1. Chapter 10 The Mole

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

3. Chemical Quantities Mole – represents a number of items - is the SI base unit used to measure the amount of a - is the SI base unit used to measure the amount of a substance substance Just like a dozen = 12, gross = 144, ream =500 Just like a dozen = 12, gross = 144, ream =500 Mole has 6.02 x10 23 particles of any substance Mole has 6.02 x10 23 particles of any substance Avogadro’s Constant 6.02 x10 23 particles = number of particles in 1 mole of substance 6.02 x10 23 particles = number of particles in 1 mole of substance We use Avogadro’s number as a grouping We use Avogadro’s number as a grouping Particles can be anything – Cars, people, atoms, molecules Particles can be anything – Cars, people, atoms, molecules

4. How big is a mole? 100 average-sized peas 100 average-sized peas Roughly the volume of a cubic inch Roughly the volume of a cubic inch 1,000,000 peas (10 6 ) 1,000,000 peas (10 6 ) Fill a refrigerator Fill a refrigerator Billion (10 9 ) peas Billion (10 9 ) peas Fill a house basement to attic Fill a house basement to attic Trillion (10 12 ) peas Trillion (10 12 ) peas Fill a 1000 houses Fill a 1000 houses

5. How big is a Mole? Quadrillion (10 15 ) peas Quadrillion (10 15 ) peas Fill all the buildings in a larger city like Minneapolis Fill all the buildings in a larger city like Minneapolis Quintillion (10 18 ) peas Quintillion (10 18 ) peas Blanket of peas 4 feet deep all over ND Blanket of peas 4 feet deep all over ND Sextillion (10 21 ) peas Sextillion (10 21 ) peas Earth covered in 4 ft Earth covered in 4 ft Septillion (10 24 ) peas Septillion (10 24 ) peas 250,000 planets covered 4ft deep 250,000 planets covered 4ft deep

6. Molecular Weight Molar mass – of any substance is the mass (in grams) of 1 mole of substance 3 different types Molar Masses Each of the following represents 1 mole: Each of the following represents 1 mole: 1)Gram Atomic mass (gam) 1 mole of atoms (single element) 1 mole of atoms (single element) Can get from periodic table Can get from periodic table write values to the tenth write values to the tenthExamples: Oxygen 1 mole O = 16.0 g Iron1 mole Fe = 55.9 g Cu +2 1 mole Cu +2 = 63.6 g

7. Molecular Weight 2)Gram molecular mass (gmm) 1 mole of molecules –> used with molecular compounds 1 mole of molecules –> used with molecular compounds Add the molar mass of each element multiplied by its subscript Add the molar mass of each element multiplied by its subscriptExample: What is the molar mass or Gram molecular mass of CO 2 ? C12.0g x 1 =12.0g O 16.0g x 2 = 32.0g 44.0g 44.0g 1 mole of CO 2 = 44.0g

8. Molecular Weight 3)Gram formula mass (gfm) 1 mole of formula units –>ionic compounds 1 mole of formula units –>ionic compounds Add the molar mass of each element multiplied by its subscript Add the molar mass of each element multiplied by its subscriptExamples: What is the molar mass or Gram Formula mass of NaCl ? Na23.0g x 1 = 23.0g Cl 35.5g x 1 =35.5g 58.5 g 1 mole of NaCl = 58.5 g

9. Molecular Mass What is the molar mass or Gram Formula mass of Na 2 SO 4 ? Na23.0g x 2 = 46.0g S32.1g x 1 = 32.1g O16.0g x 4 = 64.0g 142.1g 1 mole of Na 2 SO 4 = 142.1g How many molecules of Na 2 SO 4 is there in 1 mole? 6.02 x 10 23 molecules of Na 2 SO 4 = 1 mole Na 2 SO 4

10. Moles to mass conversions 1 mole of substance is equal to its molar mass 1 mole of substance is equal to its molar mass 1 mole of NaCl = 58.5 g NaCl 1 mole of NaCl = 58.5 g NaCl 2.54 moles of NaCl contains how many grams? 2.54 moles of NaCl contains how many grams? 2.54 moles NaCl 1 ( ( 58.5 g NaCl 1 mole NaCl ( ( = 149 g NaCl 252 grams of NaCl is how many moles? 252 grams NaCl 1 ( ( 1 mole NaCl 58.5 grams NaCl ( ( = 4.31 moles NaCl

