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Section 9.3—Analysis of a Chemical Formula

Section 9.3—Analysis of a Chemical Formula. How can we determine a chemical formula?. Percent Composition. What is Percent Composition?. Example #1. Example:

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Section 9.3—Analysis of a Chemical Formula

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  1. Section 9.3—Analysis of a Chemical Formula How can we determine a chemical formula?

  2. Percent Composition

  3. What is Percent Composition?

  4. Example #1 Example: A 8.5 g sample is composed of carbon and hydrogen. If 5.5 g of the sample is found to be carbon, what is the percent composition by mass of the sample?

  5. Example #1 Example: A 8.5 g sample is composed of carbon and hydrogen. If 5.5 g of the sample is found to be carbon, what is the percent composition by mass of the sample? If total = 8.5 g - and C = 5.5 g then H = 3.0 g

  6. Let’s Practice #1 A sample is 57.0% by mass chlorine, how many grams of chlorine are present in a 27.5 g sample?

  7. Let’s Practice #1 A sample is 57.0% by mass chlorine, how many grams of chlorine are present in a 27.5 g sample? 27.5 g   27.5 g 100 100

  8. Example #2 Percent composition can also be determined from a chemical formula Example: Find the percent composition, by mass, of CaCl2

  9. Ca 1  40.08 g/mole 40.08 g/mole =  + 70.90 g/mole Cl 2 35.45 g/mole = 110.98 g/mole Cl = 70.90 g Example #2 Percent composition can also be determined from a chemical formula Example: Find the percent composition, by mass, of CaCl2 So for 1 mole: total = 110.98 g Ca = 40.08 g

  10. Let’s Practice #2 Find the percent composition, by mass, of NaNO3

  11. Na 1  22.99 g/mole 22.99 g/mole =  14.01 g/mole N 1 14.01 g/mole =  48.00 g/mole + O 3 16.00 g/mole = 85.00 g/mole Let’s Practice #2 Find the percent composition, by mass, of NaNO3

  12. Empirical Formulas

  13. What’s an Empirical Formula? Empirical– from data Empirical Formula– Chemical formula determined from lab data. Lowest possible ratio of atoms CH2 is the lowest ratio (and empirical formula) of the molecule C3H6

  14. Ratio of Atoms in a Molecule Subscripts in a chemical formula show the ratio of atoms (or ions) in a molecule a sample of CaCl2 has 1 calcium ion : 2 chlorine ions If the subscripts give the ratio of atoms, then they also give the ratio of moles of atoms a sample of CaCl2 has 1 mole of calcium ions : 2 moles of chlorine ions We can use the unit “mole” to count things Atoms and ions can be counted by “moles”

  15. Using Mole Ratio of Atoms in a Molecule Therefore,if the ratio of moles of each atom is found… then the subscripts of the chemical formula are known 1 mole C 2 mole H CH2

  16. Example #3 If given percents, use those percents as grams (as if you assume you have a 100 g sample)—Remember %’s add up to 100! 1 2 Change grams to moles for each atom Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) 3 4 Use the ratio as subscripts for writing the chemical formula Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca

  17. Example #3 If given percents, use those percents as grams (as if you assume you have a 100 g sample)—Remember %’s add up to 100! 1 2 Change grams to moles for each atom Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) 3 4 Use the ratio as subscripts for writing the chemical formula Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca mol Ca 1 36.1 g Ca 0.901 = _____ mol Ca 40.08 g Ca CaCl2 mol Cl 1 63.9 g Cl 1.80 = _____ mol Cl 35.45 g Cl 1.80 mol Cl = 2 mol Cl 0.901 0.901 mol Ca = 1 mol Ca 0.901

  18. mol C 1 40.92 g C 3.41 = _____ mol C 12.01 g C mol O mol H 1 1 54.5 g O 4.58 g H 4.53 3.41 = _____ mol H = _____ mol O 16.00 1.01 g H g O Let’s Practice #3 Find the empirical formula if a sample contains 40.92 g C, 4.58 g H and 54.5 g O C3H4O3 4.53 mol H = 1.33 mol H 3.41 3.41 mol C = 1 mol C 3.41 Multiply the ratio (1 : 1.33 : 1) by 3 to make a whole number ratio (3 : 4: 3) 3.41 mol O = 1 mol O 3.41

