Step 3 n Divide each amount by the smallest amount of moles n Use these Ratios to determine the Empirical Formula 7. 23 moles = 1. 00 B 7. 23 moles BH 3 22 moles = 3. 0 H 7. 23 moles