2. Stoichiometry Atomic Mass Atomic Mass Molar Mass of an Element Molar Mass of an Element Molecular Mass Molecular Mass Percent Composition Percent Composition Balancing Chemical Equations Balancing Chemical Equations Mass Relationship in Chemical Reactions Mass Relationship in Chemical Reactions Limiting Reactants Limiting Reactants Theoretical, Actual, and Percent Yields Theoretical, Actual, and Percent Yields

3. Atomic Mass of An Element Unit: amu, atomic mass unit = 1/12 of a 12 C atom 12 C : 12.00 amu 7 Li : 7.016 003 amu 14 N : 14.003 074 01 amu Most elements occur in nature as a mixture of isotopes Weighted average from all isotopes Cl : 6.941 amu ( 7.42% 6 Li + 92.58 % 7 Li)

4. Molar Mass of an Element Mass of an atom is very small, 1 amu ~1.66 x10 -24 g Mass of an atom is very small, 1 amu ~1.66 x10 -24 g Avogadro’s number Avogadro’s number Number of atoms in 12.000 grams of 12 C Number of atoms in 12.000 grams of 12 C Mole Mole 1 mol = 6.022137 x 10 23 atoms/molecules (Avogadro’s number) 1 mol = 6.022137 x 10 23 atoms/molecules (Avogadro’s number) Molar mass Molar mass The mass of one mole in gram The mass of one mole in gram For any element atomic mass (amu) = molar mass (grams)

5. Molecular Masses Formula mass :the total mass for all atoms in a compound Formula mass :the total mass for all atoms in a compound Sum of the masses of all the atoms in formula Sum of the masses of all the atoms in formula Molar mass Molar mass The mass of one mole of molecule in gram The mass of one mole of molecule in gram For any molecule molecular mass (amu) = molar mass (grams)

6. Mass spectrometer

7. Mass Percentage Mass percent, percent composition Mass percent, percent composition % A = x100 % Ex. Determine the mass % of H and O in H 2 O mass of 1 mol H 2 O : 2x1.008+16.00=18.01 g % H = = 11.19 % % O = = 88.81 % mass A in sample total mass of sample 2 x 1.008 g 18.01 g 16.00 g 18.01 g

8. Experimental Determination of Empirical Formula

9. Chemical Equations Law of Conservation of Matter Law of Conservation of Matter The total mass of the products must equal that of the reactants The total mass of the products must equal that of the reactants Chemical Equations Chemical Equations Show reactants, products, the state of all substances, and conditions Show reactants, products, the state of all substances, and conditions 2 CH 4 O(l) + 3 O 2 (g) 2CO 2 (g)+ 4H 2 O(l) 2 CH 4 O(l) + 3 O 2 (g) 2CO 2 (g)+ 4H 2 O(l) Reactants (state) Products (state) Reactants (state) Products (state) spark condition

10. Balancing Chemical Equations Steps: –Count the number of atoms of each element on each side of the equation and determine which atoms are not balanced –Begin with the compounds that has the most atoms and balance one atom at a time by using coefficients in front of one or more substances –Repeat steps until everything is balanced Hints: –Balance elements that appear only once on each side of reaction first –Balance elements that appear more than once later –DO NOT change subscripts –Practice !

11. Balancing Chemical Equations – Example (1) C 2 H 6 + O 2 _____ > CO 2 + H 2 O 2 C 1 C - not balanced 2 C 1 C - not balanced 6 H 2 H - not balanced 6 H 2 H - not balanced 2 O 3 O - not balanced 2 O 3 O - not balanced (2) Balance largest molecule first, C 2 H 6 C 2 H 6 + O 2 _____ > 2 CO 2 + 3 H 2 O C 2 H 6 + O 2 _____ > 2 CO 2 + 3 H 2 O 2 C 2 C 2 C 2 C 6 H 6 H 6 H 6 H 2 O 7 O - not balanced 2 O 7 O - not balanced (3) Balance O 2 C 2 H 6 + 3.5 O 2 _____ > 2 CO 2 + 3 H 2 O C 2 H 6 + 3.5 O 2 _____ > 2 CO 2 + 3 H 2 O (4) Can’t have 3.5 O 2 so double the equation ! 2 C 2 H 6 + 7 O 2 _____ > 4 CO 2 + 6 H 2 O (Balanced !)

12. Another Example (1) Start with chemical reaction HNO 3 (aq) _____ > NO 2 (aq) + O 2 (g) + H 2 O(l) 1 N, 1H, 3 O 1 N, 2H, 5 O (H and O not balanced) 1 N, 1H, 3 O 1 N, 2H, 5 O (H and O not balanced) (2) Balance largest molecule first, HNO 3 2HNO 3 (aq) _____ > NO 2 (aq) + O 2 (g) + H 2 O(l) 2HNO 3 (aq) _____ > NO 2 (aq) + O 2 (g) + H 2 O(l) 2 N, 2H, 6 O 1 N, 2H, 5 O (N and O not balanced) 2 N, 2H, 6 O 1 N, 2H, 5 O (N and O not balanced) (3) Balance N 2HNO 3 (aq) _____ > 2 NO 2 (aq) + O 2 (g) + H 2 O(l) 2HNO 3 (aq) _____ > 2 NO 2 (aq) + O 2 (g) + H 2 O(l) 2 N, 2H, 6 O 2 N, 2H, 7 O (O not balanced) 2 N, 2H, 6 O 2 N, 2H, 7 O (O not balanced) (4) Balance O 2 2HNO 3 (aq) _____ > 2 NO 2 (aq) + 0.5 O 2 (g) + H 2 O(l) 2HNO 3 (aq) _____ > 2 NO 2 (aq) + 0.5 O 2 (g) + H 2 O(l) (5) Can’t have 0.5 O 2 so double the equation ! 4HNO 3 (aq) _____ > 4 NO 2 (aq) + O 2 (g) + 2H 2 O(l) (Balanced!)

