1. Unit 4: Formula Stoichiometry

2. What is stoichiometry? Deals with the quantitative information in chemical formula or chemical reaction. Deals with the quantitative information in chemical formula or chemical reaction. Using our reference tables we will learn how to calculate the following: Using our reference tables we will learn how to calculate the following: Mass of an atom Mass of an atom Mass of a compound Mass of a compound Moles of an atom or compound (molar mass) Moles of an atom or compound (molar mass) Percent Composition in a compound Percent Composition in a compound

3. Key Terms Review Subscripts - tell you the amount of atoms in a formula Subscripts - tell you the amount of atoms in a formula Coefficients - tell you the amount of molecules of that substance Coefficients - tell you the amount of molecules of that substance

4. How many oxygen atoms in each? NH 4 NO 3 NH 4 NO 3 C 8 H 8 O 4 C 8 H 8 O 4 O 3 O 3 C 3 H 5 (NO 3 ) 3 C 3 H 5 (NO 3 ) 3 (3) (4) (3) (9)

5. Counting Atoms Practice (NH 4 ) 2 CO 3 N: H: C: O: * First list the types of atoms and then count each. Example #1: 28132813

6. Counting Atoms Practice CaCrO 4 Ca: Cr: O: * First list the types of atoms and then count each. Example #2: 114114

7. Counting Atoms Practice Ca 3 (PO 4 ) 2 Ca: P: O: * First list the types of atoms and then count each. Example #3: 328328

8. Mass Atomic MassMolecular MassFormula Mass Mass of one atom of one element from periodic table. Mass of all of the atoms in a “Covalent” or “Molecular” compound (non-metal atoms only) Mass of all of the atoms in an ionic compound (metal + non-metal ions) - All measured in amu (atomic mass units) Ex)NeEx)CHCl 3 Ex) CaCl 2

9. Gram Formula Mass & Gram Molecular mass The same as before, but instead of “amu” Mass expressed in grams (g) The same as before, but instead of “amu” Mass expressed in grams (g) H2OH2O = OH + H + + + Molecular Mass = 15.9994 amu + 1.00794 amu +1.00794 amu Gram Molecular Mass = 18.0 g

10. How to calculate molecular or formula mass… First: Identify and count atoms in the compound. First: Identify and count atoms in the compound. Second: Locate atomic mass of each element. Second: Locate atomic mass of each element. Third: Multiply mass by number of atoms, then get the total of all elements. Third: Multiply mass by number of atoms, then get the total of all elements.

11. Molecular Mass Example (Covalent Compound) Example: H 2 O Example: H 2 O H- 2 x 1 H- 2 x 1 O- 1 x 16 O- 1 x 16 Molecular Mass = 18 g Molecular Mass = 18 g

12. Formula Mass Example (ionic compound) Example: NaCl Example: NaCl Na- 1 x 23 Na- 1 x 23 Cl- 1 x 35 Cl- 1 x 35 Formula Mass = 58 g Formula Mass = 58 g

13. Percent Composition by Mass Used to calculate how much of a substance’s mass is made up of a particular element… Used to calculate how much of a substance’s mass is made up of a particular element… Formula is given on Table T in reference table Formula is given on Table T in reference table Mass of Part Mass of Whole X 100 = % Composition “by Mass”

14. Example 1: In a sample of calcium carbonate, what is the percent composition by mass of calcium? In a sample of calcium carbonate, what is the percent composition by mass of calcium? Formula = CaCO 3 Ca: 1 C: 1 O: 3 x 40 = 40 g x 12 = 12 g x 16 = 48 g } total = 100 g % Composition Calcium = 40 g 100 g X 100= 40 % Ca

15. Example 2: A compound containing carbon and hydrogen has a mass of 16 grams. When decomposed, 12.0 grams are found to be carbon. What is the percent by mass of carbon in the compound? A compound containing carbon and hydrogen has a mass of 16 grams. When decomposed, 12.0 grams are found to be carbon. What is the percent by mass of carbon in the compound? } total = 16 g % Composition Carbon = 12 g 16 g X 100= 75 % Carbon Formula = C ? H ? C = 12 g O = __ g

16. Hydrated Crystals: Some ionic compounds have surrounding water molecules. Some ionic compounds have surrounding water molecules. These are called Hydrated Crystals. These are called Hydrated Crystals. Percent water hydration can be found using the formula for % composition… Percent water hydration can be found using the formula for % composition… Example:CuSO 4. 5H 2 O 5 H 2 O molecules are attached to 1 CuSO 4

17. Percent Water Hydration (cont.) % Hydration = Total mass of Water x 100 Formula Mass Formula = CuSO 4. 5H 2 O Cu = 1 x 64 = 64 g S = 1 x 32 = 32 g O = 4 x 16 = 64 g Separately H 2 O: O = 1 x 16 = 16 g H = 2 x 1 = 2 g _________________ H 2 O = 18 x 5 = 90 g g ✔

18. The Mole Pt. 1 A mole is essentially one complete unit of any particular substance A mole is essentially one complete unit of any particular substance Think of it like the term “dozen” Think of it like the term “dozen” Meaning 12 items of any substance make up a dozen of that substance Meaning 12 items of any substance make up a dozen of that substance 1 mole is always equal to 6.02 x 10 23 particles of a substance 1 mole is always equal to 6.02 x 10 23 particles of a substance

19. The Mole Pt. 2 Avagadro’s Number Avagadro’s Number is 6.02 x 10 23 Avagadro’s Number is the number of: Avagadro’s Number is the number of: - Atoms in 1 mole of any element - Molecules in 1 mole of any compound - Particles in 1 mole of any substance