  19. Molecular Formulas

  20. What’s a Molecular Formula? Empirical Formula– Chemical formula determined from lab data. Lowest possible ratio of atoms Molecular Formula– Actual ratio of atoms in a molecule

  21. Empirical versus Molecular Formula The empirical formula is the lowest possible ratio. The molecular formula is the actual ratio A molecule with the empirical formula: Could have one of the following molecular formulas: NO2 NO2, N2O4, N4O8… CH2 CH2, C2H4, C4H8…

  22. Example #4 1 Find the empirical formula, if not given 2 Find the molar mass of the empirical formula Find the ratio of the molecular formula’s molar mass (must be given to you) to the empirical formula’s molar mass 3 Multiply the empirical formula’s subscripts by the ratio found in step 3. 4 Example: Empirical formula = C3H4O3. The molecular formula’s molar mass = 176.14 g/mole. Find the molecular formula.

  23. Example #4 1 Find the empirical formula, if not given 2 Find the molar mass of the empirical formula Find the ratio of the molecular formula’s molar mass (must be given to you) to the empirical formula’s molar mass 3 Multiply the empirical formula’s subscripts by the ratio found in step 3. 4 Example: Empirical formula = C3H4O3. The molecular formula’s molar mass = 176.14 g/mole. Find the molecular formula. 176.14 g/mole = 2 88.07 g/mole  C 3 12.01 g/mole 36.03 g/mole =  H 4 1.01 g/mole 4.04 g/mole C3H4O3  2 =  + 48.00 g/mole O 3 16.00 g/mole = C6H8O6 88.07 g/mole

  24. Hydrate Formulas

  25. What’s a Hydrate? Hydrate– Molecule that has water physically attached to it It’s not dissolved in water…hydrates can be solid, liquid or gas!

  26. Hydrates are inorganic salts "containing water molecules combined in a definite ratio as an integral part of the crystal“ • either bound to a metal center or have crystallized with the metal • When writing a hydrate, a dot is used to separate the formula compound from the number of water molecules • Examples) magnesium sulfate heptahydrate cobalt (II) chloride hexahydrate tin (II) chloride dihydrate

  27. Hydrate anhydride + water heat Hydrate & anhydride • Hydrate = molecule with water molecules physically attached • anhydride = molecule with water removed • The water can be removed by heating the hydrate Finding the ratio of anhydride molecules to water molecules gives you the hydrate formula

  28. Example #5 1 Find the mass of anhydride & water if not given 2 Change mass of anhydride & water to moles 3 Find the ratio of the moles water to moles anhydride 4 Write the hydrate formula Example: 2.46 g MgSO4 hydrate is heated and 1.20 g MgSO4 anhydride is left. Find the hydrate formula.

  29. Example #5 1 Find the mass of anhydride & water if not given 2 Change mass of anhydride & water to moles 3 Find the ratio of the moles water to moles anhydride 4 Write the hydrate formula Example: 2.46 g MgSO4 hydrate is heated and 1.20 g MgSO4 anhydride is left. Find the hydrate formula. Hydrate = anhydride + water 2.46 g = 1.20 g + water Water = 1.26 g 0.0699 mol H2O = 7.01 0.00997 mole MgSO4 mol MgSO4 1 1.20 g MgSO4 0.00997 = _______ mol MgSO4 120.38 g MgSO4 mol H2O 1 1.26 g H2O 0.0699 = _______ mol H2O MgSO4 7 H2O 18.02 g H2O

  30. Calculating % of Water • Determine the mass of the hydrated compound • Determine the mass of JUST the water • Then divide % H2O = Mass of water x 100 Mass of hydrate Example) CuSO4 ⋅ 5H2O CuSO4 + 5H2O Blue white

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