Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this C is determined from the mass of CO 2 produced H is determined from the mass of H 2 O produced O is determined by difference after the C and H have been determined

Elemental Analyses Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products

Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams C 6 H 12 O O 2  6 CO H 2 O

3.6 Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O What is the percent composition of an organic substance called ethanol? Real Life Analysis

3.6 Combust 11.5 g ethanol Collect 22.0 g CO 2 and 13.5 g H 2 O

How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredients Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount In other words, it’s the reactant you’ll run out of first (in this case, the H 2 )

Limiting Reactants In the example below, the O 2 would be the excess reagent

Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3 formed. 3.9

Theoretical Yield The theoretical yield is the amount of product that can be made In other words it’s the amount of product possible as calculated through the stoichiometry problem This is different from the actual yield, the amount one actually produces and measures

Percent Yield A comparison of the amount actually obtained to the amount it was possible to make Actual Yield Theoretical Yield Percent Yield =x 100

Percent Yield Problem When benzene (30.0 g) reacts with bromine (65.0 g), bromobenzene (42.3 g) and hydrogen bromide gas are produced. What is the percent yield? C 6 H 6 + Br 2  C 6 H 5 Br + HBr

7 Principles of Green Chemistry 1) Maximize Atom Economy 2) Maximize Safety 3) Renewable Feedstocks 4) Use Catalysts 5) Use Environmentally Neutral Solvents 6) Increase Energy Efficiency 7) Design in Biodegradability

Maximize Atom Economy Old two step method: 1) C 2 H 4 + Cl 2 + H 2 O  C 2 H 4 ClO + HCl 2) C 2 H 4 ClO + Ca(OH) 2  C 2 H 4 O + CaCl 2 + H 2 O New One step method (reusable Ag cat): 1) 2 C 2 H 4 + O 2  2 C 2 H 4 O 100% atom efficiency Single step requires less time No waste to get ride of Calculate the percent atom efficiency for the old process.

Use Catalysts Speed up reactions Reduce energy input requirements Reduce waste Can be used to detoxify exhaust Lets talk about cars!

Automobile Exhaust High compression favors NO formation N 2 + O 2  2 NO Speed of combustion favors un-burnt HC 100’s of partially oxidized materials Many are highly reactive Catalytic converter must reduce reactive materials but needs reactants to do it. Balancing air/fuel mixture needed – stoichiometry!

What is needed? Extra oxygen helps reduce un-burnt fuel but increases NO … A balance of NO and CO production can make the catalytic converter work best: 2 CO + 2 NO  N CO 2 And excess NO can be used to oxidize other organic fuel components. What mass ratio is needed for complete reaction of both NO and CO?

Octane boosters Increase the rate of fuel burning with octane boosters reduces amount of un-burnt fuel. Ethanol is least toxic and is renewable MTBE is cheap and very efficient but accumulates in water supply. AdditiveOctane Number Methanol107 Ethanol109 MTBE116 ETBE118

MTBE Production The process for making MTBE is 100% atom efficient. What mass of isobutene is required to make grams of MTBE? C 4 H 8 + CH 4 O  C 5 H 12 O(MTBE)