1. Formula Stoichiometry

2. What is stoichiometry? Deals with the specifics of QUANTITY in chemical formula or chemical reaction. Deals with the specifics of QUANTITY in chemical formula or chemical reaction.

3. Review: Counting Atoms Remember subscripts tell you how many atoms… Remember subscripts tell you how many atoms… … Coefficients tell you how many TOTAL molecules compounds … Coefficients tell you how many TOTAL molecules compounds

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 **Do the rest on your own!**

8. Atomic Mass Mass of an atom of one element based on percent abundances and masses of all isotopes (on PT) Mass of an atom of one element based on percent abundances and masses of all isotopes (on PT) Units: amu Units: amu

9. Molecular Mass Mass of a molecular compound (ex: H 2 O, CO 2 ) Mass of a molecular compound (ex: H 2 O, CO 2 ) Units: amu Units: amu

10. Formula Mass Mass of an ionic compound (ex: NaCl, MgI 2 ) Mass of an ionic compound (ex: NaCl, MgI 2 ) Units: amu Units: amu

11. Gram Formula Mass Mass expressed in grams Mass expressed in grams

12. Gram Molecular Mass H2OH2O = OH + H + + + Molecular Mass = 15.9994 amu + 1.00794 amu +1.00794 amu Gram Molecular Mass = 18.0 g

13. How do I 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 x number of atoms and total all elements. Third: Multiply mass x number of atoms and total all elements.

14. 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

15. Formula Mass Example (ionic compound) Example: NaCl Example: NaCl Na- 1 x 23 Na- 1 x 23 Cl- 1 x 35.5 Cl- 1 x 35.5 Formula Mass = 58.5 g Formula Mass = 58.5 g

16. Percent Composition by Mass Experimentally- Experimentally- Use masses given in problem Use masses given in problem Mass of Element x 100 Mass of Compound

17. Example: 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?

18. Theoretically- Theoretically- Use Atomic Masses Use Atomic Masses Formula Formula (#atoms of element) x (atomic mass) x100 Formula mass of compound Percent Composition by Atomic Mass

19. Hydrated Crystals Some ionic compounds are found to have surrounding water molecules. Some ionic compounds are found to have surrounding water molecules. These compounds are called Hydrated Crystals. These compounds are called Hydrated Crystals. The percent water hydration can be found. The percent water hydration can be found.Example CuSO 4. 5H 2 O The dot shows that 5 H 2 O molecules are attached to 1 CuSO 4 molecule. The dot shows that 5 H 2 O molecules are attached to 1 CuSO 4 molecule.

20. Percent Water Hydration % Hydration = Total mass of Water x 100 Formula Mass CuSO 4. 5H 2 O Cu – 1 x 63.5 = 63.5 S – 1 x 32 = 32 O – 4 x 16 = 64 H 2 O – 5 x 18 = 90 Formula Mass = 249.5 % Hydration = (90/249.5) x 100 = 36% H 2 O

21. The Mole Avagadro’s Number is the number of atoms in 1 mole of any substance. Avagadro’s Number is the number of atoms in 1 mole of any substance. Avagadro’s Number is 6.02 x 10 23 Avagadro’s Number is 6.02 x 10 23 That means that in 1 mole of any sample of matter there are 6.02x10 23 particles. That means that in 1 mole of any sample of matter there are 6.02x10 23 particles.

22. The Mole There are 6.02 x 10 23 particles in one mole of any substance There are 6.02 x 10 23 particles in one mole of any substance One particle can be one atom or one molecule. One particle can be one atom or one molecule. The mass of 1 mole will be different for all substances. The mass of 1 mole will be different for all substances. Atomic / Formula / Molecular Mass = Molar Mass because it is the mass of 1 mole of any substance. Atomic / Formula / Molecular Mass = Molar Mass because it is the mass of 1 mole of any substance. Example: 1 mole of H 2 O = 18g Example: 1 mole of H 2 O = 18g

23. 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) 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

24. Example Conversions Converting Grams to Moles given grams X GFM 1moles grams Converting Moles to Grams given moles X moles grams 1 GFM

25. Table T For problems that involve mass mole conversions you can use the “Mole Calculation” formula on Table T. For problems that involve mass mole conversions you can use the “Mole Calculation” formula on Table T. You will usually have to calculate the gram formula mass of the substance unless it is given. You will usually have to calculate the gram formula mass of the substance unless it is given. See Table T See Table T

26. Examples 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 Convert from moles to grams using factor label method Factor Label Setup 1.75 mol x __32 g _ = ? g 1 mol 56 g

27. Examples Example #2: How many moles are in 42 g of water? 1 st find the molecular mass of H 2 O. H: 2 x 1 = 2 O: 1 x 16 = 16 2 + 16 = _18 g_ 1 mol Convert from grams to moles… 42 g x __ 1 mol _ = ? mol 18 g 2.3 mol

28. Examples Example #3: How many moles of potassium chromate are in a 500 g sample? 1 st find the molecular mass of potassium dichromate (K 2 CrO 4 ). K:2 x 39 = 78 Cr: 1 x 52 = 52 O: 4 x 16 = 64 78+104+112 = _194 g_ 1 mol Convert from grams to moles… 500 g x __ 1 mol _ = ? mol 194 g 2.6 mol

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