1. Chemical Quantities

2. Chapter Overview

3. Measuring Matter Mass ________ Counting ________ Volume _____________ Length ________

4. Counting but not really counting If you know the mass of ______ item you can find out how many items there would be in a greater quantity.

5. How many apples are there in 11.0 kg? Step 1 11.0 kg X conversion = factor Given 1 dozen apples = 2kg (mass)

6. How many apples are there in 11.0 kg? Step 1 11.0 kg X _ dozen apples = _ kg Given 1 dozen apples = 2kg (mass) ____ dozen apples

7. Answer ____ apples This method is counting but not really counting because it is an approximation of the answer. Step 2 5.5 dozen apples * 12 apples = 1 dozen apples

8. Stoichiometry The study of relationships that can be derived from chemical formulas and equations.

9. Formula Mass formula mass if given in ________ or ______ molecular mass if substance is molecular molar mass if given in ______ Sum of the ________ ________ also called gram atomic mass gram formula mass

10. THE MOLE Number of particles –like the problem with the apples, particles are also represented in this way except they are called moles instead of dozens. Avogadro’s number –________ –this number is the unit particles per mole unit referring to molecules, atoms, and formula units

11. Molar Volume ___ _____by one mole of particles –If it is a solid or liquid find the mass of one mole and divide by _______ –If it is a gas: One mole occupies _______ at STP »STP = 0 °C and 101.3 kPa

12. Problems involving moles Example –What is the molar mass of H 2 O? Calculating molar mass _______ the masses of each of the separate elements in the formula

13. EXAMPLE FIND THE MOLAR MASS OF WATER Step One –first find the molar mass of hydrogen. –Each hydrogen atom contains 1.0 gram per mole. Multiply one by two because there are two hydrogen atoms. Answer to step one: 2.0 grams

14. –The mass of one oxygen is 15.99. But you can round up to 16. –There is only one oxygen molecule so multiply 16 by one. –Answer to step two: 16 grams Step Two – find the mass of the oxygen molecules.

15. Step Three – add the masses of the oxygen and the hydrogen together. –16 g + 2g = _______ g –Therefore the molar mass of H 2 O is _______ grams per mole.

16. “Moles” is central MolesMass X ÷ Molar mass

17. Convert mass to moles To convert mass to moles you need to ______ the mass by the molar mass. Example How many moles are in 4.0 grams of H 2 O?

18. How many moles are in 4.0 grams of H 2 O? Step one –find the molar mass of the formula –as in the previous example take 1.0 times 2 for the hydrogen and 16.0 for the oxygen. –Add them together to get 18.0 grams

19. Step 2 take the 4 grams of H 2 O and _______ them by the molar mass of water to convert mass into moles.

20. The math you would use for step 2 is as follows: 4.0g H 2 O X 1 mole H 2 O =. ___ mole 18 g H 2 O This answer can also be written as _________ mole

21. Converting moles to mass Converting moles to mass is similar to converting mass to moles because you are taking the equation backward To convert, multiply mass by molar mass.

22. Example: How many grams are in 4 moles of H 2 O? Step One –find the molar mass of the formula. –From previous examples we know that the molar mass of H 2 O is 18 grams.

23. Step Two –once you find the 18 grams for the molar mass of water, take the 4 moles of water and _______ by the molar mass.

24. The math equation for step two is shown below: 4 moles H 2 O X 18 grams H 2 O = 1mole H 2 O The answer is ____ grams. Notice in the equation above the moles cancel out to leave the grams for the answer.

25. Convert Molecules To Moles To convert _______ to moles divide the number of molecules by Avogadro’s number ( _______ ). One mole of any substance has the same number of molecules per mole

26. Example –How many moles are in 1.806 X 10 24 molecules of water? Take the number of molecules and divide by Avogadro’s number

27. The math equations would look as follows: 1.806X10 24 atoms X 1 mole water = _______ The answer : _______ moles of water. Once again notice that the atoms cancel each other out.

28. “Moles” is central MolesMass Molecules X X ÷ ÷ Avogadro’s number Molar mass

29. Converting Moles to Molecules To convert moles to molecules _______ the moles by Avogadro’s number (6.02X10 23 )

30. Converting Moles to Molecules Example –How many molecules are in 173 moles of H 2 O? Step One –Take the number of moles and _______ by Avogadro’s number

31. The math equation would look as follows: 173mole X 6.02X10 23 molecules = 1 mole The answer is: ____________ molecules

32. Gas Volume at STP to Moles STP = 0°C and 101.3 kPa To convert gas volume to moles –_______ the number of Liters by 22.4L/mole (Molar volume of any gas at STP)

33. EXAMPLE HOW MANY MOLES ARE IN 34.6 L OF H 2 O(g)? 34.6 L ÷ 22.4 L/mole = 1.54 mole Volume ÷ molar volume = moles At STP

34. “Moles” is central Moles Volume, gas Mass Molecules X X X ÷ ÷ ÷ Molar volume Avogadro’s number Molar mass

35. Moles to Gas Volume To convert moles to gas volume you have to take the number of moles and _______ it by 22.4 L/mole if conditions are at STP.

