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Chemistry Unit 8a

Chemistry Unit 8a. The Mole Molar mass, percent composition and empirical formulas. Atomic Mass. A single atom has a very small mass. (on the order of 10 -23 grams per atom)

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Chemistry Unit 8a

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  1. Chemistry Unit 8a The Mole Molar mass, percent composition and empirical formulas

  2. Atomic Mass • A single atom has a very small mass. (on the order of 10-23 grams per atom) • Because this mass is so small, we use a unit called amu to describe the mass of a single atom. (for example, 1 atom of carbon has a mass of 12.011 amu) • Average atomic mass is the weighted average of the masses of all isotopes of an element.

  3. Atomic Mass • atomic mass units are not practical for use in the lab (too small a quantity) • grams are the preferred unit of mass • Therefore, scientists needed a way to determine the number of atoms in a given mass of an element.

  4. The Mole • Carbon-12 was selected as the standard. The number of atoms in 12.0g of C-12 was determined experimentally using sophisticated equipment. • The number came out to be 6.02 x 1023 • This number is called a mole (mol)

  5. The Mole • The amount of a pure substance that contains 6.02 x 1023 particles of that substance. • Huh? Try it this way… • There are 6.02 x 1023 carbon atoms in 12 grams of carbon. • There are 6.02 x 1023 hydrogen atoms in 1.0grams of hydrogen. • There are 6.02 x 1023 oxygen atoms in 16.0 grams of oxygen. • There are 6.02 x 1023 gold atoms in 197 grams of gold.

  6. The Mole • The mole establishes a relationship between the atomic mass unit and the gram. • 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. • How would you feel about inheriting a mole of pennies??? $6.02 x 1021 or $6,020,000,000,000,000,000,000 • If you gave $1million a day to every person on Earth, it would take you > 3000 years to run out of money.

  7. The Mole • 6.02 x 1023 is known as Avogadro’s Number. Named for an Italian chemist and physicist, Amadeo Avogadro.

  8. From Moles to Molecules • How many molecules in 2.5 moles? 2.5 mol x 6.02x1023 = 1.505 x 1024 1 mole There are 1.5 x 1024 particles in 2.5mol

  9. From Molecules to Moles • How many moles in 9.7 x 1015 molecules? 9.7x1015 molecues x 1 mole = 6.02 x 1023 molecules 1.6 x 10-8 moles

  10. Formula Unit • The lowest whole-number ratio of elements in an ionic compound. (ex. NaCl – 1 sodium to 1 chlorine atom) • One mole of sodium chloride contains 6.02 x 1023 formula units of NaCl.

  11. Formula Mass 1.0 amu 1.0 amu H H Formula Mass = 1.0 amu 1.0 amu + 16 amu 18 amu O 16 amu

  12. Formula Mass • The sum of the atomic masses of all the atoms in a compound. • For example: NaCl Na 1 x 22.98977 = 22.98977amu Cl 1 x 35.453 = 35.453amu 58.44277amu The formula mass of NaCl = 58.443 amu / formula unit

  13. Formula Mass One More Time Example: CaSO4 Ca 1 x 40.08 = 40.08 amu S 1 x 32.06 = 32.06 amu O 4 x 15.9994 = 63.9976 amu 136.1376 amu The formula mass of CaSO4 = 136.14 amu / fomula unit

  14. Molar Mass • The mass in grams of 1 mole of a substance. • For example: H2O H 2 x 1.00794 = 2.01588 g O 1 x 15.9994 = 15.9994 g 18.01528g The molar mass of H2O = 18.0153 g/mol

  15. Formula vs Molar Mass • The only difference between a formula mass and a molar mass is the unit! • Formula mass represents the mass of one unit of a compound and is measured in amu. (Small amount uses a small unit.) • Molar mass represents the mass of one mole of substance and is measured in grams. (Larger quantity uses a larger mass.)

