1. Last week… Review of ionic and net ionic equations. Review of water solubilities.

2. Table of Water Solubilities i= nearly insoluble ss=slightly soluble s=soluble d=decomposes n=not isolated acetate bromide carbonate chloride hydroxide iodide nitrate oxide phosphate sulfate sulfide sodium sssssssdssS silver ssiiinisii i Being able to use this table is essential for both writing ionic equations and ranking the solubility of compounds.

3. Ionic and Net Ionic Equations i= nearly insoluble ss=slightly soluble s=soluble d=decomposes n=not isolated acetate bromide carbonate chloride hydroxide iodide nitrate oxide phosphate sulfate sulfide sodium sssssssdssS silver ssiiinisii i Last week we mixed solution #2 (NaCl) with solution #4 (AgNO 3 ). This equation is written as: NaCl + AgNO 3  NaNO 3 + AgCl

4. Ionic and Net Ionic Equations i= nearly insoluble ss=slightly soluble s=soluble d=decomposes n=not isolated acetate bromide carbonate chloride hydroxide iodide nitrate oxide phosphate sulfate sulfide sodium sssssssdssS silver ssiiinisii i To write an ionic equation from the given equation: NaCl + AgNO 3  NaNO 3 + AgCl We first need to check the table to see which compounds disassociate in water…

5. Ionic and Net Ionic Equations i= nearly insoluble ss=slightly soluble s=soluble d=decomposes n=not isolated acetate bromide carbonate chloride hydroxide iodide nitrate oxide phosphate sulfate sulfide sodium sssssssdssS silver ssiiinisii i Since the compounds in red disassociate, we need to split them up: NaCl + AgNO 3  NaNO 3 + AgCl becomes Na + + Cl - + Ag + + NO 3 -  Na + + NO 3 - + AgCl Notice the insoluble AgCl stays together.

6. Ionic and Net Ionic Equations i= nearly insoluble ss=slightly soluble s=soluble d=decomposes n=not isolated acetate bromide carbonate chloride hydroxide iodide nitrate oxide phosphate sulfate sulfide sodium sssssssdssS silver ssiiinisii i To write a net ionic equation from the newly created ionic equation: Na + + Cl - + Ag + + NO 3 -  Na + + NO 3 - + AgCl Simply eliminate any ion that appears on both the reactant and product side of the equation (shown in orange). This leaves us with: Ag + + Cl -  AgCl

7. Table of Water Solubilities i= nearly insoluble ss=slightly soluble s=soluble d=decomposes n=not isolated acetate bromide carbonate chloride hydroxide iodide nitrate oxide phosphate sulfate sulfide sodium sssssssdssS silver ssiiinisii i To rank solubility of compounds, simply use the table. For example, in order of increasing solubility: silver chloride (i)  silver sulfate (ss)  silver nitrate (s)

8. Questions? If anything from the last two labs is still unclear, be sure to see me today BEFORE you go take the quiz!

9. Lab #6 Stoichiometry Chemistry 108 Instructor: Kristine Cooper

10. Stoichiometry Chemical math  The branch of chemistry that quantifies the substances in a chemical reaction The art of figuring how much stuff you'll make in a chemical reaction from the amount of each reagent you start with

11. Atomic weight Atomic weight of an element (found on the periodic table) is given in AMUs. For example, the atomic weight of carbon is 12 amu. This is the mass of ONE atom of carbon.

12. Gram atomic weight The gram atomic weight of an element is the same NUMBER as the element’s atomic weight, the UNITS change. For example, the gram atomic weight of carbon is 12 g. This is the mass of 6.02x10 23 atoms of carbon (an Avagardo’s number or one mole).

13. Formula weight / Molecular weight Formula weight is the combined weight of all atoms in a compound. Example: – Propanol CH 3 (CH 2 ) 2 OH What is the formula weight? Count the number of each atom, multiply by it’s atomic weight of that atom, then total.

14. Percent composition Calculated by determining the percent (by mass) of each individual atom of the formula weight of the compound. Example: – Propanol CH 3 (CH 2 ) 2 OH We determined the formula weight earlier to be 60 amu. Now determine what % of 60 amu the total mass of each atom comprises. Use the total mass of each type of atom determined before, then divide by the formula weight.

15. Simplest empirical formula Example: A compound was found to contain 88.8% oxygen and 11.2% hydrogen, what is the empirical formula? 1. Set the amount of the compound as 100g. In a 100g sample, there would be 88.8g oxygen and 11.2g hydrogen. 2. Determine the moles of each kind of atom. 88.8 g O / 16 g/mol=5.55 mol O 11.2 g H / 1 g/mol = 11.1 mol H 3. Set these numbers as subscripts, then simplify the ratio (divide by the smallest subscript present). H 11.1 O 5.5  H 2 O

16. True molecular formula Simple empirical formulas only tell us the ratio of atoms present in a compound. True molecular formulas tell us the true number of each atom present. THESE ARE ALWAYS EVEN MULTIPLES OF THE EMPIRICAL FORMULA!

17. True molecular formula For example, we determined water’s empirical formula to be H 2 O, which has a formula weight of 18g for each water molecule. What is the true molecular formula for 90g of water?

18. True molecular formula First, divide 90g by the molecular weight of water (18g). 90g/18g= 5 Next, multiply each subscript in the empirical formula (H 2 O) by this number (5). H=2x5=10 and O=1x5=5, so the true molecular formula is H 10 O 5.

19. Law of Definite Composition The chemical composition of a substance is fixed, it never varies with sample size. Simply stated, water is still H 2 O whether you have a mL vs. 5L, etc.

20. Stoichiometry practice Calculate the molecular weight of methane (CH 4 ).

21. Stoichiometry practice 12 moles of a certain compound weighs 702 g. What is the molecular weight of the compound?

22. Stoichiometry practice Calculate the percent composition of each element in hydrogen peroxide (H 2 O 2 ).

23. Stoichiometry practice Calculate the mass (in grams) of 1.15 moles of vandium sulfide (VS 2 ).

24. Stoichiometry practice You want to manufacture sodium monoxide following the process: 2Na + O 2 → 2Na 2 O. If you begin with 11.5g Na, how much O 2 will be necessary to ensure all the Na has reacted?

25. Today in Lab Given an unknown oxygen-containing potassium salt, we will determine its chemical formula. By removing the oxygen from the compound and then calculating the difference in weight with and without the oxygen, we can determine the original composition. Check out the demo compounds, compare volumes.