# AS1 MOLES. oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1

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1 Moles 1 MOLES The mole the standard unit of amount of a substance the number of particles in a mole is known as Avogadro s constant (L) Avogadro s constant has a value of x mol -1. Example If one atom has a mass of x g one mole of atoms will have a mass of x g x x = 7.432g Q.1 Calculate the mass of one mole of carbon-12 atoms. [ mass of proton x g, mass of neutron x g, mass of electron x g ] MOLE CALCULATIONS Substances mass g or kg molar mass g mol -1 or kg mol -1 moles = mass molar mass Example Calculate the number of moles of oxygen molecules in 4g oxygen molecules have the formula O 2 the relative mass will be 2 x 16 = 32 so the molar mass will be 32g mol -1 moles = mass = 4g ANS moles molar mass 32g mol -1 Q.2 Calculate the number of moles in 10g of Ca atoms 10g of CaCO 3 36g of water molecules 4g of hydrogen atoms 4g of hydrogen molecules Calculate the mass of... 2 moles of CH moles of NaNO 3 6 moles of nitrogen atoms 6 moles of nitrogen molecules 20 moles of NH 3

2 2 Moles Solutions molarity concentration / mol dm -3 volume dm 3 or cm 3 moles = concentration x volume = molarity x volume in dm 3 = molarity x volume in cm The 1000 takes into account that there are 1000 cm 3 in 1dm 3 Example 1 Calculate the number of moles of sodium hydroxide in 25cm 3 of 2M NaOH moles = molarity x volume in cm = 2 mol dm -3 x 25cm 3 ANS moles 1000 Example 2 What volume of 0.1M H 2 SO 4 contains moles? volume = 1000 x moles (re-arrangement of above) in cm 3 molarity = 1000 x ANS. 20 cm mol dm -3 Example g of Na 2 CO 3 are dissolved in water and the solution made up to 250 cm 3. What is the concentration of the solution in mol dm -3? molar mass of Na 2 CO 3 = 106g mol -1 no. of moles in 250cm 3 = 4.24g / 106g mol -1 = 0.04 moles no. of moles in 1000cm 3 (1dm 3 ) = 0.16 moles ANS mol dm -3. Q.3 Calculate the number of moles in 1dm 3 of 2M NaOH 250cm 3 of 2M NaOH 5dm 3 of 0.1M HCl 25cm 3 of 0.2M H 2 SO 4 Calculate the concentration (in moles dm -3 ) of solutions containing 0.2 moles of HCl in 2dm moles of NaOH in 25cm 3

3 Moles 3 Empirical Formula EMPIRICAL FORMULAE AND MOLECULAR FORMULAE Description Expresses the elements in a simple ratio (e.g. CH 2 ). It can sometimes be the same as the molecular formula (e.g H 2 O and CH 4 ) Calculations You need mass, or percentage mass, of each element present relative atomic masses of the elements present Example 1 Calculate the empirical formula of a compound containing C (48%), H (4%) and O (48%) C H O 1) Write out percentages (by mass) 48% 4% 48% 2) Divide by the relative atomic mass 48/12 4/1 48/16... this gives a molar ratio ) If not whole numbers then scale up 4) Express as a formula C 4 H 4 O 3 Example 2 Calculate the empirical formula of a compound containing C (1.8g), O (0.48g), H (0.3g) C H O 1) Write out ratios by mass ) Divide by relative atomic mass 1.8 / / / 16 (this gives the molar ratio) ) If not whole numbers then scale up - try dividing by smallest value (0.03) ) Express as a formula C 5 H 10 O Molecular Formula Description Exact number of atoms of each element in the formula (e.g. C 4 H 8 ) Calculations Compare the empirical formula with the relative molecular mass. The relative molecular mass of a compound will be an exact multiple (x1, x2 etc.) of its relative empirical mass. Example Calculate the molecular formula of a compound of empirical formula CH 2 and relative molecular mass 84. mass of CH 2 unit = 14 divide molecular mass (84) by 14 = 6 molecular formula = empirical formula x 6 = C 6 H 12

4 4 Moles MOLAR MASS CALCULATIONS RELATIVE MASS Relative Atomic Mass (A r ) The mass of an atom relative to that of the carbon 12 isotope having a value of or average mass per atom of an element x 12 mass of an atom of carbon-12 Relative Molecular Mass (M r ) The sum of all the relative atomic masses present in an entity - even if it is ionic and so not a molecule! or average mass of an entity x 12 mass of an atom of carbon-12 MOLAR MASS Description The mass of one mole of substance. It has units of g mol -1 or kg mol -1. e.g. the molar mass of water is 18 g mol -1 molar mass = mass of one particle x Avogadro's constant (i.e x mol -1 ) Calculations methods include using the ideal gas equation PV = nrt the Molar Volume at stp For 1 mole of gas for n moles of gas PV = R T PV = n R T PV = n R T also PV = m R T M PV = m R T M where P pressure Pascals (Pa) or N m -2 V volume m 3 (there are 10 6 cm 3 in a m 3 ) n number of moles of gas R gas constant 8.31 J K -1 mol -1 T temperature Kelvin (K = C + 273) m mass g or Kg M molar mass g mol -1 or Kg mol -1 Old units 1 atmosphere is equivalent to 760 mm/hg or x 10 5 Pa (Nm -2 ) 1 litre (1 dm 3 ) is equivalent to 1 x 10-3 m 3 1 Joule is equivalent to 1 Nm

