1 / 145

Stoichiometry: Chemical Calculations

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

rahim-freeman
Télécharger la présentation

Stoichiometry: Chemical Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stoichiometry: Chemical Calculations Chemistry is concerned with the properties and the interchange of matter by reaction i.e. structure and change. In order to do this, we need to be able to talk about numbers of atoms. The key concept is the mole and the relationship between the mole and the mass of the atom.

  2. Stoichiometry: Chemical Calculations Each element has a distinct atomic mass – based on the natural abundances of the various isotopes present. Atoms combine to form molecules in fixed proportions which are usually small integers for simple molecular or ionic compounds

  3. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit

  4. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 SF6 NaCl Na2S2O3

  5. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH412.0115 + 4 x 1.0079 SF6 NaCl Na2S2O3

  6. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH412.0115 + 4 x 1.0079 SF632.066 + 6 x 18.9984 NaCl Na2S2O3

  7. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH412.0115 + 4 x 1.0079 SF632.066 + 6 x 18.9984 NaCl22.9898 + 35.453 Na2S2O32 x 22.9898 + 2 x 32.066 + 3 x 15.9994

  8. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH412.0115 + 4 x 1.0079 SF632.066 + 6 x 18.9984 NaCl22.9898 + 35.453 Na2S2O32 x 22.9898 + 2 x 32.066 + 3 x 15.9994

  9. Stoichiometry: Chemical Calculations The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH412.0115 + 4 x 1.0079 SF632.066 + 6 x 18.9984 NaCl22.9898 + 35.453 Na2S2O32 x 22.9898 + 2 x 32.066 + 3 x 15.9994

  10. 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.

  11. 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

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

  13. 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?

  14. 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

  15. 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

  16. 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

  17. 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

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

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

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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

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

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

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

  29. 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.

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

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

  32. 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

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

  34. 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

  35. 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

  36. Stoichiometry: Chemical Calculations Balancing the equation

  37. 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

  38. 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

  39. 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

  40. 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.

  41. Solutions and concentration If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O2 would be left over.

  42. 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

  43. 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

  44. 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

  45. 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 NaCl(s) melts Here copper is the solvent, zinc the solute. Zn Cu

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

  47. 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.

  48. 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.

  49. 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

  50. 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.

More Related