  16. Molar Volume • One mole of any gas at standard temperature and pressure (STP = 0oC and 1 atm) has a volume of 22.4 L or 22.4 dm3. • The volume of 1 mole of gas at STP = 22.4 L

  17. 1.12 moles O2 = Molar Volume Sample Problem • A canister with a volume of 25.0 L contains how many moles of oxygen at STP?

  18. 17.4 L = Molar Volume Sample II • 0.775 moles of gas at STP would occupy what volume?

  19. Multi-Step Conversions Particles Mass Number of Particles (6.02 x 1023) Molar Mass (calculate) Moles Molar Volume (22.4 L) Volume At STP

  20. = 7.87 x 1022 formula units of CaCl2 Multi-Step Conversions • How many formula units are in 14.5 g of CaCl2?

  21. Percent Composition • The mass of each element in a compound compared to the entire mass of the compound multiplied by 100. • The percent of the mass made up by each element in the compound. • %Comp can be determined by: • calculation from the chemical formula • by experimental analysis

  22. Calculation of % Composition CaSO4 Ca 1 x 40.078 = 40.078 S 1 x 32.066 = 32.066 O 4 x 15.9994 = 63.3998 29.568% 40.078 = 135.544 23.657% 32.066 =135.544 46.7745 % 63.39998=135.544 +______135.544 100%

  23. % Composition by Experimental Analysis • 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

  24. Empirical Formula • A formula that gives the simplest whole-number ratio of the atoms of in a compound. Molecular Empirical • Examples: C6H12O6 = CH2O C5H10O5 = CH2O H2O2 = HO

  25. Empirical Formulas • You can determine % composition from a chemical formula. (What you just calculated.) • You can also determine a chemical formula from the percent composition.

  26. Finding Empirical Formulas • 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). • Multiply the mass of each element by the molar mass of the element. • Find the whole number ratio of these calculated amounts by dividing each mole value by the smaller (est) one calculated. • Round to whole numbers and use the ratio to determine the empirical formula.

  27. You’re dying to try one, right? • A compound was found to contain 29.6% Calcium, 23.7% Sulfur and 46.8% Oxygen. What is the empirical formula for the compound?

  28. 29.6%Ca, 23.7%S, 46.8%O • Assume 29.6g Ca, 23.7g S and 46.8g O. • 29.6g x 1 mole = 0.739 mol Ca 40.078g 23.7g x 1 mole = 0.739 mole S 32.066 46.8 x 1 mole = 2.93 mole O 15.9994 • 0.739 = 1 0.739 = 1 2.93 = 3.96 0.739 0.739 0.739

  29. 29.6%Ca, 23.7%S, 46.8%O • Round to whole number 1 : 1 : 4 The empirical formula is… CaSO4

  30. Empirical Formulas • If the compound is molecular, you may have one more job to do… yeah!

  31. Molecular Empirical Formulas 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?

  32. Empirical Formula for Ribose • 40.0g x 1 mole = 3.33 mole C 12.011 6.67g x 1 mole = 6.62 mole H 1.00794 53.3g x 1 mole = 3.33 mole O 15.9994

  33. Empirical Formula for Ribose • 3.33 = 1 6.62 = 1.99 3.33 = 1 3.33 3.33 3.33 • Empirical Formula for Ribose CH2O Ready for the new step??? (Here it is anyway!)

  34. Molecular Formula for Ribose • Covalent bonds can form in many different ratios. (Ionic only form in one set ratio based on charges) • Use the empirical formula and the ratio of molar mass to empirical formula mass to determine the molecular formula.

  35. Molecular Formula for Ribose • Empirical Formula: CH2O • Molar mass (from original problem): 150g/mole • Empirical mass: 30.0g/mole (1x12.011 + 2x1.00794 + 1x15.9994=30.0) 150 g/mole (Molar mass of cmpd) = 5 30.0 g/mole (Empirical mass)

  36. Molecular Formula for Ribose • Ratio of molar mass to empirical mass = 5 • Empirical formula CH2O • Multiply empirical formula by 5 Molecular Formula of Ribose: C5H10O5 (fun, right?)

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