1 / 34

Chapter 7 “ Chemical Formulas and Chemical Compounds ”

Chapter 7 “ Chemical Formulas and Chemical Compounds ”. Yes , you will need a calculator for this chapter!. Using Formulas. How many oxygen atoms in the following? CaCO 3 Al 2 (SO 4 ) 3 How many ions in the following? CaCl 2 NaOH Al 2 (SO 4 ) 3. 3 atoms of oxygen.

hazellloyd
Download Presentation

Chapter 7 “ Chemical Formulas and Chemical Compounds ”

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 7“Chemical Formulas and Chemical Compounds” Yes, you will need a calculator for this chapter!

  2. Using Formulas • How many oxygen atoms in the following? CaCO3 Al2(SO4)3 • How many ions in the following? CaCl2 NaOH Al2(SO4)3 3 atoms of oxygen 12 (3 x 4) atoms of oxygen 3 total ions (1 Ca2+ ion and 2 Cl1- ions) 2 total ions (1 Na1+ ion and 1 OH1- ion) 5 total ions (2 Al3+ + 3 SO42- ions)

  3. Review Problems 2.75 x 1024 molecules • How many molecules of CO2 are in 4.56 moles of CO2? • How many moles of water is 5.87 x 1022 molecules? • How many atoms of carbon are in 1.23 moles of C6H12O6? • How many moles is 7.78 x 1024 formula units of MgCl2? 0.0975 mol (or 9.75 x 10-2) 4.44 x 1024 atoms C 12.9 moles

  4. More Practice 28.1 grams C • How many grams is 2.34 moles of carbon? • How many moles of magnesium is 24.31 g of Mg? • How many atoms of lithium is 1.00 g of Li? • How many grams is 3.45 x 1022 atoms of U? 1.000 mol Mg 8.68 x 1022 atoms Li 13.6 grams U

  5. What about compounds? • In 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms (think of a compound as a molar ratio) • To find the mass of one mole of a compound • determine the number of moles of the elements present • Multiply the number times their mass (from the periodic table) • add them up for the total mass

  6. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 16.00 g 24.31 g 12.01 g 84.32 g/mol 24.31 g/mol + 12.01 g/mol + 3 x (16.00 g/mol) = Thus, 84.32 grams/mol is the formula mass for MgCO3.

  7. Molar Mass • Molar mass is the generic term for the mass of one mole of any substance (expressed in grams/mol) • The same as: 1) Gram Molecular Mass (for molecules) 2) Gram Formula Mass (ionic compounds) 3) Gram Atomic Mass (for elements) • molar mass is just a much broader term than these other specific masses

  8. Examples • Calculate the molar mass of the following: Na2S N2O4 C Ca(NO3)2 C6H12O6 (NH4)3PO4 = 78.05 g/mol = 92.02 g/mol = 12.01 g/mol = 164.1 g/mol = 180.156 g/mol = 149.096 g/mol

  9. Since Molar Mass is… • The number of grams in 1 mole of atoms, ions, or molecules, • We can make conversion factors from these. - To change between grams of a compound and moles of a compound.

  10. For example • How many moles are in 5.69 g of NaOH?

  11. For example • How many moles are in 5.69 g of NaOH?

  12. For example • How many moles are in 5.69 g of NaOH? • We need to change 5.69 grams NaOH to moles

  13. For example • How many moles are in 5.69 g of NaOH? • We need to change 5.69 grams NaOH to moles • 1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g

  14. For example • How many moles are in 5.69 g of NaOH? • We need to change 5.69 grams NaOH to moles • 1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g • 1 mole NaOH = 39.998 g

  15. For example • How many moles are in 5.69 g of NaOH? • We need to change 5.69 grams NaOH to moles • 1mole Na = 22.99 g 1 mol O = 16.00 g 1 mole of H = 1.008 g • 1 mole NaOH = 39.998 g

