1. 6 How do we measure matter? Chemist can measure matter by counting, weight, mass or even volume…but one common “unit” that chemist use to measure matter is “the mole (mol)”.

2. Amedeo Avogadro 1776-1856 6 The amount of representative particles in 1 mol of any substance is equal to “avogadro’s number” of that particular representative particle. Avogadro’s Number is defined as: 6.02 x 10 23

3. 6 Substance Representative Particle REMEMBER: A representative particle is the smallest possible particle that retains the properties of the substance that it represents. Element Atom Ionic Compound Formula Unit Molecular Compound Molecule

4. 6 So….. If one mol of a substance is “Avogadro’s Number” of representative particles, then…. 1 mol Al = 6.02 x 10 23 atoms

5. 6 So….. If one mol of a substance is “Avogadro’s Number” of representative particles, then….

6. 6 So….. If one mol of a substance is “Avogadro’s Number” of representative particles, then…. 1 mol CO 2 = 6.02 x 10 23 molecules

7. 6 So….. How many moles of magnesium are in 3.01 x 10 22 atoms of Mg ? 3.01 x 10 22 atoms of Mg 6.02 x 10 23 atoms of Mg 1 mol of Magnesium = 0.5 x 10 - 1 mols of Mg

8. 6 Now solve these: 1. How many moles are in 1.20 x 10 25 atoms of carbon? 2. How many atoms are in 0.750 mols of zinc? 3. How many molecules are in 0.400 mol of N 2 O 5 ? Answer: 19.9 mol C Answer: 4.52 x 10 23 atoms of Zn Answer: 2.41 x 10 23 molecules of Dinitrogen Pentoxide Click here to see the solution process CLICK HERE TO PROCEED

9. 6 SOLUTION: 1. How many moles are in 1.20 x 10 25 atoms of carbon? 1.20 x 10 25 atoms of C 6.02 x 10 23 atoms of C 1 mol of Carbon = 19.9 mols of C RETURN NEXT

10. 6 SOLUTION: 2. How many atoms are in 0.750 mols of zinc? 0.750 mols of Zn 6.02 x 10 23 atoms of Zn 1 mol of Zn = 4.52 x 10 23 atoms of Zn RETURN NEXT

11. 6 SOLUTION: 3. How many molecules are in 0.400 mol of N 2 O 5 ? 0.400 mol of N 2 O 5 6.02 x 10 23 molecules of N 2 O 5 1 mol of N 2 O 5 = 2.41 x 10 23 molecules of N 2 O 5 RETURN NEXT

12. 6 Mastery Level Problems: 4. How many fluoride ions are in 1.46 mol of aluminum fluoride? Press the spacebar to view the solution. To view the solution, press the spacebar. 1.46 mol AlF 3 6.02 x 10 23 formula units of AlF 3 1 mol of AlF 3 1 formula unit of AlF 3 3 F - ions = 26.4 x 10 23 fluoride ions Al 3+ F-F- F-F- F-F-

13. 6 Mastery Level Problems: 5. How many ammonium ions are in 0.036 mol of ammonium phosphate, (NH 4 ) 3 PO 4 ? Press the spacebar to view the solution. To view the solution, press the spacebar. 0.036 mol (NH 4 ) 3 PO 4 6.02 x 10 23 formula units of (NH 4 ) 3 PO 4 1 mol of (NH 4 ) 3 PO 4 1 formula unit of (NH 4 ) 3 PO 4 3 NH 4 + ions = 6.50 x 10 22 NH 4 + ions

14. Mastery Level Problems: 6. How many carbon atoms are in a mixture of 3.00 mol acetylene, C 2 H 2, and 0.700 mol carbon monoxide? Press the spacebar for a hint to get started. To view the solution, press the spacebar. A problem like this actually requires two calculations. First, calculate the number of carbon atoms in acetylene. Secondly, calculate the number of carbon atoms in carbon monoxide. Finally, add these two values together to get the number of carbon atoms in the mixture! If you do this correctly, your answer should be 4.03 x 10 24 atoms of carbon. 6

15. Mastery Level Problems: 6. How many carbon atoms are in a mixture of 3.00 mol acetylene, C 2 H 2, and 0.700 mol carbon monoxide? To view the solution, press the spacebar. A problem like this actually requires two calculations. First, calculate the number of carbon atoms in acetylene. Secondly, calculate the number of carbon atoms in carbon monoxide. Finally, add these two values together to get the number of carbon atoms in the mixture! If you do this correctly, your answer should be 4.03 x 10 24 atoms of carbon. 3.00 mol C 2 H 2 6.02 x 10 23 molecules Of C 2 H 2 1 mol of C 2 H 2 1 molecule Of C 2 H 2 2 Carbon atoms 3.612 x 10 24 Carbon atoms =.700 mol CO 6.02 x 10 23 molecules Of CO 1 mol of CO 1 molecule Of CO 1 Carbon atoms 4.214 x 10 23 Carbon atoms =

16. 6 So far… 1 mol of any substance = 6.02 x 10 23 representative particles A representative particle is the smallest particle of a substance that retains the physical and chemical properties of that substance. The representative particle of an element is an atom. The representative particle of an ionic compound is a formula unit. The representative particle of a molecular compound is a molecule.

