Elemental Analysis

Combustion Analysis

Compounds containing C, H and O are routinely analyzed through combustion in a chamber like thisC is determined from the mass of CO2 producedH is determined from the mass of H2O producedO is determined by difference after the C and H

have been determined

Elemental AnalysesCompounds 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 C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O

3.6

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

• 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 CO2 and 13.5 g H2O

Limiting Reactants

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 amountIn other words, it’s the reactant you’ll run out

of first (in this case, the H2)

Limiting Reactants

In the example below, the O2 would be the excess reagent

Do You Understand Limiting Reagents?In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

3.9

Theoretical YieldThe theoretical yield is the amount of product

that can be madeIn 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 YieldA comparison of the amount actually obtained to the amount it was possible to make

Actual YieldTheoretical YieldPercent Yield = x 100

Percent Yield ProblemWhen 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?C6H6 + Br2 C6H5Br + HBr

Green Chemistry

7 Principles of Green Chemistry1) Maximize Atom Economy2) Maximize Safety3) Renewable Feedstocks4) Use Catalysts5) Use Environmentally Neutral Solvents6) Increase Energy Efficiency7) Design in Biodegradability

Maximize Atom EconomyOld two step method:

1) C2H4 + Cl2 + H2O C2H4ClO + HCl2) C2H4ClO + Ca(OH)2 C2H4O + CaCl2 + H2O

New One step method (reusable Ag cat):1) 2 C2H4 + O2 2 C2H4O

100% atom efficiency Single step requires less time No waste to get ride of

Calculate the percent atom efficiency for the old process.

Use CatalystsSpeed up reactionsReduce energy input requirementsReduce wasteCan be used to detoxify exhaust

Lets talk about cars!

Automobile ExhaustHigh compression favors NO formation

N2 + O2 2 NO

Speed of combustion favors un-burnt HC100’s of partially oxidized materialsMany 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 N2 + 2 CO2

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 boostersIncrease the rate of

fuel burning with octane boosters reduces amount of un-burnt fuel.Ethanol is least toxic

and is renewableMTBE is cheap and

very efficient but accumulates in water supply.

Additive Octane Number

Methanol 107

Ethanol 109

MTBE 116

ETBE 118

MTBE ProductionThe process for making MTBE is 100%

atom efficient. What mass of isobutene is required to make 510.00 grams of MTBE?

C4H8 + CH4O C5H12O(MTBE)