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Moles

Moles. Quantitative chemistry. Most chemical reactions involve two or more substances Substances react with each other in certain ratios The study of these ratios is called stoichiometry

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Moles

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  1. Moles

  2. Quantitative chemistry • Most chemical reactions involve two or more substances • Substances react with each other in certain ratios • The study of these ratios is called stoichiometry • To determine how much of a substance is required to react with another we need to know the exact number of atoms/molecules/ions present in a specific amount of that substance. • How many gold atoms are in 1g of gold? 3.057x1021 • What is the mass of 1 gold atom? 3.271x10-22g • Because the number of atoms in a small sample is so large, and the mass of an atom is so small, an alternative method was developed HCl + NaOHNaCl + H2O 1 : 1 1 : 1

  3. RELATIVE ATOMIC MASS (Ar) • Because the mass of a single atom is so small, we use scales of relative mass • The masses of atoms/molecules are compared to the mass of an atom of carbon-12, which has an assigned mass of 12.00 • Because Ar is relative it has no units • Isotopes are factored in when assigning an Arvalue • Example: Chlorine has 2 isotopes: chlorine-35 and chlorine 37 Abundance: 75.78% 24.22% Ar= (35x0.7578) + (37x0.2422) = 35.48 Calculate the Arfor Lithium (Li). Isotopes: Lithium-6 = 7.59%, Lithium-7 = 92.41%

  4. RELATIVE MOLECULAR MASS (Mr) • The relative molecular mass is the sum of the Ar for the atoms making up a molecule • The Mrof methane (CH4) is: • 12.01 (Ar of C) + 4x1.01 (Ar of H) = 16.05 • Calculate Mr for glucose (C6H12O6) • Ar of C = 12.01 • Ar of H = 1.01 • Ar of O = 15.99

  5. The mole and Avogadro’s constant Y U Smi? • What is a mole? • A mole is the amount of substance that contains the same number of particles (atoms/molecules/ions) as there are carbon atoms in 12g of carbon-12. • This number is called Avogadro’s constant and has the value 6.02x1023 mol-1 • Avogadro’s constant is sometimes given the symbol L • The following formula can be used to calculate the number of moles: • Calculate the number of moles for 6.422g of glucose Number of moles (n) = Mass of substance (g) Molecular Mass (Aror Mr) (g mol-1)

  6. The mass of a molecule • Sulphuric acid (H2SO4) has a molar mass of 98.079g/mol • What is the mass of one H2SO4 molecule? • Recall that a one mole of a substance contains 6.02x1023 atoms/molecules • Mass of one molecule = Molar mass / Avogadro’s constant (g mol-1) (mol-1) • 98.079 / 6.02x1023 = 1.629x10-22g

  7. The number of particles • In 1 mole of O2 there are 6.02x1023 O2 molecules • Each O2 molecule contains 2 oxygen atoms (2x 6.02x1023 atoms) • This means that 1 mole of O2 contains 2 moles of oxygen atoms • How many moles of oxygen atoms are in 0.65 moles of H3PO4? • How many moles of hydrogen atoms are in 0.085 moles of benzoic acid (C7H6O2)? To calculate multiply the number of moles of the substance by the subscript of the atom in question Moles of O atoms = 0.65 x 4 = 2.6

  8. Empirical formulas • An empirical formula is the simplest whole number ratio of the elements present in a compound. • Example CH2 – this ratio tells us that for every carbon atom there are 2 hydrogen atoms • An empirical formula can be calculated based on the %/mass of each element present in a compound. ( ) The % by mass of an element = number of atoms of the element x Ar Mr x 100 Mr C6H5NO2 = 123.10 Ar C = 12.01 Ar O = 15.99 Ar H = 1.01 Ar N = 14.01 C = 58.54% O = 25.98% H = 4.10% N = 11.38%

