1. Molar Mass, Empirical & Molecular Formulas Cartoon courtesy of NearingZero.net

2. Molecular Mass & Formula Mass A chemical formula is a convenient way of showing how many and what type of atoms make up a particular molecule of formula unitA chemical formula is a convenient way of showing how many and what type of atoms make up a particular molecule of formula unit

3. Molecular Mass & Formula Mass While the mass of an atom can be found by looking at the periodic table, the mass of a molecule or formula unit must be calculated.While the mass of an atom can be found by looking at the periodic table, the mass of a molecule or formula unit must be calculated.

4. Molecular Mass & Formula Mass The molecular mass of a substance can be found by finding the sum of the atomic masses that make up a particular moleculeThe molecular mass of a substance can be found by finding the sum of the atomic masses that make up a particular molecule CH 2 Cl 2 = 12 + 2(1) + 2(35.45)CH 2 Cl 2 = 12 + 2(1) + 2(35.45) = 84.9 amu = 84.9 amu

5. Molecular Mass & Formula Mass The formula mass, which is used for ionic compounds, is found in the same fashion as molecular compounds.The formula mass, which is used for ionic compounds, is found in the same fashion as molecular compounds. NaCl = 22.99 + 35.45NaCl = 22.99 + 35.45 = 58.44 amu = 58.44 amu

6. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g

7. The Avogadro Constant Atoms are much too small to count individuallyAtoms are much too small to count individually The mass of a single molecule is so small that it is impossible to measure by ordinary means in the laboratory.The mass of a single molecule is so small that it is impossible to measure by ordinary means in the laboratory.

8. The Avogadro Constant For this reason, scientists had to come up with a way to convert atomic mass units to something that could be used in the laboratory, gramsFor this reason, scientists had to come up with a way to convert atomic mass units to something that could be used in the laboratory, grams

9. The Avogadro Constant Chemists have found that 6.02 x 10 23 atoms of an element has a mass in gram equivalent to the mass of one atom in atomic mass unitsChemists have found that 6.02 x 10 23 atoms of an element has a mass in gram equivalent to the mass of one atom in atomic mass units

10. The Avogadro Constant This number, 6.02 x 10 23 is known as the Avogadro ConstantThis number, 6.02 x 10 23 is known as the Avogadro Constant

11. The Mole Just like a dozen donuts is equivalent to twelve donuts, one mole of donuts would be 6.02 x 10 23 donuts. Unlike the term dozen, however, moles are used to quantify very small objects, like atoms.Just like a dozen donuts is equivalent to twelve donuts, one mole of donuts would be 6.02 x 10 23 donuts. Unlike the term dozen, however, moles are used to quantify very small objects, like atoms.

12. The Mole A mole is just a particular number of atoms, ions, molecules, or formula unitsA mole is just a particular number of atoms, ions, molecules, or formula units The mole is the SI base unit representing the chemical quantity of a substance, and has been given the symbol N AThe mole is the SI base unit representing the chemical quantity of a substance, and has been given the symbol N A

13. Review: The Mole  The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C.  1 mole of anything = 6.022  10 23 units of that thing

14. The Mole The mass of one mole of molecules, atoms, ions, or formula units is called the molar mass of that speciesThe mass of one mole of molecules, atoms, ions, or formula units is called the molar mass of that species

15. Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO 2 = 44.01 grams per mole H 2 O = 18.02 grams per mole Ca(OH) 2 = 74.10 grams per mole

16. Mass-Mole and Mole- Mass Conversions In order to convert the mass of an object to the number of moles, you divide by the molar mass of the substanceIn order to convert the mass of an object to the number of moles, you divide by the molar mass of the substanceEXAMPLE: 4.56 g CO 2 x 1 mole CO 2 = 0.104 mole CO 2 44 g CO 2 44 g CO 2

