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Stoichiometry notes

Stoichiometry notes. 1 mole = 6.022 x 10 23 particles. The mole. The mass of one mole of atoms of a substance. The average atomic mass from the chart. For diatomic molecules, don’t forget to multiply by 2. Molar mass of nitrogen atoms (N) = 14.01

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Stoichiometry notes

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  1. Stoichiometry notes

  2. 1 mole = 6.022 x 1023 particles The mole

  3. The mass of one mole of atoms of a substance. • The average atomic mass from the chart. • For diatomic molecules, don’t forget to multiply by 2. • Molar mass of nitrogen atoms (N) = 14.01 • Molar mass of nitrogen molecules (N2), which is how they appear most frequently in nature = 28.02 Molar mass

  4. Moles x 6.022 x 2023 = particles Particles / 6.022 x 1023 = moles Particle mole conversions

  5. Moles x molar mass = grams Grams / molar mass = moles Mass mole conversions

  6. Must first convert to moles Then may convert moles to either mass or particles This will require 2 steps. Particle mass conversion

  7. Balance the equation • Coefficients represent relative number of moles • For example: Sn + 2Cl2 SnCl4 means that it takes twice as many moles of chlorine molecules as moles of tin atoms to react to form the product. Steps in a stoichiometry problem

  8. Identify your known and unknown substances. • Your unknown is what you are looking for. • Your known is what you know how much of it you have. You can calculate the number of moles of your known substance. Steps in a stoichiometry problem

  9. Calculate moles of known substance • For mass mass stoichioimetry, you will divide the grams of known by its molar mass. Steps in a stoichiometry problem

  10. Adjust molar ratio • Multiply moles of known by the unknown coefficient/known coefficient. • This gives you the moles of unknown. Steps in a stoichiometry problem

  11. In a mass mass problem, solve for mass of unknown • Moles of unknown x molar mass = grams of unknown Steps in a stoichiometry problem

  12. How many grams of tin metal are required to react completely with 100 g of Cl2? Sn + Cl2 SnCl4 Example

  13. This is a mass mass problem because I am given the mass of one substance (Cl2) and asked the mass of another substance (Sn) Sn + 2Cl2 SnCl4 100 g Cl2 / 71 g/mol Cl2 = 1.41 moles Cl2 1.41 moles Cl2 x (1 /2) = 0.71 moles Sn 0.71 moles Sn x 119 g/mol Sn = 84.5 g Sn Solution

  14. Essentially the same thing. • If you do it lab: • Divide what you actually got by what you calculated you should have gotten. • If it’s a word problem: • Divide what the problem tells you that you actually got by what you calculated you should have gotten. Percent recovery or percent yield

  15. React 45 g of tin with excess chlorine gas and get 90 grams of tin IV chloride. What is my percent yield? Example

  16. Sn + 2Cl2 SnCl4 45 g Sn/119 g/mol Sn = 0.378 moles of Sn 0.378 moles Sn x (1/1) = 0.378 moles of SnCl4 0.378 moles SnCl4 x 261 g/mol SnCl4 = 98.7 g You should have produced 98.7 g SnCl4 You only produced 90.0 g. ( 90.0 / 98.7) x 100 = 91.2% Solution

  17. The excess reactant is what you will always have enough of. • The limiting reactant is what when its runs out, the reaction stops. • Candle in a room will burn until the candle burns out. Oxygen is excess, candle is limiting. • Candle in a closed jar will burn until oxygen is exhausted. Candle is excess, oxygen is limiting. Limiting reactants

  18. Identified by 2 known substances. Work the problem twice, once with each known. The correct answer is the smaller mole value for the unknown. Limiting reactant problems

  19. If I have 50 g of tin and 87 g of chlorine, how many grams of tin IV chloride will I produce? Sn + Cl2 SnCl4 Example

  20. Sn + 2Cl2 SnCl4 • With tin as the known; • 50 g Sn/119 g/mol Sn = 0..42 moles of Sn • 0.42 moles Sn x (1/1) = 0.42 moles of SnCl4 • With chlorine as the known: • 87 g Cl2 / 71 g/mol Cl2 = 1.23 moles Cl2 • 1.23 moles Cl2 x (1 /2 ) = 0.62 moles SnCl4 • 0.42 < 0.62, so tin is my limiting reactant • 0.42 moles SnCl4 x 261 g/mol = 109.62 g SnCl4 Solution

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