5 Moles 5 Example 1 Calculate the number of moles of gas present in 500cm 3 at 100 KPa pressure and at a temperature of 27 C. P = 100 KPa = Pa V = 500 cm 3 x 10-6 = m 3 T = = 300 K R = 8.31 J K -1 mol -1 = 8.31 PV = nrt n = PV = x = 0.02 moles RT 300 x 8.31 Example 2 Calculate the relative molecular mass of a vapour if 0.2 g of gas occupy 400 cm 3 at a temperature of 223 C and a pressure of 100 KPa. P = 100 KPa = Pa V = 400 cm 3 x 10-6 = m 3 T = = 500 K m = 0.27g = 0.27g R = 8.31 J K -1 mol -1 = 8.31 PV = mrt M = mrt = 0.27 x 500 x 8.31 = M PV x Q.4 Convert the following volumes into m 3 a) 1dm 3 b) 250cm 3 b) 0.1cm 3 Convert the following temperatures into Kelvin a) 100 C b) 137 C b) 123 C Calculate the volume of 0.5 mol of propane gas at 298K and 10 5 Pa pressure Calculate the mass of propane (C 3 H 8 ) contained in a 0.01 m 3 flask maintained at a temperature of 273K and a pressure of 250kPa.

6 6 Moles MOLAR VOLUME The molar volume of any gas or vapour at stp is 22.4 dm 3 mol -1 ( m 3 mol -1 ) stp Standard Temperature and Pressure ( 273K and x 10 5 Pa ) The volume of a gas varies with temperature and pressure. To convert a volume to that which it will occupy at stp (or any other temperature and pressure) one uses the following relationship which is derived from Boyle s Law and Charles Law. P 1 V 1 = P 2 V 2 T 1 T 2 where P 1 initial pressure V 1 initial volume initial temperature (in Kelvin) T 1 P 2 V 2 T 2 final (in this case, standard) pressure final volume (in this case, at stp) final (in this case, standard) temperature (in Kelvin) Calculations Experiment Methods Convert the volume of gas to that at stp then scale it up to the molar volume. The mass of the gas occupying 22.4 dm 3 ( i.e litres, 22400cm 3 ) is the molar mass. It is possible to calculate the molar mass of a gas by measuring the volume of a given mass of gas and applying the above equations. Gas syringe method Victor Meyer method Dumas bulb method Example A sample of gas occupies 0.25 dm 3 at 100 C and 5000 Pa pressure. Calculate its volume at stp [273K and 100 kpa]. P 1 initial pressure = 5000 Pa P 2 final pressure = Pa V 1 initial volume = 0.25 dm 3 V 2 final volume =? T 1 initial temperature = 373K T 2 temperature = 273K thus 5000 x 0.25 = x V therefore V 2 = 273 X 5000 x 0.25 = dm 3 (9.15 dm 3 ) 373 x

7 Moles 7 Gay-Lussac s Law of Combining Volumes Statement When gases combine they do so in volumes that are in a simple ratio to each other and to that of any gaseous product(s) N.B. all volumes must be measured at the same temperature and pressure. Avogadro s Theory Statement Equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules Calculations example 1 Gay-Lussac s Law and Avogadro s Theory can be used for reacting gas calculations. What volume of oxygen will be needed to ensure that 250cm 3 methane undergoes complete combustion at 120 C? How much carbon dioxide will be formed? CH 4(g) + 2O 2(g) > CO 2(g) + 2H 2 O (g) 1 molecule 2 molecules 1 molecule 2 molecules 1 volume 2 volumes 1 volume 2 volumes (a gas at 120 C) 250cm 3 500cm 3 250cm 3 500cm 3 ANS. 500cm 3 of oxygen and 250cm 3 of carbon dioxide. Special tips An excess of one reagent is often included; e.g. excess O 2 ensures complete combustion Check the temperature, and state symbols, to check which compounds are not gases. This is especially important when water is present in the equation. example 2 20cm 3 of propane vapour is reacted with 120cm 3 of oxygen at 50 C. What will be the composition of the final mixture at the same temperature and pressure? C 3 H 8(g) + 5O 2(g) > 3CO 2(g) + 4H 2 O (l) 1 molecule 5 molecules 3 molecules 4 molecules 1 volume 5 volumes 3 volumes negligible (it is a liquid at 50 C) 20cm 3 100cm 3 60cm 3 20cm 3 will be unused ANSWER 20cm 3 of unused oxygen and 60cm 3 of carbon dioxide. example 3 1g of gas occupies 278cm 3 at 25 C and 2 atm pressure. Calculate its molar mass. i) convert volume to stp 2 x 278 = 1 x V V = 278 x 2 x 273 = 509 cm x 298 ii) convert to molar volume 1g occupies 509cm 3 at stp 1/509g 1cm x 1/509g 22400cm 3 therefore 44g occupies 22.4 dm 3 at stp ANSWER: The molar mass is 44g mol -1

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