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Chapter 10 continued

Chapter 10 continued. Getting from here to there. Multi-step conversions. If I asked you how many Cups of water are in a gallon of water how would you do that? What if you needed to give 10000 seconds in the number of hours how would you do that? These are both multi step conversions.

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Chapter 10 continued

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  1. Chapter 10 continued Getting from here to there

  2. Multi-step conversions • If I asked you how many Cups of water are in a gallon of water how would you do that? • What if you needed to give 10000 seconds in the number of hours how would you do that? • These are both multi step conversions. • Now if I gave you that you had 3.25X1023 atoms of Ne gas and asked you how many much space it takes up in L. We have the same issue as above.

  3. If 22.4 L of He is stored at STP what is the mass in grams? • If you have 275 grams of Silver. How many particles of silver do you have? • How many grams are there in 2.78x1024 particles of CO2? • How much volume does it have? Examples

  4. If 22.4 L of He is stored at STP what is the mass in grams? • 4 g He • If you have 275 grams of Silver. How many particles of silver do you have? • 1.53x1024 part. Ag • How many grams are there in 2.78x1024 particles of CO2? • 203 g CO2 • How much volume does it have? • 103 L CO2

  5. Empirical and Molecular Formulas

  6. Percent Composition • The percent by mass of each element in a compound is the percent composition of a compound.

  7. If you have one Mole of NaCHO3 • First what is the molar mass of NaCHO3? • 84g • Now we take each element mass divided by the molar mass. • Na/NaCHO3 • (23/84)x 100= 27.38% • C/NaCHO3 • (12/84) x100= 14.29% • H/NaCHO3 • 1.19% • 3O/NaCHO3 • 57.14% Calculating What is the percent composition for each element in NaHCO3?

  8. Mini Lab • Analyze chewing gum • Are the sweetening and flavoring added as a coating or mixed throughout? • Unwrap two pieces of chewing gum. Place each piece on a weighing paper. Measure and record the mass of each using the balance. • Add 150 ml of cold tap water into a 250 ml beaker. Place one piece of chewing gum in the water, and stir with stirring rod for 2 minutes. • Pat the gum dry using paper towels. Measure and record the mass of the dried gum. • Use the scissors to cut the second piece of gum into small pieces. Repeat step 3 with new tap water. Try to keep the pieces from clumping together. • Strain the water and pat the gum dry on a paper towel. Measure and record the mass of the dried gum.

  9. Analysis • Calculate: Calculate the % mass of the flavorings and sweeteners for both pieces. That would be the amount of mass lost (sweeteners and flavorings) / by the total mass of the gum. • Infer: What can you infer from the two percentages? Is the gum sugar coated or are the sweeteners mixed in? • As a Pair you and your lab partner(s) will need to do the calculation and analysis questions from the book.

  10. What is the percent by mass of Carbon in CH4? • 75% • What is the percent composition of Hydrocyanic acid (HCN)? • H= 3.7% • C= 44.44% • N= 51.85% • What is the % by mass of Zn in zinc iodide? • 20.38% • Which element has the smallest % composition in tin nitrate? • N 15.25% or 11.52% Examples

  11. If you are given % composition but not the actual formula you can use the % to find how many moles of each element are in the compound. • Example • If you are given that a compound has • 48.64% C • 8.16% H • 43.20% O • You can find out how much many carbon, to hydrogen, to oxygen are in your compound. Empirical Formula The smallest whole-number mole ratio of the elements.

  12. When finding Empirical formulas from percent composition • Look at the % given to you. • Assume that they are out of 100 grams total. • 100%= 100g • For Carbon we would then state that 48.64%= 48.64g • H=8.16 g • O= 43.20g • Next you find how many moles there are of each element • Carbon 48.64g(1mole/12g)=4.05moles • H 8.16g (1mole/ 1g) = 8.1 moles • O 43.20g (1 mole/ 16g) =2.7 moles How do we do it 48.64% C 8.16% H 43.20% O

  13. Now we simplify • C=4.05, H= 8.1, O= 2.7 • We divide them all by the lowest number of moles. • C 4.05/2.7= 1.5 moles C • H 8.1/2.7= 3 moles H • O 2.7/2.7= 1 mole O • If they had all come out in round able numbers we would be done, but 1.5 is half way so there is another step. Simplifying down to the lowest # of moles

  14. To make every thing a whole number we need to multiply to get rid of the .5. • So we multiply everyone by 2. • C 1.5(2)= 3 moles Carbon • H 3(2)= 6 moles Hydrogen • O 1(2)= 2 moles O • So we would write the empirical formula: C3H6O2 Has to be a whole number

  15. 3.08% H, 31.61% P, 65.31% O • H3PO4 • 36.11% Ca, 63.89% Cl • CaCl2 • 42% C, 6% H, 52% O • CH2O Problems Find the empirical formulas for the following

  16. Sometimes substances with very different properties and structures have the same empirical formula.C2H2C4H4are all very different, how do you draw each • The Molecular formula, specifies the actual number of atoms of each element in one molecule or formula unit of the substance. • Experimentally determined molar mass = n • Mass of empirical formula Molecular Formula

  17. If you have 40.68% C, 5.08% H, and 54.24% O in a chemical analysis of Succinic acid, and an experimental mass of 118.1 g/mole. • Find the empirical formula • (40.68g/12) x100…. • You should find 1 C, 1.5 H, and 1 O (multiplied by 2 to get out of the 1.5 • Gives you 2 C, 3 H, and 2 O: C2H3O2 • No we find the molar mass of the empirical formula. (59g/mole) Molecular formula = (empirical formula)n example

  18. We take the experimental molar mass divided by the empirical molar mass • 118.1/59=n~ 2 • We then multiply our empirical formula by our n2(C2H3O2)= • C4H6O4molecular formula Cont.

  19. A chemical analysis shows that a compound commonly used as a bleaching agent is composed of 13.79 g Na, 21.27g Cl, and 9.60 g O. Determine the empirical formulas for this compound. If you are given an experimental value of 222.08 grams/mole, what is the molecular formula? The mineral ilmenites is usually mined and processed for titanium. A sample of ilmenite contains 5.41 G of titanium, 4.64 G of titanium, and 4.65 G of oxygen. What is the empirical formula? Practice problems

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