1. SCIENTIFIC MEASUREMENT  CHEM IH: CHAPTER 3

2. Stating a Measurement In every measurement there is a  Number followed by a  Unit from a measuring device The number should also be as precise as the measuring device.

3. Ex: Reading a Meterstick. l 2.... I.... I 3....I.... I 4.. cm First digit (known)= 2 2.?? cm Second digit (known)= 0.7 2.7? cm Third digit (estimated) between 0.05- 0.07 Length reported=2.75 cm or2.74 cm or2.74 cm or2.76 cm

4. UNITS OF MEASUREMENT Use SI units — based on the metric system LengthMassVolumeTimeTemperature Meter, m Kilogram, kg Seconds, s Celsius degrees, ˚C kelvins, K Liter, L

5. Metric Prefixes

6. Conversion Factors Fractions in which the numerator and denominator are EQUAL quantities expressed in different units Example: 1 hr. = 60 min Factors: 1 hr. and 60 min 60 min 1 hr.

7. How many minutes are in 2.5 hours ? Conversion factor 2.5 hr x 60 min = 150 min 1 hr 1 hr cancel By using dimensional analysis / factor-label method, the UNITS ensure that you have the conversion right side up, and the UNITS are calculated as well as the numbers!

8. Sample Problem  You have $7.25 in your pocket in quarters. How many quarters do you have? 7.25 dollars 4 quarters 1 dollar 1 dollar X = 29 quarters

9. Learning Check How many seconds are in 1.4 days? Unit plan: days hr min seconds 1.4 days x 24 hr x ?? 1 day

10. Wait a minute! What is wrong with the following setup? 1.4 day x 1 day x 60 min x 60 sec 24 hr 1 hr 1 min 24 hr 1 hr 1 min

11. Significant Figures The numbers reported in a measurement are limited by the measuring tool The numbers reported in a measurement are limited by the measuring tool Significant figures in a measurement include the known digits plus one estimated digit Significant figures in a measurement include the known digits plus one estimated digit

12. Counting Significant Figures RULE 1. All non-zero digits in a measured number are significant. Only a zero could indicate that rounding occurred. Number of Significant Figures 38.15 cm4 5.6 ft2 65.6 lb___ 122.55 m 122.55 m___

13. Leading Zeros RULE 2. Leading zeros in decimal numbers are NOT significant. Number of Significant Figures 0.008 mm1 0.0156 oz3 0.0042 lb____ 0.000262 mL 0.000262 mL ____

14. Sandwiched Zeros RULE 3. Zeros between nonzero numbers are significant. (They can not be rounded unless they are on an end of a number.) Number of Significant Figures 50.8 mm3 2001 min4 0.702 lb____ 0.00405 m 0.00405 m ____

15. Trailing Zeros RULE 4. Trailing zeros in numbers without decimals are NOT significant. They are only serving as place holders. Number of Significant Figures 25,000 in. 2 25,000 in. 2 200. yr3 200. yr3 48,600 gal____ 48,600 gal____ 25,005,000 g ____

16. Learning Check A. Which answers contain 3 significant figures? 1) 0.4760 2) 0.00476 3) 4760 B. All the zeros are significant in 1) 0.00307 2) 25.300 3) 2.050 x 10 3 C. 534,675 rounded to 3 significant figures is 1) 535 2) 535,000 3) 5.35 x 10 5 1) 535 2) 535,000 3) 5.35 x 10 5

17. Learning Check In which set(s) do both numbers contain the same number of significant figures? 1) 22.0 and 22.00 1) 22.0 and 22.00 2) 400.0 and 40 3) 0.000015 and 150,000

18. Significant Numbers in Calculations A calculated answer cannot be more precise than the measuring tool. A calculated answer cannot be more precise than the measuring tool. A calculated answer must match the least precise measurement. A calculated answer must match the least precise measurement. Significant figures are needed for final answers from Significant figures are needed for final answers from 1) adding or subtracting 1) adding or subtracting 2) multiplying or dividing

19. Adding and Subtracting The answer has the same number of decimal places as the measurement with the fewest decimal places. 25.2 one decimal place + 1.34 two decimal places 26.54 26.54 answer 26.5 one decimal place

20. Learning Check In each calculation, round the answer to the correct number of significant figures. A. 235.05 + 19.6 + 2.1 = 1) 256.75 2) 256.83) 257 B. 58.925 - 18.2= 1) 40.725 2) 40.733) 40.7

21. Multiplying and Dividing Round (or add zeros) to the calculated answer until you have the same number of significant figures as the measurement with the fewest significant figures.

