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The Mole

The Mole. Molar mass, percent composition, empirical and molecular formulas. How many atoms are in here? What about weight?. Atomic Mass. A single atom has a very small mass. on the order of 10 -23 grams per atom 1 atom has a mass of about 0.000 000 000 000 000 000 000 01g

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The Mole

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  1. The Mole Molar mass, percent composition, empirical and molecular formulas

  2. How many atoms are in here? What about weight? .

  3. Atomic Mass • A single atom has a very small mass. • on the order of 10-23 grams per atom • 1 atom has a mass of about 0.000 000 000 000 000 000 000 01g • Because this mass is so small, we use a unit called amu to describe the mass of a single atom. • 1 atom of carbon has a mass of 12.011 amu • 1 atom of hydrogen has a mass of 1.00794 amu • 1 atom of gold has a mass of 196.967 amu

  4. Formula Mass • Formula Mass – the mass, in amu, of one formula unit of a compound. • To calculate the formula mass of a compound; • Count and record the number of each atom in the formula. • Multiply the number of atoms by the mass of that atom from the periodic table. • Add these products together.

  5. Definitions: • Atom: The smallest particle of an element (ex: He). • Molecule: The smallest particle of a compound (ex: H2O) • Atomic Weight: The decimal number on the periodic chart (Units: amu) • Formula Mass: add up the total atomic weights of the elements in a compound (Units: amu).

  6. Formula Mass Calculate the formula mass of Ca • Ca = 1(Ca) • Ca = 1(40.08) = 40.08 • 40.08 amu • Units = amu Cl Ca Cl

  7. Formula Mass Calculate the formula mass of CaCl2 • CaCl2 = 1(Ca) + 2(Cl) • CaCl2 = 1(40.08) + 2(35.45)= 110.98 • 110.98 • Units = amu Ca

  8. The Mole • The mass of a single atom is too small a quantity for use in lab. • We need to come up with a way to take atomic masses from the periodic table and turn them into a mass that is useable in lab. • In 1811, Amedeo Avogadro determined the number of atoms in one mole of a substance.

  9. The Mole • The mole establishes a relationship between the atomic mass unit (on periodic table) and the gram (used in lab). • Mole (mol) - The amount of a pure substance that contains 6.02 x 1023 particles of that substance. • The mole is used to describe a huge amount of any extremely small particle. A mole of gold, a mole of salt and a mole of water each contain 6.02 x 1023 individual units.

  10. The Mole • One mole of oxygen, one mole of salt and one mole of water. • Each sample contains one mole, 6.02 x 1023, particles.

  11. Do you think that one mole of gold atoms is going to weigh the same as one mole of aluminum atoms? Why?

  12. It doesn’t make much sense to weigh atoms in amu, because the units are so small…… • If you express atomic mass (amu) in GRAMS instead of amu (like we did yesterday), than you have one MOLE of atoms.

  13. Molar Mass • A molar mass is calculated the same way that you calculated formula mass (HUP, TWO, THREE, FOUR). • The only difference is the unit that will be placed on the final answer because you are calculating the mass of one mole of a substance. • If the formula mass of CaCl2 is 110.98 amu, then the molar mass of CaCl2 is 110.98 grams.

  14. Formula vs. Molecular Mass

  15. The Mole • There are 6.02 x 1023 carbon atoms in 12 grams, one mole, of carbon. • There are 6.02 x 1023 gold atoms in 197 grams, one mole, of gold. • There are 6.02 x 1023 formula units of calcium chloride in 110.98 grams, one mole, of CaCl2.

  16. So, how do we use the mole in Chemistry? • We use the mole to convert between the mass, volume and number of particles in a substance. • We can make conversions from: • Moles to particles and particles to moles. • Moles to grams and grams to moles. • Moles to volume of a gas and volume of a gas to moles. • Any combination of the above categories!

  17. Molar Conversions

  18. Converting from Moles to Particles. • How many particles of Sodium Chloride are there in 2.50 moles of Sodium Chloride (NaCl)? 2.50 mol NaCl x 6.02 x 1023part. = 1.51 x 1024 1 mole part. NaCl

  19. Converting from Moles to Particles • How many particles of gold are there in 4.25 moles of Au? 4.25 mol Au x 6.02 x 1023 part.= 2.56x1024 part 1 mole Au

  20. Converting from Particles to Moles. • How many moles of gold are there in 1.5 x 1024 atoms of gold? 1.5 x 1024 atoms Au x 1 mole = 2.5 moles Au 6.02 x 1023 atoms

  21. Converting from Particles to Moles • How many moles of calcium chloride are there in 3.6 x 1024 particles of CaCl2? 3.6 x 1024 units x 1 mole CaCl2 = 6.0 mol CaCl2 6.02 x 1023 part.

  22. Converting from Mass to Moles • How many moles of gold are there in 25.0 grams of gold? 25.0 g Au x 1 mole Au = 0.127 mol Au 196.97 g Au

  23. Converting from Mass to Moles • How many moles of calcium chloride are there in 75.5 grams of CaCl2? 75.5 g CaCl2 x 1 mole CaCl2 = 0.680mol CaCl2 110.98 g CaCl2

  24. Converting from Moles to Mass • How many grams of Carbon are there in 4.2 moles of Carbon? 4.2 mole C x 12.01 g C = 50. g C 1 mole C

  25. Converting from Moles to Mass • How many grams of water are there in 2.75 moles of water? The molar mass of water is: 16.00 + 1.01 + 1.01 = 18.02 g/mol 2.75 moles H2O x 18.02 g H2O = 49.6 g H2O 1 mole H2O

  26. Volume of Gases • One mole of gas (6.02 x 1023 particles of gas) occupies a volume of 22.4 liters at standard temperature and pressure. • This is true if the gas is monatomic like Helium (He), diatomic like Oxygen (O2) or a compound like Carbon dioxide (CO2).