20. The Mole Pt. 3 The mass of 1 mole will be different for all substances. The mass of 1 mole will be different for all substances. Just like a dozen eggs being different from a dozen tires (size and weight) Just like a dozen eggs being different from a dozen tires (size and weight) Gram Atomic / Formula / Molecular Mass = Gram Atomic / Formula / Molecular Mass = Molar Mass - because it is the mass of one Units(g/mol)mole of any substance. Example: 1 mole of H 2 O = 18g Example: 1 mole of H 2 O = 18g Molar Mass of Water = 18 g/mol Molar Mass of Water = 18 g/mol

21. Mole Relationships or Equalities 1 mole = formula mass in grams 1 mole = formula mass in grams 1 mole = 6.02 x 10 23 particles 1 mole = 6.02 x 10 23 particles 1 mole = 22.4L (for gases) 1 mole = 22.4L (for gases)So… Formula mass = 6.02x10 23 particles Formula mass = 6.02x10 23 particles 22.4L of a gas = 6.02x10 23 particles 22.4L of a gas = 6.02x10 23 particles Formula mass = 22.4L of a gas Formula mass = 22.4L of a gas

22. Conversion Problems Some problems involve mass-mole conversions Some problems involve mass-mole conversions you can use the “Mole Calculation” formula on Table T. you can use the “Mole Calculation” formula on Table T. You will usually have to calculate the gram formula mass of the compound… You will usually have to calculate the gram formula mass of the compound… # of Moles = given mass (g) gram formula mass

23. Example 1 - What is the mass of 1.75 moles of oxygen gas (Hint: oxygen is diatomic)? 1 st find the molecular mass of oxygen remember oxygen is a diatomic element (O 2 ) O: 2 x 16 g= 32 g ? = 56 g Then use formula on table T: Moles = Given Mass (g) Gram Formula Mass 1.75 = ? 32 g ✔

24. - How many moles are in 42 g of water? 1 st find the molecular mass of H 2 O: H: 2 x 1 g = 2 g O: 1 x 16 g = 16 g 2 + 16 = 18 g Convert from grams to moles… ? Moles = 42 g 18 g ? =2.3 mol Example 2 ✔

25. - How many moles of potassium chromate are in a 500 g sample? K:2 x 39 g = 78 g Cr: 1 x 52 g = 52 g O: 4 x 16 g = 64 g 194 g 78+52+64 = 194 g ? = 2.6 mol Example 3 1) Find the formula of Potassium Chromate: K 2 CrO 4 2) Calculate formula Mass: 3) Calculate the Moles using the formula on table T: ? Moles = 500 g 194 g

26. Empirical Formula Review

28. In a question you are given the molecular mass and empirical formula… In a question you are given the molecular mass and empirical formula… 1 st you must find the empirical mass 1 st you must find the empirical mass Then divide empirical mass by molecular mass Then divide empirical mass by molecular mass Multiply all subscripts in the empirical formula by your answer to get molecular formula Multiply all subscripts in the empirical formula by your answer to get molecular formula Empirical Mass - Is the mass of the empirical formula of a compound

29. Empirical Mass Example Step 1: CH 2 O Empirical Mass C – 1 x 12 = 12 H – 2 x 1 = 2 O – 1 x 16 = 16 30 g 30 g Step 2: Divide (Molecular mass/Empirical mass) 180/30 = 6 Step 3: Multiply Empirical Formula by Step 2 6(CH 2 O) = C 6 H 12 O 6 Molecular Mass = 180Empirical Formula = CH 2 O

30. Moles in Chemical Equations In problems involving chemical reactions we balance chemical reactions using coefficients In problems involving chemical reactions we balance chemical reactions using coefficients These coefficients also represent the mole ratios in a reaction These coefficients also represent the mole ratios in a reaction Ex) Ex) Moles C 2 H 6 Moles O 2 Moles CO 2 Moles H 2 O 2746 13.523

31. After balancing we can determine the amount of a reactant or product involved in the reaction After balancing we can determine the amount of a reactant or product involved in the reaction As long as your given the amount of one ingredient… As long as your given the amount of one ingredient… Ex) Ex) Moles in Chemical Equations Pt. 2 What is the Minimum # of moles of Oxygen gas needed to make 1 mole of aluminum oxide?

32. The mole ratio in a balanced reaction can also be used to calculate the mass of reaction ingredients The mole ratio in a balanced reaction can also be used to calculate the mass of reaction ingredients Ex) Based on the reaction below a student calculates 20g of H 2 will react with 20g Cl 2 to form 40g HCl, is this true? Ex) Based on the reaction below a student calculates 20g of H 2 will react with 20g Cl 2 to form 40g HCl, is this true? Moles in Chemical Equations Pt. 3

33. We can use these mole ratios to calculate the volume of reaction ingredients also (only in gas reactions) We can use these mole ratios to calculate the volume of reaction ingredients also (only in gas reactions) Ex) Ex) Moles in Chemical Equations Pt. 4 Mole ratio = 2 : 7 : 4 : 6 How many Liters of CO 2 (g) will be produced from the combustion of 30 Liters of C 2 H 6 (g) ? 30 L 2 = 15 L Then, 15 L x 4 = 60 L

34. A pound of meth is worth 40,000 $ A pound of meth is worth 40,000 $ 1 lb = ~454 g 1 lb = ~454 g Balance the reaction and figure out the amount of reactants needed to synthesize 454 g of meth… Balance the reaction and figure out the amount of reactants needed to synthesize 454 g of meth… ? g+ ? g  454 g