1. 2 Amounts of Substance Learning Objectives: 2.1 A r & M r, Avogadro’s number, and the mole 2.2 Ideal Gas Law 2.3 Empirical and Molecular Formula 2.4 Moles in solutions 2.5 Titrations and calculations 2.6 Atom economy and yields

2. 2.1 Relative Atomic Mass Relative atomic mass = mass of an atom relative to relative to 12 C It is used to compare masses as the actual masses of atoms is too small A r can be found from the periodic table

3. 2.1 Relative Molecular Mass Relative molecular mass = relative mass of one molecule M r = sum of A r of all atoms in the molecule M r is used to compare masses of molecules as the actual masses are too small

4. 2.1 Avogadro’s Number Avogadro’s constant = 6.02 x 10 23 atoms/molecules in 1 mole This is true for ALL atoms/elements/molecules. This allows us to compare them. Mass of 1 mole = M r in grams

5. 2.1 Calculations with M r Mass of 1 mole = M r in grams This can be used to convert from grams to moles. Moles can then be used to convert between different atoms/elements/molecules. (We will look at this in section 2.5).

6. How many moles are there in 0.53 g of Na 2 CO 3 ? Given:What do we need? 0.53 g M r = (2 x A r Na) + (1 x A r C) + (3 x A r O) M r = (2 x 23.0) + (12.0) + (3 x 16.0) = 106.0 g So… 1 mole of Na 2 CO 3 = 106.0 g So… 0.53 g x=0.0050 moles Na 2 CO 3 1 mole 106.0 g We use this as a conversion factor and the units we don’t want cancel out. MrMr

7. 2.2 Ideal Gas Law Ideal gas law relates pressure, volume, moles, and temperature.

8. 2.2 Ideal Gas Law: Units P = Pascals (Pa) V = meters cubes (m 3 ) n = moles R = constant (8.31 J K -1 mol -1 ) T = Kelvin (K)

9. How many moles of hydrogen are present in a 100 cm 3 at 20.0 o C and a pressure of 100 kPa? PV = nRT P = 100 kPa x = 100,000 Pa 1 kPa 1000 Pa V = 100 cm 3 x 1 m 3 10 6 cm 3 = 0.0001 m 3 n = ?R = 8.31 J K -1 mol -1 T = 20.0 o C = 293 K 100,000 x 0.0001 = n x 8.31 x 20 n = (100,000 x 0.0001) / (8.31 x 20) = 0.00411 moles

10. 2.3 Empirical Formula Empirical formula = lowest whole number ratio of atoms in a molecule Molecular formula = C 6 H 12 Empirical formula = CH 2

11. Calculating Empirical Formula If you have the masses of all elements in a compound you can calculate the empirical formula by converting to moles. Moles can be compared but grams cannot as all elements have different masses.

12. Calculating Empirical Formula 1.Find masses from experiment. 2.Work out number of moles. 3.Convert number of moles to whole number ratio.

13. Calculating Empirical Formula 10.1 g of a white solid contains 4.01 g calcium, 1.20 g of carbon, and 4.80 g of oxygen. CaCO Mass = 4.01 g1.20 g4.80 g Ar = 40.112.016.0 Moles = 0.100.100.30 / smallest = 0.100.100.10 Ratio = 113 Empirical Formula = CaCO 3

14. Calculating Molecular Formula 1.Find the ratio between the M r ’s molecular formula and the empirical formula. 2.Multiply by the ratio.

15. What is the molecular formula of the compound with empirical formula C 2 H 6 O and an M r of 92. Empirical Formula M r = (2 x 12.0) + (6 x 1) + (1 x 16) = 46.0 Molecular Formula M r = 92 Ratio = 92/46 = 2  Multiply by 2  C 4 H 12 O 2

16. 2.4 Moles in Solutions M r can be used to convert between grams, moles, and concentrations of different substances (see section 2.5). Concentration is measured in number of moles on a certain volume. Concentration = = mol dm -3 moles dm 3

17. What is the concentration of a solution made of 1.17 g NaCl dissolved in 500 cm3 of water? Given: mass NaCl = 1.17 gV = 500 cm 3 Concentration = mol / dm 3 1.17 g Mr = 23.0 + 35.5 = 58.5 x 1 mol 58.5 g = 0.0200 mol NaCl 0.0200 mol 500 cm 3 x 1000 cm 3 1 dm 3 = 0.0400 mol dm -3

18. 2.5 Calculations M r and balanced equations can be used to convert between grams, moles, and concentrations of different substances.

19. What mass of gas is produced by 0.12 g of magnesium and excess hydrochloric acid? Mg + 2HCl  H 2 + MgCl 2 Given: mass Mg = 0.12 gM r = 24.3 Needed: mass of H 2 M r = 2 x 1 = 2 0.12 g 1 mol 24.3 g x= 0.0050 mol Mg 0.0050 mol Mg 1 mol H 2 1 mol Mg x= 0.0050 mol H 2 0.0050 mol H 2 2 g 1 mol = 0.010 g H 2 x

20. Titrations Titrations are used to determine the amount of one of the substances. We can use the acid base equation to determine the molar ratio and calculate the amount of substance. HCl + NaOH --> NaCl + H 2 O

21. 25.0 cm 3 of a solution of sodium hydroxide, NaOH, of unknown concentration was neutralised by 20.0 cm 3 of a 0.100 mol dm -3 solution of HCl. What is the concentration of the NaOH? HCl + NaOH --> NaCl + H 2 O V = 25 cm3 NaOH ? = conc. NaOH V = 20.0 cm3 HCl reacted conc = 0.100 mol dm -3 HCl

23. 2.6 Atom Economy

24. 1.Equation 2.Molar ratios 3.Use Mr to calculate mass

25. Yield Atom economy tells us how much mass is wasted in theory. Yield describes the practical efficiency of a process, the actual amount wasted. Yield is always lower than the atom economy because… – Substance loss in the physical process – Reactions that do not go to completion – Competing chemical reactions

26. Calculating Yield