17. 6 So … So far all that information is great…but how many grams of a material is equal to one mole of that material?

18. 6 Total Mass (grams) = 1 mole Example: What is the mass of 1 mole of He? Answer: 1 mol of Helium has a mass of 4 grams. This answer is based upon Helium’s average atomic mass as seen on the periodic chart. So if 1 mol of Helium is 4 grams…..this means that if you’re given 4 grams of Helium, you’ll have Avogadro’s number of atoms! Hence….since 4 grams constitutes one mole of He, 4 grams will be known as the Gram atomic mass of Helium. Gram atomic mass is defined as the number of grams of an element that is numerically equal to atomic mass in amu’s.

19. 6 Example 2: What is the gram atomic mass (mass of 1 mol) for the following elements: a.Sodium ___________________________________ b. Phosphorus _______________________________ 23 grams 31 grams Question: If given 23 grams of sodium, how many atoms would be in this amount of matter? How many atoms would be in 31 grams of Phosphorus? 6.02 x 10 23 atoms would be present in each sample ( since we have 1 mol ). = 1 mol of Na = 1 mol of P

20. 6 Example 2: What is the mass of 1 mole of NaCl ? Solution : Na 1 x 23 g = 23 g Cl 1 x 35 g = 35 g TOTAL MASS = 58 g Since 58 g of sodium chloride (table salt) equates to 1 mol of this IONIC COMPOUND, we say that 58 grams of sodium chloride is it’s Gram Formula Mass. The “gfm” of NaCl will produce Avogadro’s Number of formula units !

21. 6 Example 3: What is the mass of 1 mole of CO 2 ? Solution : C 1 x 12 g = 12 g O 2 x 16 g = 32 g TOTAL MASS = 44 g Since 44 g of carbon dioxide equates to 1 mol of this MOLECULAR COMPOUND, we say that 44 grams of carbon dioxide is it’s Gram Molecular Mass. The “gmm” of CO 2 will produce Avogadro’s Number of molecules!

22. 6 So far… 1 mol of any substance = 6.02 x 10 23 representative particles 1 mole of a substance = TOTAL Mass (grams) of that substance. The “TOTAL Mass of a substance is its gram equivalent to a.m.u’s Gram Atomic Mass = Term used to describe 1 mole of an element. Gram Formula Mass = Term used to describe 1 mole of an ionic compound. Gram Molecular Mass = Term used to describe 1 mole of an molecular compound. Molar Mass

23. 6 Molar Volume of Gases STANDARD TEMPERATURE AND PRESSURE Standard Temperature = 0 o C Standard Pressure = 1 atmosphere (atm) Rule: 1 mol of any gas will occupy a volume of 22.4 L.

24. 6 1. Determine the volume, in Liters, of 0.600 mol of sulfur dioxide gas, SO 2 (g), at STP. Answer: _____________________________ 13.4 L SO 2

25. 6 2. Determine the number of moles in 33.6 L of He (g), at STP. Answer: _____________________________ 1.5 mol of He

26. 6 3.What is the volume at STP of these gases? a. 5.4 mol of O 2 b. 3.2 x 10 -2 mol CO 2 Answer: __________________________ 1.21 x 10 2 L O 2 Answer: __________________________ 0.717 L CO 2

27. 6 4.At STP, how many moles are in these volumes? a. 89.6 L of SO 2 b. 5.42 x 10 -1 mL Ne Answer: __________________________ 4 mol SO 2 Answer: __________________________ 2.42 x 10 -5 mol of Ne

28. 6 5. The density of a gaseous compound of carbon and oxygen is 1.94 g/L at STP. Determine the gram molecular mass of the compound. Is the compound carbon dioxide or carbon monoxide? SOLUTION: 1.94 g L 22.4 L 1 mole = mole The total mass of CO 2 gas is 44 grams. Therefore, CO 2 is 44 grams/mole. The gas with this density is CO 2 44 g

29. 6. The densities of gases A, B, and C are 1.25 g/L, 2.86 g/L and 0.714 g/L respectively. Calculate the molecular mass of each of the following and determine which is, sulfur dioxide, SO 2, Nitrogen gas, N 2, and Methane gas, CH 4 6 Solution: 1.25 g L 22.4 L 1 mole = 28 g mole The total mass of N 2 gas is 28 grams. Therefore, N 2 is 28 grams/mole. Gas “A” is nitrogen gas.

