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Stoichiometry: Chemical Calculations

Stoichiometry: Chemical Calculations. Chemistry is concerned with the properties and the interchange of matter by reaction – structure and change. In order to do this, we need to be able to talk about numbers of atoms. Stoichiometry: Chemical Calculations. The mole and atomic mass

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Stoichiometry: Chemical Calculations

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  1. Stoichiometry: Chemical Calculations Chemistry is concerned with the properties and the interchange of matter by reaction – structure and change. In order to do this, we need to be able to talk about numbers of atoms

  2. Stoichiometry: Chemical Calculations The mole and atomic mass The mole is defined as the number of elementary entities as are present in 12.00 g of 12C. Numerically, this is equal to Avogadro’s Number 6.022 x 1023 Therefore, in 12.00 g of 12C there are 6.022 x 1023 ‘elementary entities’, in this case atoms.

  3. Stoichiometry: Chemical Calculations The mole and atomic mass Atomic masses, in atomic units, u, are defined relative to 12C. Therefore, The formula mass of an element or compound contains 1 mole, 6.022 x 1023, of particles

  4. Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na?

  5. Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na?

  6. Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na? Atomic mass of Na = 22.9898 u

  7. Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u As 1 u = 1/12 x mass (12C) And 1 mole = 6.022 x 1023 particles = number of particles in 12 g 12C

  8. Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms – how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u Mass of 1 mole of Na = 22.9898 g

  9. Stoichiometry: Chemical Calculations Examples How many particles are there in 5.0000 g of Na? 22.9898 g Na = 1 mole Na Then 1 g Na = 1 mol Na 22.9898 5 x 1 g Na = 5 x 1 mol Na 22.9898 5 g Na = 0.2175 mol Na 5 g Na = 0.2175 x (6.022 x 1023) particles Na

  10. Stoichiometry: Chemical Calculations Examples How many particles are there in 5.0000 g of Na? 1.310 x 1023 atoms

  11. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane?

  12. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10

  13. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g

  14. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u = 58.123 u

  15. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u = 58.123 u Relative Molecular Mass of Butane = 58.123 g

  16. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g

  17. Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g 0.23 mole of butane = 13.368 g

  18. Stoichiometry: Chemical Calculations Chemical Equations These are formulæ which show the chemical change taking place in a reaction.

  19. Stoichiometry: Chemical Calculations Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Physical state

  20. Stoichiometry: Chemical Calculations Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Reactants Product Physical state

  21. Stoichiometry: Chemical Calculations Chemical Equations As matter cannot be created or destroyed in a chemical reaction, the total number of atoms on one side must be equal to the total number of atoms on the other.

  22. Stoichiometry: Chemical Calculations Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water

  23. Stoichiometry: Chemical Calculations Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C6H12 Oxygen, O2

  24. Stoichiometry: Chemical Calculations Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C6H12 Oxygen, O2 Products: Carbon Dioxide, CO2 Water, H2O

  25. Stoichiometry: Chemical Calculations Chemical Equations Example Initially, we can write the reaction as

  26. Stoichiometry: Chemical Calculations Chemical Equations Example Initially, we can write the reaction as This is NOT a correct equation – there are unequal numbers of atoms on both sides

  27. Stoichiometry: Chemical Calculations Chemical Equations Example Initially, we can write the reaction as This is NOT a correct equation – there are unequal numbers of atoms on both sides Reactants: 6 C, 12 H, 2 O Products: 1 C, 2 H, 3 O

  28. Stoichiometry: Chemical Calculations Balancing the equation

  29. Stoichiometry: Chemical Calculations Balancing the equation 6 C, 12 H, 2 O 1 C, 2 H, 3 O 6 C on LHS means there must be 6 C on the RHS 6 C, 12 H, 2 O 6 C, 2 H, 13 O 13 O on RHS means there must be 13 O on LHS 6 C, 12 H, 13 O 6 C, 2 H, 13 O

  30. Stoichiometry: Chemical Calculations Balancing the equation 6 C, 12 H, 13 O 6 C, 2 H, 13 O 12 H on RHS means there must be 12 H on LHS 6 C, 12 H, 13 O 6 C, 12 H, 18 O 18 O on RHS means there must be 18 H on LHS 6 C, 12 H, 18 O 6 C, 12 H, 18 O

  31. Stoichiometry: Chemical Calculations The final balanced equation is and the coefficients are known as the stoichiometric coefficients. These coefficients give the molar ratios for reactants and products This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction

  32. Stoichiometry: Chemical Calculations If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O2 would be left over.

  33. Stoichiometry: Chemical Calculations The final balanced equation is and the coefficients are known as the stoichiometric coefficients. These coefficients give the molar ratios for reactants and products This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction

  34. The Exam

  35. Solutions A solution is a homogenous mixture which is composed of two or more components the solvent - the majority component and one or more solutes - the minority components

  36. Solutions Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent. Some are solids where both the solvent and the solute are solids. Brass is an example

  37. Solutions Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent. Some are solids where both the solvent and the solute are solids. Brass is an example

  38. Solutions Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent. Some are solids where both the solvent and the solute are solids. Brass is an example Here copper is the solvent, zinc the solute. Zn Cu

  39. Solutions Gas-Solid solution: Hydrogen in palladium Steel

  40. Solutions Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water. Solutions in water are termed aqueous solutions and species are written as E(aq). Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water.

  41. Solutions Aqueous Solutions Water is one of the best solvents as it can dissolve many molecular and ionic substances. The properties of solutions which contain molecular and ionic solutes are very different and give insight into the nature of these substances and solutions.

  42. Solutions Ionic Solutions An ionic substance, such as NaClO4, contain ions – in this case Na+ and ClO4-. The solid is held together through electrostatic forces between the ions. In water, the solid dissolves and the particles move away from each other and diffuse through the solvent. This process is termed Ionic Dissociation

  43. Solutions Ionic Solutions In an ionic solution, there are therefore charged particles – the ions – and as the compound is electrically neutral, then the solution is neutral. When a voltage is applied to the solution, the ions can move and a current flows through the solution. The ions are called charge carriers and whenever electricity is conducted, charge carriers are present.

  44. Solutions Molecular Solutions A molecular solution does not conduct electricity as there are no charge carriers present. The bonding in a molecule is covalent and involves the sharing of atoms and there is no charge separation.

  45. Solutions Electrolytes A solute that, when dissolved, produces a solution that conducts is termed an electrolyte, which may be strong or weak. A strong electrolyte is one which is fully dissociated in solution into ions A weak electrolyte is one which is only partially dissociated.

  46. Solutions Moles and solutions When a substance is dissolved in a solvent, we relate the quantity of material dissolved to the volume of the solution through the concentration of the solution. The concentration is simply the number of moles of the material per unit volume: C = n V n = number of moles; V = volume of solvent

  47. Solutions Moles and solutions The units of concentration are: C = n = moles V L3 and we define a molar solution as one which has 1 mole per liter. Alternatively, Concentration = Molarity = number of moles volume of solution

  48. Solutions Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

  49. Solutions Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na2SO4(s): Molar Atomic Mass of Na: 22.9898 gmol-1 Molar Atomic Mass of S: 32.064 gmol-1 Molar Atomic Mass of O: 15.9994 gmol-1

  50. Solutions Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na2SO4(s): (2 x 22.9898)+32.064+(4x15.9994)=142.041gmol-1 1 mole of Na2SO4(s) = 142.041g 1/142.041 mole of Na2SO4(s) = 1 g

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