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Chapter 7: Moles and Quantities

Chapter 7: Moles and Quantities. A. Moles: Mole – a given quantity (amount) of a substance - the molar mass (gram formula mass or molecular weight) of any substance (changes for each substance) - 22.4 L of any gas - 6.02 x 10 23 particles in any substance.

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Chapter 7: Moles and Quantities

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  1. Chapter 7: Moles and Quantities A. Moles: • Mole – a given quantity (amount) of a substance - the molar mass (gram formula mass or molecular weight) of any substance (changes for each substance) - 22.4 L of any gas - 6.02 x 1023particles in any substance

  2. B. Molar Mass: • Molar mass – add up masses of all atoms in a compound from the P.T. - round mass (g) to 1 decimal (.1) ex. MgCl2 24.3 g + (35.5 g x 2) = 95.3 g • Hydrate – a compound with water molecules that are within the substance (do not see) ex. CuSO45H2O ( just means to add) 63.5 + 32.1 + (16.0 x 4) + (18.0 x 5) Answer = 249.6 grams

  3. C. Moles  Mass Conversions: • Factor-Label Method: • First, find molar mass of the substance • Start with the given from the problem • Cancel out units on the diagonal • New unit goes on top • Add #’s in • Multiply top, divide bottom to get answer ex. 2 moles of NaCl, convert to grams 2 moles x 58.5 g = 1 mole 117.0 grams

  4. Proportions Method: • Start with given • Use = sign in between • Cross-Multiply, then divide ex. 2 moles of NaCl, convert to grams 2 moles = x 1 mole 58.5 g (2 moles x 58.5 g) / 1 mole Answer = 117.0 grams

  5. ex. Convert 123.8 grams Li2SO3 to moles 123.8 g x 1 mole = 1.3 moles Li2SO3 93.9 g 123.8 g = x 93.9 g 1 mole (123.8 g x 1 mole) / (93.9 g) = 1.3 moles Li2SO3

  6. D. Moles  Liters Conversions: • 1 mole = 22.4 L of any gas at STP • STP – Standard Temperature and Pressure - at 1 atm (atmosphere) of pressure and 0 C • Start with the given • Show work (factor-label or proportions) • The compound does not matter • Do not need to find the molar mass

  7. E. Moles  Particles Conversions: • 1 mole = 6.02 x 1023particles (or molecules) of any substance • 6.02 x 1023 = Avogadro’s number • Start with the given • Show work (factor-label or proportions) • The compound does not matter • Do not need to find the molar mass

  8. F. Multi-Step Mole Conversions: • When “mole” is not in the question as either the given or the find (the question) • Convert the given to moles • Convert moles to the find • Show work (factor-label or proportions)

  9. ex. How many grams are in 2.7 L of NO2? 2.7 L x 1 mole x 46 g = 5.5 g NO2 22.4 L 1 mole 2.7 L = x 22.4 L 46 g x = 5.5 g NO2

  10. G. Percent Composition: • Percent composition – the percent by mass of each element in a compound • Formulas: % comp = grams of element x 100 grams of compound % comp = total mass of element x 100 total mass of compound

  11. ex. 8.2 grams Mg combines with 5.4 grams O to form MgO. What is the percent composition? % comp Mg = (8.2 g / 13.6 g) x 100 = 60.3 % Mg % comp O = (5.4 g / 13.6 g) x 100 = 39.7 % O Total % will always = 100 %

  12. ex. What is the percent composition of K2CrO4? K2CrO4 = 194.2 g % comp K2 = (78.2 g / 194.2 g) x 100 = 40.3 % K % comp Cr = (52.0 g / 194.2 g) x 100 = 26.8 % Cr % comp O4 = (64.0 g / 194.2 g) x 100 = 33.0 % O

  13. H. Empirical vs. Molecular Formulas: • Law of Definite Proportions – for any given compound, the elements always combine in the same proportions or ratio (subscripts) • Empirical formula – the lowest whole-number ratio of elements in a compound (reduced form) • Molecular formula – the “true” formula of a compound (not reduced) ex. H2O2 (hydrogen peroxide) – molecular HO (not a real compound) - empirical

  14. ex. Of the following compounds, which ones are empirical formulas? CH, C2H2, C6H12, CH2O, C2H4O2, C4H8O Answers:CH, CH2O, C4H8O (reduced)

  15. Calculating empirical formulas: • Convert % (comp) to grams (change units) • Convert grams to moles • Divide all mole answers by the smallest mole answer (get you the ratios of atoms) • Write empirical formula based on these mole ratios (should be in the reduced form)

  16. Calculating empirical formulas (cont.): ex. Find the E.F. of: 79.8% C and 20.2% H 79.8 g C x 1 mole = 6.65 moles C = 1 C 12.0 g 6.65 20.2 g H x 1 mole = 20.2 moles H = 3 H 1.0 g 6.65 Answer:CH3

  17. Calculating molecular formulas: • First find empirical formula (if not given) • Find molar mass of empirical formula • Divide the molar mass of molecular formula by the molar mass of empirical formula (given) • Using your answer, multiply all the atoms of the empirical formula to get the new molecular formula • Check molar mass of new found molecular formula

  18. Calculating molecular formulas (cont.): ex. What is the molecular formula with a molar mass of 180 g. if its empirical formula is CH2O? CH2O = 30 g. 180 g / 30 g = 6 (CH2O) Answer:C6H12O6

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