  16. For example • How many moles are in 5.69 g of NaOH? • We need to change 5.69 grams NaOH to moles • 1mole Na = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.008 g • 1 mole NaOH = 39.998 g

  17. Calculating Percent Composition of a Compound • Like all percent problems: part whole • Next, divide by the total mass of the compound; then x 100 • Find the mass of each of the components (the elements), x 100 % = percent

  18. Example • Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S. 29.0 g Ag X 100 = 87.09 % Ag 33.3 g total Total = 100 % 4.30 g S X 100 = 12.91 % S 33.3 g total

  19. Getting it from the formula • If we know the formula, assume you have 1 mole, • then you know the mass of the elements and the whole compound (these values come from the periodic table!).

  20. Examples • Calculate the percent composition of C2H4. • How about aluminum carbonate? 85.63% C, 14.37 % H 23.06%Al, 15.40% C, and 61.54 % O

  21. Formulas 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. • Example: molecular formula for benzene is C6H6 (note that everything is divisible by 6) • Therefore, the empirical formula = CH (the lowest whole number ratio)

  22. Formulas(continued) Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  23. Formulas(continued) Formulas for molecular compounds might be empirical (lowest whole number ratio). Molecular: H2O C6H12O6 C12H22O11 (Correct formula) Empirical: H2O CH2O C12H22O11 (Lowest whole number ratio)

  24. Calculating Empirical = CH2O • Just find the lowest whole number ratio C6H12O6 CH4N • A formula is not just the ratio of atoms, it is also the ratio of moles. • In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen. • In one molecule of CO2 there is 1 atom of C and 2 atoms of O. = this is already the lowest ratio.

  25. Calculating Empirical • We can get a ratio from the percent composition. • Assume you have a 100 g sample - the percentage becomes grams (75.10% = 75.10 grams) • Convert grams to moles. • Find lowest whole number ratio by dividing each number of moles by the smallest value.

  26. Example • 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 1mol C = 3.2198 mol C 12.01 g C • 16.22 g H x 1mol H = 16.0913 mol H 1.008 g H • 45.11 g N x 1mol N = 3.2198 mol N 14.01 g N Now divide each value by the smallest value

  27. Example • The ratio is 3.2198 mol N = 1 mol N 3.2198 mol N • The ratio is 3.2198 mol C = 1 mol C 3.2198 mol N • The ratio is 16.0913 mol H = 5 mol H 3.2198 mol N = C1H5N1 which is = CH5N

  28. Practice • A compound is 43.64 % P and 56.36 % O. What is the empirical formula? • P2O5 • Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? • C4H5N2O

  29. Empirical to Molecular • Since the empirical formula is the lowest ratio, the actual molecule would weigh more. • By a whole number multiple. • Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each subscript in the empirical formula • Caffeine has a molar mass of 194 g. what is its molecular formula? = C8H10N4O2

  30. Hydrates • Hydrate: a compound that has a specific number of water molecules bound to its atoms • You can determine the formula of a hydrate by heating it until the water is gone, and analyzing the data

  31. BaCl2?H2O • Mass of the compound before heating (hydrate): 5.00 g • Mass after heating (anhydrous): 4.26 g • Find the mass of the water lost by subtracting 5.00g – 4.26 g = 0.74 g H2O • The mass of just the BaCl2 is 4.26 g

  32. Continued • Convert the mass of BaCl2 and the mass of H2O to mol • 4.26 g BaCl21 mol BaCl2 = 0.02046 mol 208.2 g BaCl2 • 0.74 g H2O 1 mol H2O = 0.04107 mol 18.016 g H2O • Divide by the smallest molar value

  33. Continued • 0.02046 molBaCl2 = 1 0.02046 molBaCl2 • 0.04107 mol H2O = 2 0.02046 molBaCl2 • Formula of the hydrate: BaCl22H2O

  34. Another Practice • FePO4? H2O • Mass before heating: 7.61 g • Mass after heating: 5.15 g • Mass of water: 2.46 g • Formula of the hydrate: FePO4 4H2O

More Related