17. Mass-Mole and Mole-Mass Conversions In order to convert moles to mass, you multiply by the molar mass of the substanceIn order to convert moles to mass, you multiply by the molar mass of the substanceEXAMPLE: 0.58 moles NH 4 NO 3 x 80.g NH 4 NO 3 = 46 g NH 4 NO 3 1 mole NH 4 NO 3 1 mole NH 4 NO 3

18. Volume-Mole and Mole-Volume Conversions In order to do any conversions with moles and volume, the substance you are dealing with must be a gas and exist at STP which is 0°C and 1 atmIn order to do any conversions with moles and volume, the substance you are dealing with must be a gas and exist at STP which is 0°C and 1 atm To convert the volume of an object to the number of moles, you divide by 22.4 LTo convert the volume of an object to the number of moles, you divide by 22.4 L EXAMPLE: 54.2 L of N 2 x 1 mole N 2 = 2.42 moles N 2 22.4 L N 2

19. Volume-Mole and Mole-Volume Conversions To convert the number of moles to volume, you multiply by 22.4 LTo convert the number of moles to volume, you multiply by 22.4 LEXAMPLE: 0.78 moles of He x 22.4 L He = 17 L He 1 mole He 1 mole He

20. Molecules/Atoms/Formula Units - Moles and Vice Versa To convert the number of M/A/F to moles, divide by 6.02 x 10 23 M/A/FTo convert the number of M/A/F to moles, divide by 6.02 x 10 23 M/A/F To convert moles to M/A/F multiply by 6.02 x 10 23 M/A/FTo convert moles to M/A/F multiply by 6.02 x 10 23 M/A/F Remember...Remember... –molecules are for covalent compounds –formula units are for ionic compounds

21. Molecules/Atoms/Formula Units - Moles and Vice Versa EXAMPLE: 5.21 X 10 24 molecules SO 2 x 1 mole SO 2 = 8.65 moles SO 2 6.02 x 10 23 molecules SO 2 1.25 moles Cu(NO 3 ) 2 x 6.02 x 10 23 f.u. = 7.53 x10 23 f.u. Cu(NO 3 ) 2 1 mole Cu(NO 3 ) 2

22. Converting from Mass-Volume, Mass- M/A/F, Volume-M/A/F, or Anything in Between Convert the first unit of moles. Then convert from moles to the unit you’re looking for.Convert the first unit of moles. Then convert from moles to the unit you’re looking for. MOLES ARE THE MIDDLE MEN! MOLES ARE THE MIDDLE MEN!

23. The Mole

24. Percentage Composition The percentage composition of a compound is a statement of the relative mass each element contributes to the mass of the compound as a whole “PART TO WHOLE!”

25. Percentage Composition To determine percent composition of a particular element in a compound, divide the mass of the element by the molecular mass of the compound, and multiply by 100To determine percent composition of a particular element in a compound, divide the mass of the element by the molecular mass of the compound, and multiply by 100 If the percent composition were calculated for all elements of the compound, and added together, the sum of the percents should equal 100%If the percent composition were calculated for all elements of the compound, and added together, the sum of the percents should equal 100%

26. Percentage Composition FOR EXAMPLE: Calculate the % composition of carbon in ethanol (C 2 H 5 OH) 1.Calculate the molecular mass of the compound: 2(12) + 5(1) + 1(16) + 1 = 46 amu 2.Calculate the total mass of the element in question: 2(12) = 24 amu 3.Divide the mass of the element by the mass of the compound and multiply by 100: 24 amu x 100 = 52 % carbon 46 amu

27. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

28. Empirical Formulas The empirical formula of a substance is the simplest, whole-number ratio between the atoms of the elements present in a compound

29. Empirical Formulas A step-by-step method for determining empirical formulas is found below: 1.Convert all quantities to moles (if the quantity is a percent, remove the percent sign, and stick in a grams sign!) 2.Divide all numbers by the SMALLEST number of moles 3.If the quotient is NOT a whole number, convert the decimal to a fraction, then multiply by the denominator of the fraction to make it a whole number 4.Write each element with its corresponding subscript