22. Learning Check A. 2.19 X 4.2 = 1) 9 2) 9.2 3) 9.198 B. 4.311 ÷ 0.07 = 1) 61.58 2) 62 3) 60 C. 2.54 X 0.0028 = 0.0105 X 0.060 1) 11.32) 11 3) 0.041

24. MEASURING MASS  A moleis a quantity of things, just as… 1 dozen= 12 things 1 gross = 144 things 1 mole= 6.02 x 10 23 things  “Things” usually measured in moles are atoms, molecules, ions, and formula units

25.  You can measure mass, or volume, or you can count pieces  We measure mass in grams  We measure volume in liters  We count pieces in MOLES

26. A MOLE…  is an amount, defined as the number of carbon atoms in exactly 12 grams of carbon- 12  1 mole = 6.02 x 10 23 of the representative particles  Treat it like a very large dozen 6.02 x 10 23 is called: Avogadro’s number

27.  Similar Words for an amount:  Pair: 1 pair of shoelaces = 2 shoelaces  Dozen: 1 dozen oranges = 12 oranges  Gross: 1 gross of pencils= 144 pencils  Ream: 1 ream of paper= 500 sheets of paper

28. What are Representative Particles?  The smallest pieces of a substance: 1. For a molecular compound: it is the molecule. 2. For an ionic compound: it is the formula unit (made of ions) 3. For an element: it is the atom  Remember the 7 diatomic elements? (made of molecules)

29.  How many oxygen atoms in the following? 1. CaCO3 3 atoms of oxygen 2. Al2(SO4)3 12 (4 x 3) atoms of oxygen  How many ions in the following?  CaCl2  3 total ions (1 Ca2+ ion and 2 Cl1- ions)  NaOH  2 total ions (1 Na1+ ion and 1 OH1- ion)  Al2(SO4)3  5 total ions (2 Al3+ + 3 SO4 ions)

30. CONVERSION FACTOR  MOLES = representative particles x ____________1 mole_____________ 6.02 x 10 23 representative particles

31. EXAMPLES: ATOMS  MOLES  How many atoms of Al are in 1.5 mol of Al?  Conversion: 1 mole = 6.02 x 10 23 atoms 1.5 mol of Al6.02 x 10 23 atoms 1 mole = 9.03 x 10 23 atoms of Al

32. EXAMPLES: MOLECULES  MOLES How many atoms of H are there in 3 moles of H 2 O? Conversions: 1 mole = 6.02 x 10 23 molecules H 2 O molecule = 2 atoms of Hydrogen 3 moles of H 2 O 6.02 x 10 23 molec 1 mole 2 atoms H 1 H 2 O molecule = 3.612 x 10 24 atoms H

33. MOLAR MASS Determined simply by looking at the periodic chart Molar mass = Atomic Mass Ca 20 40.08 * Thus, 1 mol Ca = 40 g Atomic Mass same as Molar Mass

34. MOLAR MASS  To calculate the molar mass of a compound, find the number of grams in each element in one mole of the compound  Then add the masses within the compound Example: H 2 O H= 1.01 2 (1.01) + 1 (15.999)= 18.02 O= 15.999

35. SOME PRACTICE PROBLEMS  How many atoms of O are in 3.7 mol of O?  2.2 X 10 24 atoms of oxygen  How many atoms of P are in 2.3 mol of P?  1.4 x 10 24 atoms of phosphorus  How many atoms of Ca are there in 2.5 moles of CaCl 2 ?  1.5 x 10 24 atoms Ca  How many atoms of O are there in 1.7 moles of SO 4 ?  4.1 x 10 24 atoms of oxygen

36. Remember!!!!  The molar mass of any substance (in grams) equals 1 mole  This applies to ALL substance: elements, molecular compounds, ionic compounds  Use molar mass to convert between mass and moles  Ex: Mass, in grams, of 6 mol of MgCl 2 ? mass of MgCl 2 = 6 mol MgCl 2 92.21 g MgCl 2 1 mol MgCl 2 = 571.26 g MgCl 2

37. VOLUME AND THE MOLE  Volume of 1 mol of solid and liquids are not the same  But gases are more predictable, under the same physical conditions  Avogadro’s hypothesis helps explain: equal volume of gases, at the same temp and pressure contains equal number of particles  Volume varies with changes in temperature  Ex: helium balloon

38.  Gases vary at different temperatures, makes it hard to measure  Because of variation use STP  Standard Temperature and Pressure  Temperature = 0° C  Pressure = 1 atm (atmosphere) or 101.3 kPal

