Ch. 6: Chemical Composition

# Ch. 6: Chemical Composition

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## Ch. 6: Chemical Composition

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1. Ch. 6: Chemical Composition Dr. Namphol Sinkaset Chem 152: Introduction to General Chemistry

2. I. Chapter Outline • Introduction • Counting by Weighing • Chemical Formulas as Conversion Factors • Mass Percent Composition • Finding Formulas from Mass Data

3. I. Introduction • The FDA RDA for Na = 2.4 g. • Na usually consumed as NaCl. • But 2.4 g NaCl ≠ 2.4 g Na. • How do we figure out a problem like this?

4. II. Selling Nails • Nails can be sold two ways: as individual units or by the pound. • By the pound much more convenient if want to buy hundreds of nails. • Analogy extends to atoms/molecules.

5. II. Sample Problem • If a dozen large nails weighs 0.275 lb., how many nails are contained in 5.5 lb?

6. II. Counting Atoms • Finding the number of atoms in a sample of matter is similar to selling nails by the pound. • We used a dozen as a unit of count, but 12 is way too small for atom. • In chemistry, the mole is used instead – it’s the chemist’s “dozen.”

7. II. The Mole • The mole is the SI unit for amount and is defined as the amount of material containing 6.0221421 x 1023particles. • This # is known as Avogadro’s number. • This # is defined by the number of atoms in exactly 12 g of carbon-12. • One mole of atoms, ions, or molecules generally makes up a macroscopically meaningful size.

8. II. How Much is a Mole?

9. 6.022 x 1023 atoms 1 mole atoms 6.022 x 1023 atoms 1 mole atoms II. Avogadro’s Number • Every time you hear “mole,” you should think 6.022 x 1023. • We can use Avogadro’s number as a conversion factor to calculate # of atoms. OR

10. II. Sample Problem • How many atoms of gold are in a pure gold ring containing 8.83 x 10-2 moles of Au?

11. II. Grams and Moles • In the nail example, we had a relationship between a dozen nails and a mass. • We need the same relationship for the mole. • The mass of 1 mole of atoms of an element is its molar mass. The value of an element’s molar mass in g/mole is numerically equal to the element’s atomic mass in amu (from periodic table).

12. II. Mole of Atoms & Compounds • 1 Fe atom weighs 55.85 amu, so 1 mole of Fe atoms weighs 55.85 g. • 1 O atom weighs 16.00 amu, so 1 mole of O atoms weighs 16.00 g. • Same applies for compounds. • 1 molecule H2O weighs 18.02 amu, so 1 mole H2O weighs 18.02 g.

13. II. Sample Problem • Calculate the molar mass of calcium sulfate.

14. II. The Mass of a Mole Depends on Unit Size

15. 1 mole Fe 55.85 g Fe 55.85 g Fe 1 mole Fe II. Mole Equivalencies • Since mass and moles are related, we can set up conversion factors. • e.g. 1 mole Fe weighs 55.85 g. OR

16. II. Sample Problem • Graphite, a crystalline form of carbon, is used in pencils. How many moles of carbon are in 0.315 g of graphite?

17. II. Sample Problem • How many grams of sulfur are in 2.78 moles of sulfur?

18. II. Sample Problem • How many moles of NO2 are in 1.18 g of NO2?

19. II. Sample Problem • How many molecules of H2O are in a sample of water with a mass of 3.64 g?

20. II. Sample Problem • If a sample of molecular bromine weighs 2100 g, how many molecules of molecular bromine are in the sample?

21. III. Inherent Conversion Factors • Any object that can be broken down into parts or pieces has an inherent conversion factor. • For example, a clover.

22. III. 3 Leaves : 1 Clover • Since there are 3 leaves in one clover, we can write a conversion factor between them.

23. III. Other Examples

24. III. Scaling Up • The ratios hold as long as we keep the unit the same for both the parts and the whole. • 3 leaves : 1 clover  3 dozen leaves : 1 dozen clovers. • In chemistry, we scale up to the mole. • 2 H atoms : 1 H2O molecule  2 moles H atoms : 1 mole H2O molecules.

25. III. Breaking Down CCl4

26. III. Sample Problem • List all the possible atom : formula unit mole relationships in barium nitrate.

27. III. Mole Relationships • A chemical formula gives the relationships between moles of substances, NOT grams. • Always convert to moles then use a mole relationship to get to the substance of interest.

28. III. Sample Problem • Determine the number of moles of oxygen in 2.45 moles of nickel(II) phosphate.

29. III. Sample Problem • Calculate the mass of carbon in 25.0 g of C6H12O6.

30. III. Sample Problem • How many grams of sodium chloride contains 2.00 grams of sodium?

31. IV. Mass Percent • The mass percent expresses an element’s percentage of the total mass of the compound.

32. IV. Sample Problem • Calculate the mass percent of nitrogen in ammonium nitrate.

33. 100 g NaCl 39 g Na 100 g NaCl 39 g Na IV. Mass Percent as a Conversion Factor • The key to using mass percents is realizing that they can be expressed as conversion factors. • NaCl is 39% Na can be expressed as 39 g Na : 100 g NaCl.

34. IV. Sample Problem • If CCl2F2 is 58.64% Cl by mass, how many grams of CCl2F2 contains 22 g of Cl?

35. V. Finding Formulas from Mass Data • Given elemental mass % data of a compound, it’s possible to find the formula of the compound. • Elemental analysis is a common test performed on newly synthesized compounds.

36. V. Empirical vs. Molecular • Often experiments will yield the relative masses of each element in a compound. • With just the relative masses, the best we can do is find the empirical formula. • Additional data is needed to go from an empirical formula to a molecular formula.

37. V. The Key to Finding Empirical Formulas • In any formula, subscripts represent the number of atoms of a given element in that compound. • However, these ratios scale up! • Therefore, the subscripts can also be interpreted as mole ratios between atoms of a given element in the compound.

38. V. Finding Empirical Formulas • To find an empirical formula, you must calculate the number of moles of each atom present in a certain sample of the compound. • These moles become the temporary subscripts in the formula. • You then use math to convert to whole numbers.

39. V. Sample Problem • Calculate the empirical formula of ethyl butyrate (pineapple oil) if its mass composition is 62.04% C, 10.41% H, and 27.55% O.

40. V. Sample Problem • A certain compound is 50.66% C, 4.25% H, and 45.09% S by mass. Determine the empirical formula.

41. V. Determining Molecular Formulas • To find a molecular formula from an empirical formula, you must know the molecular molar mass. • The molecular molar mass is always a multiple of the empirical molar mass. • To find the multiple, we divide the molecular molar mass by the empirical molar mass.

42. V. Sample Problem • Butane has an empirical formula of C2H5. If its molar mass is 58.12 g/mole, determine the molecular formula of butane.

43. V. Sample Problem • The carcinogen benzo[a]pyrene (MW = 252.30 g/mole) is found to be 95.21% C and 4.79% H by mass. What is its molecular formula?