1 / 36

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)

wlowery
Download Presentation

Chemistry Unit 8a

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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?)

More Related