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) + 2H 2(g)  CH 3 OH (l) Ex: How many moles of O 2 are required to produce 10. moles of CO 2 ? 2 CO + O 2  2 CO 2 1 mol CO = 2 mol H 2 = 1 mol CH 3 OH 10. mol CO 2 x __________ mol CO 2 mol O 2 2 1 = 5.0 mol O 2

3. What other relationships do we have for the mole? 1 mol = 6.022 x 10 23 atoms / molecules / particles 1 mol = [molar mass] 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. Mass A – Mole B Ex: Calculate moles of O 2 produced if 2.50 g KClO 3 decomposes completely: 2 KClO 3  2 KCl + 3 O 2 x ______________ mol O 2 mol KClO 3 3 2 x ________________ mol KClO 3 122.6 1 g KClO 3 2.50 g KClO 3 = 0.0306 mol O 2 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

5. Mass A – Mass B Ex: Determine the mass of NaCl that decomposes to yield 355 g Cl 2 2NaCl  2 Na + Cl 2 355 g Cl 2 mol Cl 2 g Cl 2 71.0 1 x _____________ x ______________ mol Cl 2 mol NaCl 1 2 x ______________ mol NaCl g NaCl58.5 1 = 585 g NaCl Cl 2 x 35.5 = 71.0 g/mol Na 1 x 23.0 = 23.0 Cl 1 x 35.5 = 35.5 58.5 g/mol

6. Mole A – Mass B Ex: Calculate the number of grams of oxygen required to react exactly with 4.30 mol of propane, C 3 H 8, in the reaction by the following balanced equation: C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g) 4.30 mol C 3 H 8 mol O 2 mol C 3 H 8 1 5 x _____________ x __________ mol O 2 g O 2 1 32.0 = 688 g O 2 O 2 x 16.0 = 32.0 g/mol