13. Mass Relationship in Chemical Reaction Stoichiometry Stoichiometry Study quantitative relationship between substances in chemical reactions Study quantitative relationship between substances in chemical reactions Show the relationship between the quantities of reactants and products Show the relationship between the quantities of reactants and products Start with balanced chemical equations Start with balanced chemical equations Work with moles Work with moles General Steps for Stoichiometry General Steps for Stoichiometry Balance the chemical equations Balance the chemical equations Calculate formula masses and convert masses to moles Calculate formula masses and convert masses to moles Use the mole ratios in chemical equations to find the desired quantities Use the mole ratios in chemical equations to find the desired quantities Convert back to mass if required Convert back to mass if required

14. Stoichiometry – Example If 25.00 g of B 2 O 2 is present, how many grams of Mg are needed to consume it? B 2 O 3 + 3 Mg ______ > 2 B+ 3 MgO B 2 O 3 + 3 Mg ______ > 2 B+ 3 MgO (1) Determine the number of mole of B 2 O 3 (molar mass of B 2 O 3 : 2 x 10.81 g/mol+3x16.00 g/mol=69.62 g/mol) (molar mass of B 2 O 3 : 2 x 10.81 g/mol+3x16.00 g/mol=69.62 g/mol) (2) Determine the number of moles of Mg needed Moles Mg : 0.3591 mol B 2 O 3 x = 1.077 mol Mg Moles Mg : 0.3591 mol B 2 O 3 x = 1.077 mol Mg (3) Calculate grams Mg needed grams Mg : moles x MM Mg 1.077 mol x 24.31 g/mol = 26.18 grams (4) One step further: we can find out 7.762 g of B are produced. 3.000 mol Mg 1.000 mol B 2 O 3 25.00 gx = 0.3591 mol B 2 O 3 1 mol B 2 O 3 69.62g B 2 O 3

15. Limiting Reagents Before reaction has started After reaction is complete 2 + 1 = 1 Limiting reagent: The reactant that is completely consumed or is in the shortest supply The reactant that is completely consumed or is in the shortest supply

16. Limiting Regent – Example For the reaction below, if you have 5.0 g of hydrogen and 10 g oxygen, which reactant is totally consumed and which reactant will remain and how much? 2 H 2 + O 2 ______ > 2 H 2 O 2 H 2 + O 2 ______ > 2 H 2 O (1) First determine the number of mole of H 2 and O 2 (2) Determine the number of moles of H 2 O produced (3) Thus, O 2 is the limiting reactant and will be totally consumed (4) 1.9 mol of H 2 remains. 1.9 mol H 2 x 2.0 g/mol = 3.8 g H 2 We find 2.5 mol H 2 0.31 mol O 2, how? 2.5 mol H 2 = 2.5 mol H 2 O 2 mol H 2 O 1 mol O 2 0.31 mol O 2 = 0.62 mol H 2 O 2 mol H 2 O 2 mol H 2

17. Theoretical, Actual, and Percent Yield Theoretical yield Theoretical yield The amount of product that should be formed according to the stoichiometry of a chemical reaction The amount of product that should be formed according to the stoichiometry of a chemical reaction Actual yield Actual yield The amount of product actually formed The amount of product actually formed Generally actual yield is less than the theoretical yield because Generally actual yield is less than the theoretical yield because Impurity in reactants Impurity in reactants Loss of products Loss of products Side reactions: form by-products Side reactions: form by-products Percent yield Percent yield % yield = x 100 Actual yield Theoretical yield

18. Example - Empirical Formula If 0.2536-g sample of a compound of Fe and S was reacted with O2 to give 0.1688 g Fe 2 O 3 and 0.2708 g SO 2. What is the empirical formula? (1) First determine the mass of Fe and O in the sample MM SO 2 : 64.065 g/molFe 2 O 3 : 159.69 g/mol MM SO 2 : 64.065 g/molFe 2 O 3 : 159.69 g/mol (2) Convert mass to mole (3) Find the molar ratio 111.69 g Fe 159.69 g Fe 2 O 3 32.066 g S 64.065 g SO 2 0.1688 g Fe 2 O 3 = 0.1181 g Fe 0.2708 g SO 2 = 0.1355 g S 0.1811 g Fe 55.845 g/mol 0.1355 g S 32.066 g/mol = 0.002115 mol Fe = 0.004226 mol S 0.004226 mol 0.002115 mol = 1.998 FeS 2