# 1 STOICHIOMETRYSTOICHIOMETRY DeterminingFormulas.

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1 STOICHIOMETRYSTOICHIOMETRY DeterminingFormulas

2 Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, C x H y, and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O What is the empirical formula of C x H y ?

3 Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. C x H y + some oxygen ---> CO 2 + H 2 O 0.379 g0.1035 g 0.379 g 0.1035 g Puddle of C x H y 0.115 g 0.379 g CO 2 +O 2 0.1035 g H 2 O 1 H 2 O molecule forms for each 2 H atoms in C x H y 1 CO 2 molecule forms for each C atom in C x H y

4 Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1. Calculate amount of C in CO 2 8.61 x 10 -3 mol CO 2 --> 8.61 x 10 -3 mol C 2. Calculate amount of H in H 2 O 5.744 x 10 -3 mol H 2 O -- >1.149 x 10 -2 mol H C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

5 Using Stoichiometry to Determine a Formula Now find ratio of mol H/mol C to find values of x and y in C x H y. 1.149 x 10 -2 mol H/ 8.61 x 10 -3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3 H 4 C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

6 STOICHIOMETRYSTOICHIOMETRY LIMITINGREACTANTS

7 Reactions Involving a LIMITING REACTANT In a given reaction, there is not enough of one reagent to use up the other reagent completely.In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.The reagent in short supply LIMITS the quantity of product that can be formed.

8 LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

9 STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

10 Rxn 1: Zn remaining * More than enough Zn to use up the 0.100 mol HC Rxn 2: no Zn left * Not enough Zn to use up 0.100 mol HCl LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2

11 PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of AlCl 3 can form? Mass reactant Coefficients Mole ratio Moles reactant Moles product Mass product

12 Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

13 2 Al + 3 Cl 2 ---> 2AlCl 3 Reactants must be in the mole ratio Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

14 Deciding on the Limiting Reactant If There is not enough Al to use up all the Cl 2 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Lim. reagent = Al

15 If There is not enough Cl 2 to use up all the Al 2 Al + 3 Cl 2 ---> 2AlCl 3 Lim reagent = Cl 2 Deciding on the Limiting Reactant

16 We have 5.40 g of Al and 8.10 g of Cl 2 Step 2 of LR problem: Calculate moles of each reactant

17 Find mole ratio of reactants This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl 2 2 Al + 3 Cl 2 ---> Al 2 Cl 6

18 Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of AlCl 3 can form? Limiting reactant = Cl 2 Base all calcs. on Cl 2 Limiting reactant = Cl 2 Base all calcs. on Cl 2 moles Cl 2 moles AlCl 3 grams Cl 2 grams AlCl 3 2 Al + 3 Cl 2 ---> 2AlCl 3

19 CALCULATIONS: calculate mass of Al 2 Cl 6 expected. CALCULATIONS: calculate mass of Al 2 Cl 6 expected. Step 1: Calculate moles of AlCl 3 expected based on LR. Step 2: Calculate mass of AlCl 3 expected based on LR.

20 Cl 2 was the limiting reactant.Cl 2 was the limiting reactant. Therefore, Al was presentTherefore, Al was present in excess. But how much? in excess. But how much? First find how much Al was required. First find how much Al was required. Then find how much Al is in excess.Then find how much Al is in excess. How much of which reactant will remain when reaction is complete?

21 2 Al + 3 Cl 2 products 0.200 mol 0.114 mol = LR Calculating Excess Al Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess