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Exam 1b (PS2 and PS3 material) Comments

Ppt07(PS4). Exam 1b (PS2 and PS3 material) Comments. Material starting with this Ppt is not on the exam. Ppt06 is the end of PS3 (and thus Exam 1b) material).

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Exam 1b (PS2 and PS3 material) Comments

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  1. Ppt07(PS4) Exam 1b (PS2 and PS3 material) Comments • Material starting with this Ppt is not on the exam. Ppt06 is the end of PS3 (and thus Exam 1b) material) • You need to be able to distinguish between the mass of a single atom (isotopic mass) and the mass number (both of these apply to a particular isotope of an element) • You need to be able to distinguish between the mass of a single atom (isotopic mass) and the “atomic mass” of an element (i.e., weighted average of the isotopic masses) • The “atomic mass” is that decimal number on the periodic table • Learn the meaning of complete atomic symbols, subscripts, superscripts (left and right), etc. Ppt07(PS4)

  2. Exam 1b Comments (cont.) • You must memorize/learn the names of elements and ions (and charges on ions) on the handout sheet • But ALSO learn how to use the “neutrality principle” and understand the meaning of subscripts and superscripts (charges)! There is a lot more to nomenclature than memorization of “names”! • Don’t forget that binary molecular compounds have a different set of naming rules • Learn the way to recognize and name acids • This listing is not even close to being exhaustive—anything related to any part of any problem on either PS2 or PS3 is fair game. • Practice Exam for 1b & its key are posted • 1st portfolio check (grade) will be during exam

  3. Plan (Application of the Counting By Weighing Idea From Ppt05) Ppt07(PS4) • Example: Determining the ratio of atoms in a compound (subscripts in formula) • Example: Determining the number of FU in a sample of H2O • A “mole” of a chemical substance is a certain number of FUs. [later]

  4. I. Example: Determining the ratio of atoms in a compound (subscripts) A (hypothetical, teensy-weensy) sample of a compound (let’s pretend we did not know about ions yet) is separated chemically to form 258000 amu of Li(s) and 384000 amu of P(s). 0. Are there more atoms of Li or P in the sample? • If you’ve already answered, you’re probably wrong!!  Remember: amu is a mass unit; it doesn’t (by itself) tell you how many items! Ppt07(PS4)

  5. Apply “counting by weighing” idea! • “lbs divided by lbs per BB = # BB’s • “amus divided by amus per atom = # atoms”! Ppt07(PS4)

  6. ….258000 amu of Li(s) and 384000 amu of P(s) 1. How many atoms (really ions) of each? More Li atoms (even though less mass)! Why? A Li atom has much less mass than a P atom! Ppt07(PS4)

  7. 2. What is the ratio of Li atoms: P atoms in the compound? 3. What is the chemical (really empirical) formula?  Li3P  subscripts tell you exact composition of one FU, and thus the ratio in any-sized sample!! Ppt07(PS4) 7

  8. 4. How many FU’s of Li3P were in the original sample? Recall: #FUs? 4 ___ FU of FeCl3 • Answer: 12400 FU of Li3P! • The #FU’s of Li3P = # P atoms* in this case because there is one P atom* per FU • Trike analogy: # trikes = # seats (but not # wheels!) ----- * Really “ions” Ppt07(PS4) • Reminder (what we have already determined): • 37170.4… = 37200 Li atoms* • 12399.0… = 12400 P atoms* • Formula of compound is Li3P

  9. 5. Another way? (to get the number of FU of compound in the sample)  (3 x 6.941 amu) + (1 x 30.97 amu) = 51.793 amu(per FU) • Mass of original sample? • 258000 amu of Li and 384000 amu of P •  Sum = 642000 amu(of compound) • Bowling ball idea all over again! (Same as before) Ppt07(PS4) What is the (average) mass of one FU of Li3P?

  10. Take a step back  Called the “formula mass” (technically should be “FU mass”) We can: • Determine the number of atoms from total mass of atoms if we know mass per atom • If two sets of atoms came from a sample of a compound, we can find • the number of atoms of each and then find • the ratio of atoms, which tells us the • (Empirical) Formula of the compound [subscripts] • Determine the number of FUs of a substance from total mass if we know the mass per FU

  11. Summary Table (Generalized [Mines] FU and Formula Mass Examples)

  12. II. Example: Determining the number of molecules (FU) in a sample of H2O (without “moles” or “molar mass”) How many FU (molecules, here) of H2O are in 150. g of water? Careful! g  amu!!!  1 amu = 1.6605 x 10-24 g formula mass (like lb/BB) conversion Plan: g  amu  #FU (bowling balls) (lbs) Ppt07(PS4)

  13. The number of molecules (FU) in 150. g of H2O (continued) for H2O? formula mass (like lb/BB) amu  #FU (bowling balls) (lbs) That’s a LOT of molecules (in a fairly small glass of water)!!! 5,010,000,000,000,000,000,000,000 8,000,000,000,000,000,000 Two estimates of the # of sand grains on Earth!! 5,000,000,000,000,000,000,000 Ppt07(PS4)

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