36. EXAMPLE Take 73.6 moles X 22.4Liters/mole _______ Liters at STP How many Liters are there in 73.6 moles of H 2 O(g) ?

37. Combination Calculations Example How many molecules are in 29.43 Liters of oxygen gas at STP? 29.43 L X 1 mole = ______mole 22.4 L 1.31 mole X 6.02 X 10 23 molecules = 1 mole 29.43 L ÷ ______Liters/ mole = ____________ molecules

38. Mass of one atom Divide the molar mass by Avogadro’s number

39. Mass of one atom What is the mass of one oxygen atom? 16.0 grams ÷ 6.02 X 10 23 atoms mole 1 mole 16.0 grams X 1 mole__________ = mole 6.02 X 10 23 atom __________ g/atom

40. Gas Density The density of a gas is usually measured in grams per LITER It can be used to calculate the molar mass of a gas.

41. Molar mass from gas density Density = mass / volume If the mass you’re looking for is Molar Mass then use the Molar Volume The molar volume of any gas at STP is 22.4 Liters / mole

42. Molar mass from gas density What is the molar mass of CO 2 if its density at STP is 1.96 g / L ? D = Molar mass Molar volume 1.96 g / L = Molar mass 22.4 L / mole Molar mass = 1.96 g X 22.4 L Lmole Molar mass = _____ g / mole

43. Percent Composition by Mass Percent composition is the relative amounts of the elements in the compound. To calculate the percent composition, take the ___ of the required ____ and divide by the _______ of the compound.

44. Percent Composition Example Calculate the percent composition of propane, C 3 H 8. First get the total mass of the compound 3 moles C X 12 g/mole = 36 grams C 8 moles H X 1 g/mole = 8 grams H Total mass of propane = 44 grams

45. Percent Composition C 3 H 8 %C = 36.0 grams CX 100 44.0 grams propane = _______ % %H = 8.0 grams H X 100 44.0 grams propane = _______ %

46. Determining Empirical Formula The empirical formula shows the lowest whole number of _______ in a compound. –For example the empirical formula of benzene, C 6 H 6 is CH. A 1 : 1 ratio

47. Determining Empirical Formula To calculate the empirical formula first get the number of ____ of each element Then get the lowest ____ ____ ____ of those moles. –These numbers become the ________ If data is given as percent assume a 100 gram sample and just change the percent to grams.

48. Example Empirical Formula What is the empirical formula of a substance that is 80.0% C and 20.0% H by mass? –First change % to grams by assuming a 100 g sample 80.0 g C and 20.0 g H

49. Empirical Formula Example Convert grams to moles (÷molar mass) 80.0 grams ÷ 12.0 grams/mole C = 6.67 moles C 20.0 grams ÷ 1.01 grams/mole H = 19.8 moles H The ____ of moles of H to C is calculated 19.8 moles / 6.67 moles = 2.97 or 3 : 1

50. Empirical Formula Therefore the Empirical formula of the compound is CH 3

51. Molecular Formula The molecular formula shows the ____ ____ of ____ atom in the molecule. –For example: C 6 H 6 shows ____ atoms of carbon and ____ atoms of hydrogen in the benzene molecule

52. Molecular Formula To calculate the molecular formula from the empirical formula –First find the molar mass of the empirical formula –Next find the molar mass of the molecule –Then ____ the molar mass of the molecule by the ____ ____ of the empirical formula to find how many times ____ the molecule is than the empirical formula.

53. If the molar mass of ethane is 30.0 grams per mole and the empirical formula as determined in a previous example is CH 3, what is the molecular formula of ethane? First: The molar mass of CH 3 is 12.0 + 3(1.01) = 15.0 g/mole Second: The molar mass of ethane is 30.0 g/mole Third: The ratio of molecule : empirical is 30.0 g/mole ÷ 15.0 g/mole = ____

54. So the molecular formula is ____ times ____ than the empirical formula. 2 X CH 3 = C 2 H 6

55. CREDITS Produced by –Jay Seymour –Patricia Brown –1999