  27. Converting from Volume to Moles of a gas • How many moles of oxygen are there in 10.0L of oxygen? 10.0L O2 x 1 mole O2 = 0.446 mole O2 22.4 L O2

  28. Converting from Volume to Moles of a gas • How many moles of Carbon dioxide (CO2) are there in 2.50 L of CO2 gas? 2.50 L CO2 x 1 mole CO2 = 0.112 mole CO2 22.4 L CO2

  29. Converting from Moles to Volume of a gas • How many liters of Helium are there in 3.6 moles of Helium? 3.6 moles He x 22.4 L He = 81 L He 1 mole He

  30. Converting from Moles to Volume of a gas • What volume will 8.24 moles of CO2 occupy? 8.24 mol CO2 x 22.4 L CO2 = 185 L CO2 1 mole CO2

  31. Multi-Step Conversions • If you need to convert from: • Mass to particles or particles to mass, • Mass to volume or volume to mass, • Volume to particles or particles to volume, then you need to perform a 2-step conversion. • Notice the mole is not present in the starting or ending quantities. You will ALWAYS use the mole when converting between these units!

  32. Multi-Step Conversions • How many formula units of calcium chloride are there in 10.0 grams of CaCl2? 10.0 g CaCl2 x 1 mole CaCl2 x 6.02 x 1023 units = 110.98 g CaCl2 1 mole CaCl2 5.42 x 1022 formula units of CaCl2

  33. Multi-Step Conversions • What volume would 2.25 grams of hydrogen gas (H2) occupy at standard temperature and pressure? 2.25 g H2 x 1 mole H2 x 22.4 L = 25.0 L H2 2.02 g H2 1 mole

  34. Multi-Step Conversions • How many atoms of Argon gas are there in a typical light bulb (0.20 liters) at standard temperature and pressure? 0.20 L Ar x 1 mole x 6.02 x 1023 atoms = 5.4 x 1021 22.4 L 1 mole atoms Ar

  35. Percent Composition • The Percent Composition – the percent by mass of each element in a compound. • To calculate %Comp: mass of element x 100 = % by mass mass of compound

  36. Percent Composition • Determine the percent by mass of each element in calcium chloride (CaCl2). Ca 1 x 40.08 = 40.08 40.08 x 100 = 36.11% Ca 110.98 Cl 2 x 35.45 = 70.90 70.90 x 100 = 63.89% Cl 110.98 40.08 + 70.90 =110.98 (molar mass)

  37. Percent Composition • Calculate the percent by mass of each element in Calcium sulfate, CaSO4. Ca 1 x 40.08 = 40.08 40.08 x 100 = 29.44% Ca 136.15 S 1 x 32.07 = 32.07 32.07 x 100 = 23.55% S 136.15 O 4 x 16.00 = 64.00 64.00 x 100 = 47.01% O 136.15 40.08 + 32.07 + 64.00 = 136.15 molar mass

  38. Percent Composition • You can determine the %Comp of a substance experimentally. • Measure the mass of a sample • Decompose the sample (usually by heating) to separate the component substances. • Measure the mass of the substance that remains. • Calculate %Comp as before

  39. Empirical Formula • Empirical Formula – the smallest whole number mole ratio of elements in a compound. • The empirical formula for a class of molecules is often the same for each sample. • Examples: C5H10O5 = CH2O C6H12O6 = CH2O C11H22O11 = CH2O

  40. Calculating Empirical Formula Assume you are working with a 100.0g sample so the mass of each element will be the same as the percent of that element (for simplicity). Convert each mass in grams to moles using the molar mass of the element. Find the whole number ratio of these calculated amounts by dividing each mole value by the smallest number of moles calculated. Round to whole numbers and use the ratio to determine the empirical formula.

  41. Calculating Empirical Formula • A compound was found to contain 29.6% Calcium, 23.7% Sulfur and 46.8% Oxygen. What is the empirical formula for the compound? • If you had a 100 gram sample, you would have 29.6g Ca, 23.7g S and 46.8g O.

  42. Calculating Empirical Formula Ca 29.6g x 1 mole = 0.739 mol Ca 40.08g S 23.7g x 1 mole = 0.739 mole S 32.07g O 46.8 x 1 mole = 2.93 mole O 16.00g 0.739 = 1 0.739 = 1 2.93 = 3.96 0.739 0.739 0.739 Round to the nearest whole numbers, 1 : 1 : 4 Therefore the empirical formula is CaSO4

  43. Molecular Formulas • Molecular Formula – the actual number of atoms of each element in one molecule of a compound. • The ratio of molecular mass to empirical mass is the same as the ratio of molecular formula to empirical formula.

  44. Calculating Molecular Formula • Example: Molecular formula Molecular Mass C6H12O6 180.18 Empirical formula Empirical Mass CH2O 30.03 Ratio C6H12O6 = 6 Ratio 180.18 = 6 CH2O 30.03

  45. Calculating Molecular Formula • Ribose has a molar mass of 150g/mol and a chemical composition of 40.0%C, 6.67% hydrogen and 53.3% oxygen. What is the molecular formula for ribose? • Begin by calculating the empirical formula for Ribose.

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