  9. An unknown sample has been analysed and found to contain 47% potassium, 14.5% carbon and 38.5% oxygen. What is its empirical formula? Step 1 – convert to g • If you have been asked to calculate an empirical formula based on % composition then you assume that the mass is 100g. This way 1%=1g. K = 47.00% = 47.0g C = 14.50% = 14.5g O = 38.50% = 38.5g Ar= 39.09 Ar C = 12.01 Ar O = 15.99 Step 3 – Divide each value by the smallest and round to a whole number Step 2 – convert to moles Number of moles = mass of substance (g) / Ar K = 1.20/1.20 = 1 C = 1.21/1.20 = 1 O = 2.41/1.20 = 2 K = 47.0/39.09 = 1.20 moles C = 14.5/12.01 = 1.21 moles O = 38.5/15.99 = 2.41 moles Empirical formula = KCO2

  10. A compound was analysed and found to contain 53.4% Ca, 43.8% O, and 2.8% H.  What is the empirical formula of the compound? • Step 1 – convert to grams • Step 2 – calculate # of moles • Step 3 – divide each value by the smallest and round to a whole number • Step 4 – write the empirical formula Ar Ca= 40.1 Ar O = 16.00 ArH = 1.01

  11. C = 57.14% = 57.14g H = 6.16% = 6.16g N = 9.52% = 9.52g O = 27.18% = 27.18g Multiply each by the lowest factor that gives a whole number C = 4.76 / 0.68 = 7 x 2 = 14 H = 6.10 / 0.68 = 9 x 2 = 18 N = 0.68 / 0.68 = 1 x 2 = 2 O = 1.70 / 0.68 = 2.5 x 2 = 5 • A compound was analysed and found to contain 57.14% C, 6.16% H, 9.52% N and 27.18% O. What is the empirical formula of the compound? C = 57.14 / 12.01 = 4.76 mol H = 6.16 / 1.01 = 6.10 mol N = 9.52 / 14.01 = 0.68 mol O = 27.18 / 15.99 = 1.70 mol Empirical Formula = C14H18N2O5 C = 4.76 / 0.68 = 7 H = 6.10 / 0.68 = 9 N = 0.68 / 0.68 = 1 O = 1.70 / 0.68 = 2.5 Too far to round

  12. Molecular Formulas • A molecular formula is the total number of atoms of each element present in a molecule of the compound. • It is a multiple of the empirical formula (simplest whole number ratio). To calculate the molecular formula from an empirical formula you need the Mr and the empirical formula mass. Example: The empirical formula of benzene is CH, the molar mass is 78.12 g mol-1. What is the molecular formula? Mass of empirical formula unit = (12.01+1.01) 13.02 N = 78.12 / 13.02 = 6 The empirical formula unit occurs 6 times (n) Therefore, the molecular formula = C6H6

  13. Combustion reactions •  A 0.250g sample of hydrocarbon undergoes complete combustion to produce 0.845g of CO2 and 0.173g of H2O. What is the empirical formula of this compound? • Step 1 - Determine the grams of C in 0.845 g CO2 and the grams of H in 0.173 g H2O. • carbon: 0.845 g x (12.01 / 43.99) = 0.2307g • hydrogen: 0.173 g x (2.02 / 18.01) = 0.0194g MrCO2 = 43.99 MrH2O = 18.01 Step 2 - Convert grams of C and H to their respective amount of moles. • carbon: 0.2307 / 12.011 = 0.0192 mol • hydrogen: 0.0194 / 1.01 = 0.0192 mol Empirical formula = CH

  14. Step 1 - Determine the grams of carbon in 0.3664 g CO2 and the grams of hydrogen in 0.1500 g H2O carbon: 0.3664 x (12.01 / 43.99) = 0.1000 g hydrogen: 0.15 x (2.02 / 18.01) = 0.0168 g Grams of oxygen = 0.25 – (0.1 + 0.0168) = 0.1332 • A 0.250g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664g of CO2 and 0.150g of H2O. What is the empirical formula of this compound? Step 3 – Divide by the lowest value carbon: 0.0083 / 0.0083 = 1 hydrogen: 0.0166 / 0.0083 = 2 oxygen: 0.0083 / 0.0083 = 1 Step 2 – Convert grams to moles carbon: 0.1000 / 12.01 = 0.0083 mol hydrogen: 0.0168 / 1.01 = 0.0166 mol oxygen = 0.1332 / 15.99 = 0.0083 mol Empirical Formula = CH2O

  15. Balancing Equations • Mass is conserved in chemical reactions • 5g A + 10g B 15g C • The same number and type of atoms must appear on both sides of the equation • CH4 + O2 CO2 + H2O • CH4 + 2O2 CO2 + 2H2O • Reactants Products • C = 1 C = 1 • H = 4 H = 4 • O = 4 O = 4 Reactants Products C = 1 C = 1 H = 4 H = 2 O = 2 O = 3

  16. Only whole number coefficients can be used to balance equations • Example = 2O2 can be used, 1.5 O2 cannot be used • When balancing equations look for the odd and even numbers. • When balancing hydrocarbon combustion reactions it is best to first balance the carbon atoms, then the hydrogen atoms and then the oxygen atoms • N2 + H2 NH3 • C4H10 + O2 CO2 + H2O

  17. (l) = liquid • (g) = gas • (s) = solid • (aq) = aqueous (dissolved in water) • State symbols are often included in chemical equations • H2O(l) • CO2(g) • MgCl(S) • HCl(aq)

  18. Practice • NO+ O2 NO2 • C3H8 + O2 CO2 + H2O • CaCO3+ HCl CaCl2+ CO2 + H2O • HI + H2SO4 H2S + H2O + I2

  19. Ionic Equations • Just like mass, charge (electrons/protons) is conserved in a reaction • Although an equation can be balanced with regards to the number of atoms it may be unbalanced with regards to charge. • Balance the number of atoms, then balance the total charge of the products and reactants Cr2O72- + Fe2+ + H+ Cr3+ + H2O + Fe3+

  20. Total and Net ionic equations Aqueous sodium hydroxide reacts with aqueous copper(II) sulphate to precipitate copper(II) hydroxide 2NaOH(aq) + CuSO2(aq) Na2SO2(aq) + Cu(OH)2(s) To write a total ionic equation we have to break up aqueous solutions into constituent ions 2Na+ + 2OH- + Cu2+ + SO22- 2Na+ + SO22- + Cu(OH)2(s) To write a net ionic equation to remove the spectator ions from the equation Cu2+ + 2OH- Cu(OH)2(s)

  21. Practice Problems K2CO3(aq) + 2 AgNO3(aq) 2 KNO3(aq) + Ag2CO3(s) H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2H2O(l) 3 HCl(aq) + Al(OH)3(s) 3 H2O(l) + AlCl3(aq) 2 Na3PO4(aq) + 3 CaCl2(aq) 6 NaCl(aq) + Ca3(PO4)2(s)

  22. Calculations using chemical equations • Chemists need to be able to calculate the mass of each reactant required to produce a certain amount of a product. • Example: 4 Na(s) + O2(g) 2 Na2O(s) • How much sodium reacts with 3.20g of oxygen? What mass of Na2O(s) is produced? Step 1 – use the mass of oxygen to find the moles of oxygen 3.2 / 32.00 = 0.100 mol Step 2 – 1 mole of oxygen reacts with 4 moles sodium 0.100 * 4 = 0.400 mol Step 3 – Calculate the mass of sodium 0.400 * 22.99 = 9.20g Step 4 – calculate the mass of Na2O Conservation of mass Using moles: 0.200 * 61.98 = 12.4g

  23. 2 NH3 + 3CuO N2 + 3 H2O + 3Cu If 2.56g of NH3 is reacted with excess CuO, calculate the mass of Cu produced Step 1 – Calculate the number of moles of NH3 2.56 / 17.04 = 0.150 mol Step 2 – Calculate number of moles of Cu 2 moles of NH3 produces 3 moles of Cu 0.150 * (3/2) = 0.225 mol Step 3 – Calculate the mass of Cu 0.225 * 63.55 = 14.3g CuO is in excess (more than needed), so we don’t need to worry about the number of moles of CuO.