39. Standard Temperature Pressure  At STP: 1 mole, 6.02 x 10 23 atoms, of any gas has a volume of 22.4 L  Called Molar Volume  Used to convert between # of moles and vol of a gas @ STP  Ex: what is the vol of 1.25 mol of sulfur Vol of S= 1.25 mol S 22.4 L = 28 L 1 mol

40. MOLAR MASS FROM DENSITY  Different gases have different densities  Density of a gas measured in g/L @ a specific temperature  Can use the following formula to solve : grams = grams X 22.4 L mole L 1 mole  Ex: Density of gaseous compound containing oxygen and carbon is 1.964 g/ L, what is the molar mass?  grams = 1.964 g X 22.4 L then you solve mole 1 L 1 mole= 44.o g/mol

42. Calculating Percent Composition of a Compound  Like all percent problems: a part ÷ the whole 1. Find the mass of each of the components (the elements) 2. Next, divide by the total mass of the compound 3. Then X 100 % = percent Formula: % Composition = Mass of element X 100% Mass of compound

43. Example: A compound is formed when 9.03 g of Mg combines completely with 3.48 g of N. What is the percent composition of the compound? 1. First add the 2 mass of the 2 compounds to reach the total mass 9.03 g Mg + 3.48 g N = 12.51 g Mg 3 N 2 1. Find the % of each compound % Mg= 9.03 g Mg X 100% = 72.2 % 12.51 g Mg 3 N 2 % N= 3.48 g N X 100%= 27.8 % 12.51 g Mg 3 N 2

44. % Composition from Chemical Formula  Can find the percent composition of a compound using just the molar mass of the compound and the element  % mass=mass of the element 1 mol cmpd X100% molar mass of the compound  Example: Find the percent of C in CO 2 12.01 g C X 100% = 27.3% C 44.01 g CO 2 Can find O % by subtracting 27.3% from 100%

45. Using % Composition  Can use % composition as a conversion factor just like the mole  After finding the % comp. of each element in a cmpd. can assume the total compound = 100g  Example: C= 27.3% 27.3 g C O= 72.7 % 72.7 g O  In 100 g sample of compound there is 27.3 g of C & 72.7 g of O How much C would be contained in 73 g of CO 2 ? 73 g CO 2 27.3 g C= 19.93 g C 100 g CO 2

46. EMPIRICAL FORMULAS  Empirical formulas are the lowest WHOLE number ratios of elements contained in a compound

47. REMEMBER…  Molecular formulas tells the actual number of of each kind of atom present in a molecule of the compound  Ex: H 2 O 2 HO MolecularEmpiricalFormula CO 2 MolecularEmpirical For CO 2 they are the sameFormula

48.  Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = can not be reduced)  Examples: NaCl MgCl2 Al2(SO4)3 K2CO3 Simplest whole number ratio for NaCl

49.  A formula is not just the ratio of atoms, it is also the ratio of moles  In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen  In one molecule of CO 2 there is 1 atom of C and 2 atoms of O  Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio)  Molecular: H 2 O C 6 H 12 O 6 C 12 H 22 O 11 (Correct formula)  Empirical: ( Lowest whole H 2 O CH 2 O C 12 H 22 O 11 number ratio)

50. CALCULATING EMPIRICAL  We can get a ratio from the percent composition 1. Assume you have a 100 g sample the percentage become grams (75.1% = 75.1 grams) 2. Convert grams to moles 3. Find lowest whole number ratio by dividing each number of moles by the smallest value

51. Example calculations  Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N  Assume 100 g sample, so 38.67 g C x 1 mol C = 12.0 g C 16.22 g H x 1 mol H = 1.0 g H 45.11 g N x 1 mol N = 14.0 g N *Now divide each value by the smallest value 3.22 mole C 3.22 mole N 16.22 mole H

52. …Example 1  The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N  The ratio is 16.22 mol H = 5 mol H 3.22 mol N 1 mol N C 1 H 5 N 1 which is = CH 5 N

53. MORE PRACTICE  A compound is 43.64 % P and 56.36 % O What is the empirical formula? PO3  Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O What is its empirical formula? C 4 H 5 N 2 O

54. EMPIRICAL TO MOLECULAR  Since the empirical formula is the lowest ratio, the actual molecule would weigh more  Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

55. EXAMPLE  Caffeine has a molar mass of 194 g, what is its molecular formula? 1. Find the mass of the empirical formula, C 4 H 5 N 2 O 2. Divide the molar mass by the empirical mass: 194.0 g/mol = 97.1 g/mol 3. Now multiply the entire empirical formula by 2 2(C 4 H 5 N 2 O) = final molecular formula 2 C 8 H 10 N 4 O 2