30. Empirical Formulas Find the empirical formula for a compound containing 43.9% C, 7.32 % H, 48.78 % O Step One: 43.9 g C x 1 mole = 3.66 moles 12 g 12 g 7.32 g H x 1 mole = 7.32 moles 1 g 1 g 48.78 g O x 1 mole = 3.05 moles 16 g 16 g

31. Empirical Formulas Step 2: C: 3.66/3.05 =1.2 H: 7.32/3.05 =2.4 O: 3.05/3.05 =1 Step 3: C: 1.2 = 6/5 x 5 = 6 H: 2.4 = 12/5 x 5 = 12 O: 1 x 5 = 5 Step 4: C 6 H 12 O 5

32. Molecular Formula Molecular formulas indicate the actual number of atoms of each element making up a moleculeMolecular formulas indicate the actual number of atoms of each element making up a molecule To determine a molecular formula, one must know both the empirical formula and the molecular mass of the substanceTo determine a molecular formula, one must know both the empirical formula and the molecular mass of the substance

33. Molecular Formula What is the molecular formula of a substance that has an empirical formula of AgCO 2 and a formula mass of 304 g? Step 1: Determine the formula mass of the empirical formula: 108+12+2(16) = 152 g Step 2: Divide the formula mass by the mass of the empirical formula: 304 g/152 g = 2 Step 3: Multiply the subscripts of the empirical formula by the ratio found in step 2: Ag 2 C 2 O 4

34. Formulas  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

35. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

36. Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

37. Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

38. Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

39. Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

40. Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

41. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

42. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the mass given by the emipirical formula.

43. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

44. Hydrates Hydrates are compounds that have crystallized from a water solution. In these compounds, the water molecules adhere to the ions in the compound, and become part of the crystalHydrates are compounds that have crystallized from a water solution. In these compounds, the water molecules adhere to the ions in the compound, and become part of the crystal

45. Hydrates The formula for a hydrate indicates the number of water molecules attached with a coefficientThe formula for a hydrate indicates the number of water molecules attached with a coefficient CuSO 4 5H 2 0 would have 5 water molecules attached to it To name hydrates, name the regular compound and use a Greek prefix to indicate the number of waters presentTo name hydrates, name the regular compound and use a Greek prefix to indicate the number of waters present CuSO 4 5H 2 0 would be named copper (II) sulfate pentahydrate CuSO 4 5H 2 0 would be named copper (II) sulfate pentahydrate

46. Hydrates When calculating the formula mass for a hydrate, the mass of the water is added to the mass of the formula unitWhen calculating the formula mass for a hydrate, the mass of the water is added to the mass of the formula unit CuSO 4 5H 2 0 = 63.5 + 32 + 4(16) + 5(18) CuSO 4 5H 2 0 = 63.5 + 32 + 4(16) + 5(18) = 249.5 amu = 249.5 amu To determine the ratio of compound to water in a hydrate, the compounds can be heated to drive off the water. By comparing the mass of the sample to the mass of the water, one can determine the formula of the hydrateTo determine the ratio of compound to water in a hydrate, the compounds can be heated to drive off the water. By comparing the mass of the sample to the mass of the water, one can determine the formula of the hydrate

47. Hydrates EXAMPLE: A 10.407 g sample of hydrated barium iodide is heated to drive off the water. The mass of the dry sample is 9.520 g. What is the formula of the hydrate? Step 1: Find the mass of the water in the compound: 10.407 g – 9.520 g = 0.887 g water Step 2: Convert mass of both compounds to moles: 9.520 g BaI 2 x 1 mole BaI 2 = 0.0243 moles BaI 2 391 g BaI 2 391 g BaI 2 0.887 g H 2 O x 1 mole H 2 O = 0.0493 moles H 2 O 18 g H 2 O 18 g H 2 O

48. Hydrates Step 3: Divide both results by the smaller number of moles: 0.0243 moles / 0.0243 moles = 1 0.0493 moles / 0.0243 moles = 2 Step 4: Write the formula for the hydrate: BaI 2 2H 2 0