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CH 3: Stoichiometry

CH 3: Stoichiometry. Moles. Mole Defined. Mole = number equal to the # of carbon atoms in exactly 12 grams of pure C-12. Mole = 6.022 x 10 23 Called Avogadro’s number One mole of protons = 1g One mole of neutrons = 1g One mole of an element = atomic mass in grams.

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CH 3: Stoichiometry

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  1. CH 3: Stoichiometry Moles

  2. Mole Defined • Mole = number equal to the # of carbon atoms in exactly 12 grams of pure C-12. • Mole = 6.022 x 1023 • Called Avogadro’s number • One mole of protons = 1g • One mole of neutrons = 1g • One mole of an element = atomic mass in grams

  3. Molar Mass and related terms • Molar mass – mass in grams of one mole of a substance – molecular weight, mass of 1 molecule in amu • Formula weight, mass of 1 formula unit for an ionic compound in amu • Calculate by summing the masses of the component atoms, units: • molar mass – grams or grams/mole • molecular wt & formula wt - amu

  4. Moles, mass and # particles • Molar mass links the mass of a substance to the number of particles present 1 mole Mg = 24.31g = 6.022 x 1023 Mg atoms

  5. Molar Mass Related Calculations: • Molar mass of a substance. • Moles present in a given mass of substance. • Mass of a given number of moles of a substance. • Number of particles in a given mass or a given number of moles of a substance.

  6. What is the molar mass of glucose. The formula for glucose is C6H12O6. • What is the mass of 0.023 moles of glucose?

  7. How many glucose molecules are present in 0.023 moles of the compound? • How many moles of glucose are present in 3.5 x 10-3 grams of the substance?

  8. Mass Percent • Compounds are typically described by either their chemical formula or their percent by mass of the component elements. Mass % X = n (molar mass X) x 100% molar mass of compound

  9. Mass Percent Calculations • Calculate the mass percent of each element in C6H12O6 • Assume one mole of the substance.

  10. Empirical and Molecular Formulas • Molecular formula – ratio of atoms in a molecule • Empirical formula – simplest ratio of elements in a molecule • Glucose: • Molecular formula: C6H12O6 • Empirical Formula: • Molecular formula = n (empirical formula)

  11. Mass %  Empirical Formula Given mass percent data: • “Calculate” mass in grams of each element in 100 g of compound. • Convert each mass into moles of the element. • Divide each molar answer by the smallest of the values. • If the numbers obtained in step 3 are not whole numbers, multiply all by an integer so the results are whole numbers • Whole numbers obtained in step ¾ are the subscripts in the empirical formula.

  12. Empirical Formula  Molecular Formula Molar mass needed • Calculate mass of one mole of the empirical formula. Molar Mass = integer (n) empirical formula weight (n)(empirical formula) = molecular formula

  13. Caffeine • Molar mass = 194 g/mol • 49.49 % C • 5.19 % H • 28.85 % N • 16.48 % O What is the empirical and molecular formula of caffeine?

  14. Acetaminophen (Tylenol) Molar mass = 151 g/mol • 63.56 % C • 6.00 % H • 9.27 % N • 21.17 % O What is the empirical and molecular formula of acetaminophen?

  15. CHC 3: #86 Urea: Molar Mass 60 g/mol (I looked this up – not in question) 1.121 g N 0.161 g H 0.480 g C 0.640 g O What is the empirical and molecular formula of urea?

  16. 3.7/3.8 Chemical Reactions • Writing chemical reactions • Write the correct chemical formula for each reactant and product. • Use a subscript after each formula to indicate the state of each substance • (s) – solid • (l) – liquid • (g) – gas • (aq) – aqueous (dissolved in water)

  17. Chemical equations must obey the law of conservation of matter – balancing does this. • Balance the equation by adding coefficients in front of reactants and products as needed • DO NOT CHANGE THE FORMULAS FOR THE COMPOUNDS

  18. One of my favorite chemical reactions: Mg(s) + HCl(aq) MgCl2 (aq) + H2 (g) • As written the reaction does not obey the law of conservation of matter – reaction needs to be balanced

  19. Mg(s) + 2 HCl(aq) MgCl2 (aq) + H2 (g) • Now the reaction is balanced!

  20. Balance the Reactions N2 (g) + H2 (g) NH3 (g) Na(s) + Cl2 (g) NaCl(s) Pb(NO3)2(aq) + NaCl(aq) PbCl2(s) + NaNO3(aq)

  21. Meaning of the Balanced Reaction • Atomic level • Molar level • Stoichiometry - relationship between moles, the balanced reaction and mass • Page 10? outlines the needed steps

  22. 2 Na + Cl2 2 NaCl 2 mol Na + 1 mol Cl2 2 mol NaCl

  23. Cookie example! 2 eggs + 1 bag chips  50 cookies

  24. Molar Ratios • Balanced reactions lead to molar ratios N2 + 3 H2  2 NH3

  25. N2 (g) + 3 H2 (g) 2 NH3 (g) • If 3.5 grams of N2 to react, how many moles and how many grams of NH3 would be made?

  26. N2 (g) + 3 H2 (g) 2 NH3 (g) • How many grams of H2 are needed to make37.8 grams of NH3?

  27. 2 LiOH + CO2 Li2CO3 + H2O • How many grams of LiOH are needed to react 750. grams of CO2? • How many grams of Li2CO3 will be made if 750. grams of CO2 react?

  28. Limiting Reagent • Reactants are not always combined in stoichiometric quantities. • Often one reactant is used up before others. • This reactant is said to be the limiting reagent (or reactant.) • this reactant limits how much product can be made.

  29. Limiting Reagent Calculations • When you are given masses for each reactant you must determine which reactant is limiting. • Convert grams of each reactant to moles. • Use the molar ratios to determine which reactant is limiting. • Base the amount of product that can be made on the limiting reagent.

  30. Another Approach… • Convert grams of each reactant to moles. • Calculate the moles (or grams) of a product that could be made from each reactant. • Whichever reactant results in the smaller quantity of product is the limiting reagent. • Base the amount of product made on the limiting reagent.

  31. N2 (g) + 3 H2 (g) 2 NH3 (g) • 12.5 grams of nitrogen is combined with 2.30 grams of hydrogen. • Which reactant is in excess and which is limiting? • How many grams of ammonia will be made?

  32. Percent Yield % Yield = actual yield x 100 % theoretical yield

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