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Stoichiometry – Chemical Quantities Notes

Explore stoichiometry - the study of quantitative relationships from chemical formulas and equations. Learn about mole-mole relationships, mole ratios, balancing chemical equations, and more. Practice problems included.

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Stoichiometry – Chemical Quantities Notes

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  1. Stoichiometry – Chemical Quantities Notes

  2. Stoichiometry Stoichiometry – Study of quantitative relationships that can be derived from chemical formulas and chemical equations Mole ─ Mole Relationship need a balanced equation Mole Ratio – the ratio of moles of one substance to moles of another substance in a balanced chemical equation The coefficients in a balanced equation give the relative numbers of molecules, as well as, the relative number of moles. CO(g) + 2H2(g) CH3OH(l) 1 mol CO = 2 mol H2 = 1 mol CH3OH Ex: How many moles of O2 are required to produce 10. moles of CO2? 2 CO + O2 2 CO2 10. mol CO2 1 mol O2 x __________ 5.0 mol O2 = 2 mol CO2

  3. What other relationships do we have for the mole? • 1 mol = 6.022 x 1023 atoms / molecules / particles • 1 mol = [molar mass] g • 1 amu = 1.66x10-24 g We can add these mole relationships on either end of the mole ratio: # unit A x 1 mol A x mol B x __ unit B = # unit B _ unit A _ mol A 1 mol B mole relationship mole ratio mole relationship (switch units) (switch substances) (switch units) *A is the GIVEN substance & B is the WANTED substance

  4. K 1 x 39.1 = 39.1 Cl 1 x 35.5 = 35.5 O 3 x 16.0 = 48.0 122.6 g/mol Mass A – Mole B Ex: Calculate moles of O2 produced if 2.50 g KClO3 decomposes completely: 2 KClO3 2 KCl + 3 O2 2.50 g KClO3 1 mol KClO3 3 mol O2 x ________________ x ______________ = 122.6 g KClO3 2 mol KClO3 0.0306 mol O2

  5. Na 1 x 23.0 = 23.0 Cl 1 x 35.5 = 35.5 58.5 g/mol Mass A – Mass B Ex: Determine the mass of NaCl that decomposes to yield 355 g Cl2 2NaCl  2 Na + Cl2 1 mol Cl2 2 mol NaCl 58.5 g NaCl 355 g Cl2 x _____________ x ______________ x ______________ = 71.0 g Cl2 1 mol Cl2 1 mol NaCl Cl 2 x 35.5 = 71.0 g/mol 585 g NaCl

  6. Mole A – Mass B Ex: Calculate the number of grams of oxygen required to react exactly with 4.30 mol of propane, C3H8, in the reaction by the following balanced equation: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 5 mol O2 32.0 g O2 4.30 mol C3H8 x _____________ x __________ = 1 1 mol O2 mol C3H8 688 g O2 O 2 x 16.0 = 32.0 g/mol

  7. Learning Check How many grams of water are needed to produce 9.23 moles of oxygen? ____ Na2O2 + ____ H2O ____ NaOH + ____ O2

  8. 2. Limiting Reactant & Percent Yield

  9. Background Knowledge Check Label the reactant(s) and product(s) in the following reaction: 2 Mg + O2 2MgO Reactant(s): Product(s): Mg and O2 MgO

  10. Limiting Reactant Manufacturers of cars and bicycles order parts in the same proportion as they are used in their product. Car manufacturers order four times as many wheels as engines and bicycle manufacturers order twice as many pedals as seats. In the same manner, when chemicals are mixed together so they can undergo a reaction, they are often mixed in stoichiometric quantities – exactly the correct amounts so that all the reactants “run out” at the same time. If the chemicals aren’t mixed to run out at the same time, one of the chemicals will limit or halt the reaction from taking place any further. The reactant that “runs out” or limits the reaction is called the limiting reactant. The reactant that still remains or is extra is called the excess reactant. In any stoichiometric problem, where reactants are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting in order to calculate correctly the amounts of products that will be formed.

