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Empirical formula:

Empirical formula:. Hydroquinone, used as a photographic developer is 65.4% C, 5.5% H, and 29.1% O, by mass. What is the empirical formula?. If we work with 100 grams, we would have 65.4 g C, 5.5 g H and 29.1 g O. Then: Change g to moles and convert to whole numbers. 65.4 g. = 5.42 mol C.

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Empirical formula:

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  1. Empirical formula:

  2. Hydroquinone, used as a photographic developer is 65.4% C, 5.5% H, and 29.1% O, by mass. What is the empirical formula? If we work with 100 grams, we would have 65.4 g C, 5.5 g H and 29.1 g O. Then: Change g to moles and convert to whole numbers. 65.4 g = 5.42 mol C C: H: 5.5 g C3H3O1 = 5.45 mol H O: 29.1 g = 1.82 mol O Divide by smallest to get whole number

  3. A compound was sent out for analysis and was found to contain 6.7% H, 39.9% C, and 53.4% O. It was also found to have a molecular mass of 60.0 g/mol. Determine both the empircal and the molecular formulas. If we work with 100 grams, we could call the %’s grams!

  4. A compound was sent out for analysis and was found to contain 6.7% H, 39.9% C, and 53.4% O. It was also found to have a molecular mass of 60.0 g/mol. Determine both the empirical and the molecular formulas.  H: 6.7 g empirical formula: = 6.64 = 2.00 3.32 CH2O empirical formula wt.: C: 39.9 g  = 3.32 = 1.00 3.32 30.03 molecular formula:  O: 53.4g = 3.33 3.32 = 1.00 so: C2H4O2 We need whole numbers

  5. Stoichiometry: “working with ratios”

  6. When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) a. How many moles of NO2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) 4.3 mol ? mol 4.3 mol N2O5 = moles NO2 8.6 b. How many moles of O2 can be produced from 4.3 moles of N2O5? 2N2O5(g) 4NO2(g) + O2(g) 4.3 mol ? mol 4.3 mol N2O5 = mole O2 2.15

  7. When N2O5 is heated, it decomposes: 2N2O5(g) 4NO2(g) + O2(g) 210 g a. How many moles of N2O5 were used if 210g of NO2 were produced? 210 g NO2 = moles N2O5 2.283 b. How many grams of N2O5 are needed to produce 75.0 grams of O2? 75.0 g O2 = grams N2O5 506.25

  8. Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) First copy down the the BALANCED equation!

  9. Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) Now place numerical the information below the compounds.

  10. Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide one Two starting amounts? Where do we start?

  11. Limiting/excess/theoretical yield Reactant Problems: Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide Based on: KO2 0.15 mol KO2 = mol O2 0.1125

  12. Limiting/excess /theoretical yield Reactant Problems: Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles Hide 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 0.10 mol H2O Based on: H2O = mol O2 0.150

  13. Limiting/excess /theoretical yield Reactant Problems: Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Based on: KO2 0.15 mol KO2 = mol O2 0.1125 it was limited! 0.10 mol H2O Based on: H2O = mol O2 0.150 What is the theoretical yield? Hint: which reactant was the limiting reactant? H2O = XS reactant!

  14. Theoretical yield vs.Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

  15. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide one 120.0 g KO2 Based on: KO2 = g O2 40.51

  16. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g Hide 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Question if only 35.2 g of O2 were recovered, what was the percent yield?

  17. If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced? 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g) ? g 120.0 g 47.0 g 120.0 g KO2 Based on: KO2 = g O2 40.51 47.0 g H2O Based on: H2O = g O2 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H2O. 125.3 - 40.51 = 84.79 g of O2 that we have XS water to form. 84.79 g O2 31.83 = g XS H2O

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