  24. Formula for solving moles questions involving masses • An alternative method for solving these types of problems • 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l) • If 10.00g of butane is used, calculate: • The mass of oxygen required • The mass of CO2 produced m2 n2 M2 m1 n1 M1 m1 = mass of first substance n1 = coefficient of first substance M1 = molar mass of first substance = 10 2 * 58.14 m2 13 * 32.00 = 10.00 * 13 * 32.00 2 * 58.14 = 35.78g

  25. m2 n2 M2 m1 n1 M1 = • Mass of CO2 produced: Substance 1 – butane Substance 2 – carbon dioxide 10 2 * 58.14 m2 8 * 44.01 = m2 = 10 x 8 x 44.01 2 x 58.14 = 30.28g

  26. Practice Problems • How many moles of hydrogen gas are produced when 0.4 moles of sodium react with excess water? Na + H2O NaOH + H2 • How many moles of O2 react with 0.01 mol C3H8? C3H8 + O2 CO2 + H2O • Calculate the mass of arsenic(III) chloride produced when 0.150g of arsenic reacts with excess chlorine. As + Cl2 AsCl3 • What mass of SCl2 must be reacted with excess NaF to produce 2.25g of NaCl? SCl2 + NaF S2Cl2 + SF4 + NaCl

  27. Calculating the yield of a chemical reaction • It is important to know the yield of a chemical reaction, for instance there may be two possible methods for synthesising a compound. • Route 1 = 95% yield Route 2 = 88% yield • Yield is usually quoted as a percentage • Example: synthesis of 1,2-dibromoethane C2H4(g) + Br2(l) C2H4Br2(l) 10g of ethene will react exactly with 56.95g of bromine The theoretical yield is 66.95g The actual yield may be 50.00g % yield = (actual yield / theoretical yield) x 100 = (50.00 / 66.95) x 100 = 74.68 %

  28. If the yield of ethyl ethanoate obtained when 20.0g of ethanol is reacted with excess ethanoic acid is 30.27g, calculate the % yield. C2H5OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O Calculate the theoretical yield molecular mass of ethanol = 46.08 g mol-1 # of moles = 20.00 / 46.08 = 0.434 mol 1:1 ratio of ethanol and ethyl ethanoate molecular mass of ethyl ethanoate = 88.12 g mol-1 Mass of ethyl ethanoate = 0.434 x 88.12 = 38.24 g ethanol ethanoic acid ethyl ethanoate (30.27 / 38.24) x 100 = 79.16%

  29. Quiz 24th August • Empirical and molecular formulas • Combustion analysis • Ionic equations • Calculations using mass and moles • Yield

  30. Limiting reactants • In most chemical reactions exact quantities are not used, instead an excess of one or more reagents is used. • Therefore, one reactant is consumed before the others and is called the limiting reactant (also known as the limiting reagent) 2Al(s) + 3Cl2(g) 2AlCl3(s) 114g of Al is reacted with 186g of Cl2 Which is the limiting reactant? Al = 114/26.98 = 4.23 moles Cl2 = 186/70.9 = 2.62 moles To use all Al we need 4.23 x = 6.35 mol Cl2 To use all Cl2we need 2.62 x = 1.75 mol Al We have enough Al to use all the Cl2. We don’t have enough Cl2 to use all the Al. 2.62moles Cl2 x = 1.75 mol Limiting Reagent

  31. Identify the limiting reactant in each of the following reactions: • 0.1mol Sb4O6 reacts with 0.5mol H2SO4 Sb4O6 + H2SO4 Sb2(SO4)3 + H2O