  11. Analogy: Baking Cookies A recipe calls for 2 cups of flour for every egg. You have 5 cups of flour and 3 eggs. What is your limiting ingredient? What is your excess ingredient? Steps for Solving Stoichiometry Problems Involving Limiting Reactants • Write and balance the equation for the reaction, if necessary. • For each reactant, convert grams reactant to grams product. • Compare grams of product: • The smaller grams of product is the theoretical yield and is the amount of product made • The smaller grams of product came from the limiting reactant • The larger grams of product came from the excess reactant flour eggs

  12. Mg 1 x 24.3 = 24.3 O 1 x 16.0 = 16.0 40.3 g/mol Ex: 7.24 moles of Mg and 3.86 moles of O2 react to form MgO. 2 Mg + O2 2MgO How many grams of MgO are formed ? What is the limiting reactant ? What is the excess reactant ? Mg O2 7.24 mol Mg 2 mol MgO 40.3 g MgO x ______________ x ____________ 292 g MgO = 2 mol Mg 1 mol MgO 2 mol MgO 3.86 mol O2 40.3 g MgO x _____________ 311 g MgO x ____________ = 1 mol O2 1 mol MgO

  13. N 1 x 14.0 = 14.0 H 3 x 1.0 = 3.0 17.0 g/mol Ex: Suppose 2.50 x 104 g of N2 and 5.00 x 103 g of H2 are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when the reaction is run to completion. N2 + 3 H2 2 NH3 What is the limiting reactant? What is the excess reactant? H2 N2 2.50 x 104 g N2 1 mol N2 mol NH3 17.0 g NH3 2 x ___________ x ______________ x ___________ 28.0 g N2 mol NH3 1 mol N2 1 = 3.04 x 104 g NH3 17.0 g NH3 5.00 x 103 g H2 1 mol H2 2 mol NH3 x ______________ x ___________ x ___________ mol NH3 2.0 g H2 3 mol H2 1 = 2.8 x 104 g NH3 N 2 x 14.0 = 28.0 g/mol H 2 x 1.0 = 2.0 g/mol

  14. Percent Yield Theoretical yield – amount of product predicted from the amounts of reactants used, calculated from the limiting reactant Actual yield – amount of product actually obtained through experiment Percent yield – comparison of actual and theoretical yield Percent Yield = Actual yield X 100 Theoretical yield

  15. C 1 x 12.0 = 12.0 H 4 x 1.0 = 4.0 O 1 x 16.0 = 16.0 32.0 g/mol Example: Methanol, CH3OH, can be produced by the reaction between carbon monoxide and hydrogen. Suppose 6.85 x 104 g of CO is reacted with 8.60 x 103 g of hydrogen. CO + 2H2 CH3OH • Calculate the theoretical yield of methanol. • If 3.57 x 104 g of CH3OH is actually produced, what is the percent yield of methanol? 6.9 x 104 g CH3OH 6.85 x 104 g CO 1 mol CO 1 mol CH3OH 32.0 g CH3OH x ___________ x ______________ x ___________ 28.0 g CO 1 mol CH3OH 1 mol CO 7.83 x 104 g CH3OH = 8.60 x 103 g H2 1 mol H2 1 32.0 g CH3OH mol CH3OH x ___________ x ___________ x ______________ 2.0 g H2 2 mol CH3OH mol H2 1 H 2 x 1.0 = 2.0 g/mol 6.9 x 104 g CH3OH = C 1 x 12.0 = 12.0 O 1 x 16.0 = 16.0 28.0g/mol 3.57 x 104 g 52 % X 100 = % Yield = 6.9 x 104 g

  16. Learning Check Learning Check - If 1.30 grams of oxygen and 3.10 grams of iron are reacted, then what is the theoretical yield of iron (III) oxide. What is the limiting reactant? If 3.00 grams of iron (III) oxide is produced, then what is the percent yield? Balanced equation: 4 Fe + 3 O2 2 Fe2O3 Theoretical Yield? Limiting Reactant? Excess Reactant? Percent Yield?

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