Mathematics of Mathematics of Chemical EquationsChemical Equations

By using “mole to mole” By using “mole to mole” conversions and conversions and balancedbalanced

equationsequations, we can calculate , we can calculate the exact amounts of the exact amounts of

substances that will be used substances that will be used up or produced during a up or produced during a

reaction.reaction.

•What you’ll learn: • Write mole ratios from a balanced chemical

equation

• Calculate the number of moles and the mass of a reactant or product when given the number of moles or the mass another reactant.

• Identify the limiting reactant in a chemical reaction.

• Determine the percent yield of a chemical reaction.

Basic ExampleBasic Example

NN22 + 3H + 3H22 2NH 2NH33

• From this balanced equationFrom this balanced equation, we know that for , we know that for every 1 mole of Nevery 1 mole of N22 we need ___ moles of H we need ___ moles of H22 to to react in order for ___ moles NHreact in order for ___ moles NH3 3 to be producedto be produced..

• How many moles of NHow many moles of N22 would we need to react would we need to react with 6 moles of Hwith 6 moles of H22??

6 moles H2 moles H23

moles N2

1= 2 moles of N2x

3

2

NN22 + 3H + 3H22 2NH 2NH33

• How many moles of Ammonia (NHHow many moles of Ammonia (NH33) ) are produced if you have an excess are produced if you have an excess of Nitrogen gas, and a sample of 0.43 of Nitrogen gas, and a sample of 0.43 moles of Hydrogen?moles of Hydrogen?

0.43 moles H2

moles H23

moles NH32 = 0.287 moles

of NH3

x

Mass – Mass ConversionsMass – Mass Conversions

• You MUST BALANCE the equation first!You MUST BALANCE the equation first!

NaClONaClO33 NaCl + O NaCl + O22

How many grams of OHow many grams of O22 would be produced if would be produced if the final mass of NaCl produced was 375.6 the final mass of NaCl produced was 375.6 g?g?

375.6 g NaCl375.6 g NaCl

2 2 3

x58.5 g NaCl

mole NaCl

1 x x mole NaCl

mole O2

2

3

mole O21

32 Grams O2

= 308.18g O2

Stoichiometric Stoichiometric Conversions Conversions A Three Step ProcessA Three Step Process

• Step 1 Convert the value of the given Step 1 Convert the value of the given substance to moles of that substance.substance to moles of that substance.

• Step 2 Convert from moles of the Step 2 Convert from moles of the given substance to moles of the given substance to moles of the unknown substance (use the mole unknown substance (use the mole ratio from the balanced equation).ratio from the balanced equation).

• Step 3 Convert from moles of the Step 3 Convert from moles of the unknown to proper unit. unknown to proper unit.

Limiting ReactantsLimiting Reactants

• When calculating your theoretical When calculating your theoretical products, it is important to factor in products, it is important to factor in which reactant you will run out of which reactant you will run out of firstfirst! !

• This is called the “Limiting Reactant”. This is called the “Limiting Reactant”. The other reactant(s) is called the The other reactant(s) is called the “Excess Reactant”.“Excess Reactant”.

Ex:Ex: If you have 10grams of OIf you have 10grams of O2 2 and 10 Liters and 10 Liters of Hof H22, how many molecules of water can you , how many molecules of water can you

create?create?

2H2H22 + O + O22 2H 2H22OO

• Step 1: Convert both to moles.Step 1: Convert both to moles.

10gO10gO2 x 2 x 1 mole O 1 mole O22 = = 0.31 moles O0.31 moles O22

1 32g O1 32g O22

10L 10L HH2 x 2 x 1 mole H 1 mole H2 = 2 = 0.45 moles H0.45 moles H22

1 22.4L H1 22.4L H22

• Step 2: Complete a mole to mole Step 2: Complete a mole to mole conversion, from one reactant to the conversion, from one reactant to the other reactant to find the theoretical other reactant to find the theoretical need.need.

2H2H22 + O + O22 2H 2H22OO

0.31 moles O0.31 moles O22 x x 2 moles H2 moles H22 = = 0.62 moles H0.62 moles H22

1 mole O1 mole O2 2 NEEDEDNEEDED

YOU KNOWYOU KNOW that you have 0.45 moles H that you have 0.45 moles H22 from your given values. So, that means from your given values. So, that means you do not have enough Hyou do not have enough H22. It is . It is therefore the Limiting Reactant.therefore the Limiting Reactant.

• IF YOU COMPLETED STEP 2 FINDING IF YOU COMPLETED STEP 2 FINDING MOLES OF OMOLES OF O22::

2H2H22 + O + O22 2H 2H22OO

0.45 moles H0.45 moles H22 x x 1 moles O1 moles O22 = = 0.225 moles 0.225 moles OO22

2 mole H2 mole H2 2 NEEDEDNEEDED

You would come to the same conclusion, You would come to the same conclusion, because this shows that you have more because this shows that you have more than enough Othan enough O22. (You have .31 moles). (You have .31 moles)

• Step 3: Decide from Step 2 which Step 3: Decide from Step 2 which reactant is the L.R. and use its moles reactant is the L.R. and use its moles to calculate your answer. (MUST to calculate your answer. (MUST have a balanced Chemical Equation!)have a balanced Chemical Equation!)

2H2H22 + O + O22 2H 2H22OO

• 0.45mol H0.45mol H22 x x 2 mol H2 mol H22OO x x 6.02x106.02x102323 molecules H molecules H22OO = =

11 2 mol H 2 mol H22 1 mole H1 mole H22OO

2.71x102.71x102323 molecules H molecules H22OO

Percent YieldPercent Yield

• Most chemical reactions never succeed Most chemical reactions never succeed in producing the predicted amount of in producing the predicted amount of product. product.

• So, the actual amount of product is less So, the actual amount of product is less than expected, due to experimental error than expected, due to experimental error

• This is generally due to experimental This is generally due to experimental error such as evaporation, product left error such as evaporation, product left on filter paper etc.on filter paper etc.

Percent YieldPercent Yield

• Percent yield is the ratio of actual Percent yield is the ratio of actual yield to the theoretical yield yield to the theoretical yield expressed as a percent. expressed as a percent.