  32. 0.20mol AsCl3 reacts with 0.25mol H2O AsCl3 + H2O As4O6 + HCl

  33. 5g Cu reacts with 8g HNO3 according to the equation: Cu + HNO3 Cu(NO3)2 + H2O + NO

  34. Ideal Gases • Gases are complicated. They're full of billions and billions of energetic gas molecules that can collide and possibly interact with each other. • Since it's hard to exactly describe a real gas, people created the concept of an Ideal gas as an approximation that helps us model and predict the behaviour of real gases. • The term ideal gas refers to a hypothetical gas composed of molecules which follow these rules: Ideal gas molecules do not attract or repel each other. The only interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container. A collision wherein no kinetic energy is converted to other forms of energy during the collision. In other words, kinetic energy can be exchanged between the colliding objects (e.g. molecules), but the total kinetic energy before the collision is equal to the total kinetic energy after the collision Ideal gas molecules themselves take up no volume. The gas takes up volume since the molecules expand into a large region of space, but the Ideal gas molecules are approximated as point particles that have no volume in and of themselves.

  35. There are no gases that are exactly ideal, but there are plenty of gases that are close enough that the concept of an ideal gas is an extremely useful approximation for many situations. • In fact, for temperatures near room temperature and pressures near atmospheric pressure, many of the gases we care about are very nearly ideal. • If the pressure of the gas is too large (e.g. hundreds of times larger than atmospheric pressure), or the temperature is too low (e.g. -200°C) there can be significant deviations from the ideal gas law. 

  36. Calculations involving gases Standard temperature and pressure (STP) = 0°C (273k), 1 atm (100kPa) The volume of a gas at STP is constant : 1 mole = 22.4 dm3 mol-1 Avogadros Law Volumes of gases are often given in dm3 (litres) or cm3 1 dm3 = 1000 cm3 Volume in dm3 Calculate the mass of 250cm3 of O2 at STP Molar volume (22.4dm3 mol-1) Number of moles 250cm3 = 0.25dm3 0.25 / 22.4 = 0.0112 mol 0.0112 x 31.98 = 0.36

  37. What mass of KClO3 decomposes to produce 100cm3 of O2 at STP? Step 1 – moles of O2 100cm3/ 1000 = 0.100dm3 0.1 / 22.4 = 4.464x10-3mol Step 2 – moles of KClO3 Step 3 – mass of KClO3 Mass = 2.976x10-3 x 122.55 = 0.3647g 2 KClO3(s) 2KCl(s) + 3O2(g) Volume in dm3 Molar volume (22.4dm3 mol-1) Number of moles x 4.464x10-3 = 2.976x10-3 mol

  38. Alternative Formulas When a mass is given and the volume of another substance is required m1 = mass of first substance (g) n1 = coefficient of first substance M1 = molar mass of first substance V2 = volume of second substance if it is a gas (dm3) n2 = coefficient of second substance Mv= molar volume of a gas (22.4dm3 at STP) V2 n2 Mv m1 n1 M1 = When a volume is given and the volume of another substance is required V2 n2 V1 n1 No need to convert units of volume to dm3 with this equation. V2 will have the same units as V1 =

  39. What volume of SO2 (measured at STP) is obtained when 1kg of As2S3 is combusted? 2As2S3(s) + 9O2(g) 2As2O3(s) + 6SO2(g) Substance 1 = As2S3 m1 = n1 = M1 = Substance 2 = SO2 V2 = n2 = Mv=

  40. Ar: N: 14.00 H: 1.01 O: 15.99 Calculate the volume of NO(g) and H2O(g) produced when 0.6dm3 of O2(g) is reacted with excess ammonia. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

  41. Macroscopic properties of ideal gases Macroscopic means ‘on a large scale’ • Microscopic properties of a gas are the properties of the particles that make up the gas • So far we have dealt with questions involving the volumes of gases at STP. In order to work out volumes of gases under other conditions, we must understand a little about the properties of gases.

  42. Boyle’s law P1V1 = P2V2 • The relationship between pressure and volume • At a constant temperature, the volume of a fixed mass of an ideal gas is inversely proportional to its pressure • Example – if the pressure of a gas is doubled at constant temperature, then the volume will be halved and vice versa (double volume = halve the pressure)

  43. V1 V2 T1 T2 = Charles’ law • The relationship between volume and temperature • The volume of a fixed mass of an ideal gas at constant pressure is directly proportional to its temperature in kelvin. • Example – If an ideal gas has a volume of 200cm3 at 120K, it will have a volume of 400cm3 at 240K if the pressure remains constant

  44. P1 P2 T1 T2 = Gay-Lussac's Law • The relationship between pressure and temperature • For a fixed mass of an ideal gas at constant volume, the pressure is directly proportional to its absolute temperature (K). • Example – If the temperature (in Kelvin) of a fixed volume of an ideal gas is doubled, the pressure will also double

  45. Which gas law states that at a constant temperature, the volume of a fixed mass of an ideal gas is inversely proportional to its pressure? • Which gas law states that when a fixed mass of an ideal gas has a constant volume, the pressure is directly proportional to its absolute temperature (K)? • Which gas law states that the volume of a fixed mass of an ideal gas at constant pressure is directly proportional to its temperature in kelvin? Boyle’s Law Charles’ law Gay-Lussac’s law Boyle’s Law Charles’ law Gay-Lussac’s law • Which gas law requires constant volume? • Which gas law requires constant temperature? • Which gas law requires constant pressure? Boyle’s Law Charles’ law Gay-Lussac’s law Boyle’s Law Charles’ law Gay-Lussac’s law Boyle’s Law Charles’ law Gay-Lussac’s law Boyle’s Law Charles’ law Gay-Lussac’s law

  46. Combined gas law If the volume of an ideal gas collected at 0°C and 1.00 atmospheric pressure (1.01x105 Pa) is 50cm3, what would the volume be at 60°C and 1.08x105 Pa ? • Boyle’s law, Charles’ law and Gay-Lussac’s law can be combined to produce the following equation: Temperature must be in Kelvin Any units may be used for P and V, as long as they are the same on both sides P1 = 1.01 x105 Pa V1 = 50cm3 T1 = 0°C = 273K P2 = 1.08x105 Pa V2 = ? T2 = 60°C = 60 + 273 = 333K 1.01x105 x 50 1.08x105 x V2 273 333 = 1.01x105 x 50 x 333 1.08x105 x 273 = 57cm3 V2 =

  47. P2 = 1.00atm V2 = 756cm3 T2= 273K P1 = 2.05atm V1 = 1.34dm3 T1 = ? What temperature in °C is required for an ideal gas to occupy 1.34dm3 at a pressure of 2.05atm if it occupies 756cm3 at STP? = (756/1000) = 0.756dm3 2.05 x 1.34 1.00 x 0.756 T1 273 = 2.05 x 1.34 x 273 1.00 x 0.756 T1 = = 992 K 992 – 273 = 719°C

  48. 1 mole of any gas = 22.4L at STP Ideal gas law • If the relationship between P, V and T are combined with Avogadro’s law the ideal gas equation is obtained. Boyle’s Law Charles’ Law Avogadro’s Law Ideal Gas Law

  49. Units What is 456cm3 in m3 ? What is 707000 Pa in atm ? What is 0.86m3 in dm3 ? What is 2 atm in kPa? 4.56x10-4 m3 P = Pa Volume = m3 R = 8.31 J K-1 mol-1 T = K 7 atm 860 dm3 202 kPa Converting Units Pressure: Volume: atm to Pa = atm x 1.01x105 dm3 to m3 = divide by 1000 kPa to Pa = kPa x 1000 cm3 to m3 = divide by 1,000,000 cm3 to dm3 = divide by 1000 1L = 1dm3 1ml = 1cm3 1atm = 760mmHg 1mmHg = 133.322Pa

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