1. Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

2. Chemical Equations  Are sentences.  Describe what happens in a chemical reaction.  Reactants  Products  Equations should be balanced.  Have the same number of each kind of atoms on both sides because...

3. Abbreviations  (s)   (g)   (aq)  heat    catalyst

4. Atomic Mass  Atoms are so small, it is difficult to discuss how much they weigh in grams  Use atomic mass units  The average atomic mass is found on the periodic table  This is not a whole number because it is an average!!

5. Atomic Mass  The average atomic mass is determined by using the masses of its various isotopes and their relative abundances  Amu = (abundance x mass) + (abundance x mass) + … x mass) + …

6. Molar Mass  Mass of 1 mole of a substance  Often called molecular weight  To determine the molar mass of an element, look on the table  To determine the molar mass of a compound, add up the molar masses of the elements that make it up

7. Find the molar mass of  CH 4  Mg 3 P 2  Ca(NO 3 ) 2  Al 2 (Cr 2 O 7 ) 3  CaSO 4 · 2H 2 O

8. Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms)(atomic weight) (molar mass of compound) x 100

9. Percent Composition So the percentage of carbon in ethane (C 2 H 6 ) is… %C = (2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu = x 100 = 80.0%

10. The Mole  The mole is a number  A very large number, but still, just a number  6.022 x 10 23 of anything is a mole  Scientists call this number Avogadro’s number

11. Avogadro’s Number  6.02 x 10 23  1 mole of 12 C has a mass of 12 g

12. Using Moles Moles provide a bridge from the molecular scale to the real-world scale

13. Empirical Formulas  The empirical formula for a substance shows the lowest ratio of atoms of each element it contains  This is based on mole ratios.

14. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition.

15. Calculating Empirical Formulas 1.%  g 2.g  mol 3.Divide all by smallest mole value ☺ multiply ‘til whole

16. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

17. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

18. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

19. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

20. Empirical To Molecular Formulas  Empirical describes the lowest ratio  Molecular describes the actual molecule  Need molar mass  Ratio of empirical to molar mass will tell you the molecular formula.  Must be a whole number

21. Calculating Molecular Formulas 1.Calculate empirical formula 2.Calculate empirical mass 3.Molar Mass / Empirical Mass ☺ multiply ‘til whole 4. Multiply all subscripts in empirical formula

22. Example  A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 184 g/mol. Determine the molecular formula of this molecule. Determine the molecular formula of this molecule.

23. Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

24. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

25. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

26. Example One way of producing O 2 (g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.0 g sample of potassium chlorate is decomposed.  How many moles of O 2 (g) are produced?  How many grams of oxygen?  How many grams of potassium chloride are produced?

27. How Many Cookies Can I Make?  You can make cookies until you run out of one of the ingredients  Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

28. How Many Cookies Can I Make?  In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

29. Limiting Reactants  The limiting reactant is the reactant present in the smallest stoichiometric amount  In other words, it’s the reactant you’ll run out of first (in this case, the H 2 )

30. Limiting Reactants In the example below, the O 2 would be the excess reagent

31. Limiting Reactants 1.Convert both reactants to moles - HAVE 2.Mole ratio between reactants - NEED 3.Compare need to have - LR 4.Mole ratio between LR and product 5. Convert product to grams

32. Example Ammonia is produced by the following reaction N 2 + H 2  NH 3 What mass of ammonia can be produced from a mixture of 100.g N 2 and 500.g H 2 ? How much unreacted material remains?

33. Theoretical Yield  The theoretical yield is the amount of product that can be made  In other words it’s the amount of product possible as calculated through the stoichiometry problem  This is different from the actual yield, the amount one actually produces and measures in the lab

34. Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100

35. Examples  Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.