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Ch. 10: The Mole. * Part of Sec. 13.2: Avogadro’s Principle (pages 452-453) Sec. 10.3: Moles of Compounds (pages 333 – 334 & 338 - 339). Moles of Compounds Avogadro’s Principle. Relate numbers of particles and volumes by using Avogadro’s principle.
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Ch. 10: The Mole *Part of Sec. 13.2: Avogadro’s Principle (pages 452-453) Sec. 10.3: Moles of Compounds (pages 333 – 334 & 338 - 339)
Moles of CompoundsAvogadro’s Principle • Relate numbers of particles and volumes by using Avogadro’s principle. • Recognize the mole relationships shown by a chemical formula. • Determine the number of atoms or ions in a mass of a compound.
Avogadros’ Principle O O O C O The particles making up different gases can vary greatly in size.
Avogadros’ Principle • Yet, the particles in all gases are far enough away from each other that their size has no effect on the amount of space (volume) a sample occupies. • All gases are mostly empty space.
Avogadros’ Principle • That means, at the same temperature and pressure, 1000 oxygen molecules will occupy the same space as 1000 carbon dioxide molecules. They both occupy the same space as 1000 helium atoms. • This is Avogadro’s Principle.
Avogadros’ Principle • Avogadro’s Principle states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. • Further, one mole (or 6.02 x 1023particles) of a gas (no matter what it is) will always have the same volume.
Molar Volume • The molar volume of a gas is the volume occupied by one mole of a gas at STP. • STP stands for standard temperature and pressure: 0o C and 1 atm. pressure. • Avogadro showed that 1 mole of any gas will occupy a volume of 22.4 liters at STP.
Molar Volume OR 1 mole of any gas will occupy 22.4 L of space.
Mole/Volume Conversions • Calculate the volume that 0.881 mol of helium will occupy at STP. 0.881 mol x 22.4 L = 19.7 L 1 mole • Determine the volume of a container that holds 2.4 mol of gas at STP. • How many moles of nitrogen gas will be contained in a 2 L flask at STP?
Molar Volume • Now it is possible to relate moles, mass, # of particles and volume for all gases. • 1 mole = 6.02 x 1023 particles = atomic or formula mass in grams = 22.4 L
MASS PARTICLES Use: 1 mole = atomic mass in grams Use: 1 mole = 6.02 x 1023 particles MOLES Use: 1 mole = 22.4 L VOLUME
Mass/Volume Conversions • Calculate the volume that 50 g of methane (CH4) will occupy at STP. 50 g x 1 mole = 3.1 mol x 22.4 L = 69 L 16 g 1 mol • What volume will 0.416 g of krypton gas occupy at STP? • How many grams of carbon dioxide gas are in a 1.0 L balloon?
Chemical Formulas & the Mole • The formula CCl2F2 (freon) indicates that 1 molecule of CCl2F2 contains 1 C atom, 2 Cl atoms, and 2 F atoms. • What if we multiply all the numbers in this expression by 6.02 x 1023?*** • This means, 1 mole of CCl2F2 will contain 1 mole of C atoms, 2 moles of Cl atoms, and 2 moles of F atoms. • These equivalencies that we find in the formula can be written as conversion factors. They are . . • How many moles of fluorine atoms are in 5.5 moles of freon?
Chemical Formulas & the Mole 5.5 moles CCl2F2 x 2 moles F atoms 1 mole CCl2F2 = 11.0 moles F atoms
Chemical Formulas & the Mole • Conversion factors can be determined for any element or ion from the formula of a compound. • In the conversion factor, the subscript for an element or ion in the formula isthe number of moles of that element or ion in one mole of the compound.
Chemical Formulas & the Mole • Determine the moles of aluminum ions (Al+3) in 1.25 moles of aluminum oxide (Al2O3). • How many moles of oxygen atoms are present in 5.0 mol diphosphorus pentoxide (P2O5)? • Determine the number of moles of oxygen atoms in 0.34 mol of oxygen gas (O2).
Converting Mass to # of Particles Remember: # of Particles Mass
Converting Mass to # of Particles • Now, in addition to the conversions pictured, you can determine the # of atoms or ions in a compound by using a conversion factor that is written from the formula of a compound.
Converting Mass to # of Particles • A sample of silver chromate (Ag2CrO4) has a mass of 25.8 g. a. How many Ag+ ions are present? b. How many CrO42- ions are present? Part A Step 1: mass given to moles compound 25.8 g Ag2CrO4 x 1 mole = 0.0778 mol Ag2CrO4 331.8 g
Converting Mass to # of Particles Step 2: moles compound to moles particles 0.0778 mol Ag2CrO4 x 2 moles Ag+ = 0.156 moles Ag+ 1 mol Ag2CrO4 Step 3: moles particles to # of particles 0.156 moles Ag+ x 6.02 x 1023 Ag+ ions 1 mole Ag+ = 9.39 x 1022 Ag+ ions
Converting Mass to # of Particles • A sample of silver chromate (Ag2CrO4) has a mass of 25.8 g. b. How many CrO42- ions are present? 1. 25.8 g Ag2CrO4 x 1 mole = 0.0778 mol Ag2CrO4 331.8 g 2. 0.0778 mol Ag2CrO4 x 1 mol CrO4-2 ions 1 mol Ag2CrO4 = 0.0778 mol CrO4-2 ions 3. 0.0778 mol CrO4-2 ions x 6.02 x 1023 CrO4-2 ions 1 mol CrO4-2 ions = 4.68 x 1022 CrO4-2 ions
Practice Problems • A sample of aluminum chloride has a mass of 35.6 g. How many Cl- ions are present in the sample? • How many moles of calcium ions are present in 1 g of CaCO3? How many calcium ions are present? • What mass of sodium chloride (NaCl) contains 4.59 x 1024 Cl- ions?