11. Conversions – Mass/grams/particles How many particles (Formula Units) of NaCl are there in 252 grams of NaCl How many particles (Formula Units) of NaCl are there in 252 grams of NaCl 252 grams NaCl 1 ( ( 1 mole NaCl 58.5 grams NaCl ( ( = 2.59 x 10 24 FU NaCl 6.02 x 10 23 FU of NaCl 1 mole NaCl ( ( 3.50 x 10 24 molecules of NO 3 has a mass of how many grams? 3.50 x 10 24 molecules of NO 3 has a mass of how many grams? 3.50 x 10 24 molecules of NO 3 1 ( ( 6.02 x 10 23 molecules of NO 3 1 mole NO 3 ( ( 62.0 grams NO 3 ( ( = 360. grams NO 3

12. Flow chart for conversions Number of particles Atoms Atoms Molecules Molecules Formula Units Formula Units Mass of Mass ofCompound Number of Moles of compound of compound MolarMass Avagadro’s Number Number

13. Volume of a Mole of a Gas Volume of 1 mole of gas varies with Volume of 1 mole of gas varies with Δ in temperature Δ in temperature Δ in pressure Δ in pressure Gases usually measured @ Standard temp and pressure (STP) Gases usually measured @ Standard temp and pressure (STP) 0° and 101.3 KPa or 1 ATM 0° and 101.3 KPa or 1 ATM 1 Mole of any gas occupies a volume of 22.4L 1 Mole of any gas occupies a volume of 22.4L 22.4 L Know as Molar Volume of Gas 22.4 L Know as Molar Volume of Gas

14. Mole to a gas Example: 25 g of Neon = ? L 25 g Ne 1 ( ( 1 mole Ne 20.2 g Ne ( ( 22. 4 L Ne 1 mole Ne ( ( = 28 L Ne

15. 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 you are provided amounts you are provided amounts % mass of element = grams of element x 100 grams of compound % mass of element = grams of element x 100 grams of compoundExample: Find the percent composition of the compound that is formed from 128.0 g of sulfur and 192.0 g of oxygen.

16. 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 192.0 g +128.0 g 320.0 g Total mass of compound % mass of S = 128.0 g x 100 = 320.0 g 40.00 % S % mass of O = 192.0 g x 100 = 320.0 g 60.00 % O

17. 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 Find the mass Find the mass Divide it grams of element by molar mass of compound Divide it grams of element by molar mass of compound Multiply by 100 Multiply by 100 %mass = mass of 1 mole of element x 100 molar mass of compound molar mass of compoundExample Find the percent composition of Cu, SO 4 and H 2 O in CuSO 4  2H 2 O Find the percent composition of Cu, SO 4 and H 2 O in CuSO 4  2H 2 O

18. Empirical Formula Empirical Formula – lowest whole-number ratio of elements in a compound Example C 3 H 6 OEmpirical formula C 3 H 6 OEmpirical formula C 6 H 12 O 2 Molecular formula C 6 H 12 O 2 Molecular formula The percentage composition of diborane is 78.1% B and 21.9% H Determine the empirical formula.

19. Empirical Formula Step 1 Step 1 Change % to grams Change % to grams 78.1% B = 78.1 g B 21.9% H = 21.9 g H 78.1% B = 78.1 g B 21.9% H = 21.9 g H Step 2 Step 2 Change the grams to moles for each element Change the grams to moles for each element (78.1 g B ) (1 mole B ) = 7.23 mole B 1 (10.8 g B) 1 (10.8 g B) (21.9 g H) (1 mole H ) = 22 mole H 1 (1.0 g H ) 1 (1.0 g H )

20. Empirical Formula Step 3 Step 3 Divide each amount by the smallest amount of moles Divide each amount by the smallest amount of moles Use these Ratios to determine the Empirical Formula Use these Ratios to determine the Empirical Formula 7.23 moles = 1.00 B 22 moles = 3.0 H 7.23 moles7.23 moles BH 3