Unit 1: 1 st Year Review Part 6 Assignments Due End of AM Woodstown: AP 2005 #2 or not. Your choice Topics of the Day: Stoichiometry Limiting Factors Empirical Formulas Wake-Up Activity: In Your Notebook: Draw an atom level reaction showing 6 hydrogen gas molecules and 2 oxygen gas molecules as reactants. On the other side of the arrow draw as many complete water molecules as could be made as well as any remaining molecules that were not used.

Logic Behind the Wake-Up Sometimes there aren’t enough chemicals for everything to react completely and perfectly. The one that runs out first limits the amount of products that can form. – Called the limiting reagent or limiting reactant Which molecule was the limiting reactant in the Wake-Up?

More Explanation Write the balanced reaction equation ignoring how much you really have to work with 2H 2(g) + O 2(g)  2H 2 O (g) Coefficients  You need two hydrogen molecules for each oxygen molecule. Use logic – 6 H 2(g) would need 3 O 2(g) – There are only 2, so the O 2(g) runs out first  oxygen is the limiting reagent – 2 O 2(g) would need 4 H 2(g) which leaves 2 H 2(g) left over  confirms O 2(g) is the limiting reagent

Using Stoichiometry to Find Limiting Reagent What is the limiting reagent when 14.2g of carbon is combusted in the presence of 20.0g of oxygen gas to make carbon monoxide? Equation: 2C (s) + O 2(g)  2CO (g) 14.2g C1 mol C2 mol CO= 1.18 mol CO produced 12.01g C2 mol C 20.0g O 2 1 mol O 2 2 mol CO= 1.25 mol CO produced 32.00g O 2 1 mol O 2 LIMITING REAGENT

Empirical Formulas Reduced version of the formula The greatest common factor of all of the subscripts should be one – If not, reduce it again because it isn’t an empirical Often obtained from experimental data such as masses or percentages

What is the empirical formula of a compound that is 40% carbon, 6.7% hydrogen, and 53.3% oxygen? Step 0: If given grams directly, skip to step 2 Step 1: Assume 100g sample – Why? Because it makes the math easy. 53.3% of 100g is 53.3g of O – 40.0g of C and 6.7g of H Step 2: Divide the mass of each element by its molar mass – Why? Because this will give us moles. Subscripts are ratios of moles within the substance. – 40 / = 3.3 mol C – 6.7 / 1.01 = 6.6 mol H – 53.3 / 16 = 3.3 mol O Step 3: Divide all of the moles by the smallest mole value – Why? We’re trying to figure out the ratio. The smallest is usually the best place to start to eliminate fractions.

What is the empirical formula of a compound that is 40% carbon, 6.7% hydrogen, and 53.3% oxygen? Step 3: Divide all of the moles by the smallest mole value – Why? We’re trying to figure out the ratio. The smallest is usually the best place to start to eliminate fractions. – 3.3 mol is the smallest – C 3.3 / 3.3 = 1 – H 6.6 / 3.3 = 2 – O 3.3 / 3.3 = 1 Step 4: Write a formula and use the result from Step 3 as the subscripts. – CH 2 O – Remember, you don’t write 1’s. They are understood. Step 5: Adjust as appropriate to have all subscripts be whole numbers. – If one ends in.5 try multiplying all subscripts by 2.

Example #2 – Silver Oxide Lab You knew the mass of the substance. You heated away the oxygen to get the mass of the silver. You then subtracted to the the oxygen. You then divided the silver only and oxygen only masses by their molar masses to get moles. Then you got the ratio. It’s neat when the labs align with what you’re learning in class.

Molecular Formula Related to the Empirical formula This is the REAL formula It is the empirical formula with the subscripts multiplied by the same positive integer – Remember, 1 is a positive integer – In that case the empirical and molecular formulas will be the same

Example: CH 2 O Possible molecular formulas 1x CH 2 O = CH 2 O 2x CH 2 O = C 2 H 4 O 2 3x CH 2 O = C 3 H 6 O 3 Etc. How do you know which is correct? You need more information

Target Mass Without a target mass or target mass range, you can’t figure out the molecular formula. All you could get is the empirical. Example: Target range is between 110g/mol and 130g/mol Step 1: determine the molar mass of the empirical formula – 60.06g/mol in our example. Step 2: Divide the average number of the target range by the molar mass of the empirical formula – 120g/mol / 60.06g/mol = about 2 times (okay to round if close) Step 3: Multiply the subscripts by the result from step 2. – 2x CH 2 O = C 2 H 4 O 2 Step 4: make sure subscripts are whole numbers.

Homework Presentation If we have time…

Homework Solve Problems from HP. (preferred) Solve AP 2006 #3 (if time crunched)