30. 6. The densities of gases A, B, and C are 1.25 g/L, 2.86 g/L and 0.714 g/L respectively. Calculate the molecular mass of each of the following and determine which is, sulfur dioxide, SO 2, Nitrogen gas, N 2, and Methane gas, CH 4 6 Solution: 2.86 g L 22.4 L 1 mole = 64 g mole The total mass of SO 2 gas is 64 grams. Therefore, SO 2 is 64 grams/mole. Gas “B” is Sulfer dioxide gas.

31. 6. The densities of gases A, B, and C are 1.25 g/L, 2.86 g/L and 0.714 g/L respectively. Calculate the molecular mass of each of the following and determine which is, sulfur dioxide, SO 2, Nitrogen gas, N 2, and Methane gas, CH 4 6 Solution: 0.714 g L 22.4 L 1 mole = 16 g mole The total mass of CH 4 gas is 16 grams. Therefore, CH 4 is 16 grams/mole. Gas “C” is CH 4 gas.

32. Compounds are made of at least 2 elements. To determine the percent composition of a compound use the following formula: 6 grams of element Total mass of compound % mass =X 100% Example: Calculate the percent sodium in Sodium Hydroxide. Solution: The formula for Sodium Hydroxide is NaOH so….. Na 1 x 23 g = 23 g O 1 x 16 g = 16 g H 1 x 1 g = 1 g TOTAL MASS = 40 g % Na = 23 g 40 g X 100 = 58 % Na

33. An empirical formula is a formula that shows the LOWEST WHOLE NUMBER ratio of the elements in a compound. 6 Molecular Formula Empirical Formula C6H6C6H6 CH CH 4 Example 2: Example 1: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen. Solution: Step 1 – Assume you have a 100g sample. Remember the percentage composition is independent of sample size. This means that the percent of nitrogen and percent of oxygen in this compound WILL NOT CHANGE….regardless of sample size! Solution: So….if we have a 100 g sample, then we can convert the percentages directly to grams….

34. An empirical formula is a formula that shows the LOWEST WHOLE NUMBER ratio of the elements in a compound. 6 Molecular Formula Empirical Formula C6H6C6H6 CH CH 4 Example 2: Example 1: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen. 25.9 g N 74.1 g O Step 2: Convert grams to moles 14 g N 1 mol N 16 g O 1 mol O = 1.85 mol N = 4.63 mol O

35. An empirical formula is a formula that shows the LOWEST WHOLE NUMBER ratio of the elements in a compound. 6 Molecular Formula Empirical Formula C6H6C6H6 CH CH 4 Example 2: Example 1: What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen. Step 3: Now divide each value by the smallest number. = 1 mol N = 2.5 mol O 25.9 g N 74.1 g O 14 g N 1 mol N 16 g O 1 mol O = 1.85 mol N = 4.63 mol O 1.85 Step 4: Now, use these values as subscripts…. N 1 O 2.5 Step 5 : Since empirical formulas have to contain WHOLE NUMBERS, double each subscript to determine this compounds empirical formula…. N 2 O 5 Question: Is this compound dinitrogen pentoxide? Answer: It certainly looks like it based upon the empirical formula……but…….empirical formulas only show: * What elements are being combined * The lowest whole number ratio of the combining elements Empirical formulas DO NOT tell us the compounds identity.

36. A molecular formula is either the same as the empirical formula or…a whole number multiple of the empirical formula. 6 Molecular Formula Empirical Formula C6H6C6H6 CH CH 4 Example 2: Example 1:

37. A compounds molecular formula can be determined if we know: Empirical formula The molar mass of the molecular compound (whose formula we are trying to find. ) 6

38. Molar Mass Empirical Formula 6 a. 60 g CH 4 N b. 78 g NaO

39. Empirical efm Molar Mass Molecular formula 6 efm a.) CH 4 N 30g 60g/30g C 2 H 8 N 2 formula b.) NaO 39g 78g/39g Na 2 O 2

40. 6 % Molar mass efm Divide by the smallest number and establish the empirical formula. Use factor label method to convert to moles. Mass (grams) Determine the chemical formula

41. 6 The water in a crystal is called water of hydration. MgSO 4 7H 2 O. heat MgSO 4 + 7H 2 O One formula unit of Magnesium Sulfate Heptahydrate contains 7 water molecules.