Chemical QuantitiesChemical Quantities

Chapter 9Chapter 9

9.2 Mole-Mole Relationships9.2 Mole-Mole Relationships

2H2H22OO(l)(l) → 2H→ 2H2(g)2(g) + O + O2(g)2(g)

2 moles of water breaks down into 2 moles 2 moles of water breaks down into 2 moles of Hydrogen and 1 mole of Oxygenof Hydrogen and 1 mole of Oxygen

How many moles of products are formed How many moles of products are formed when we break down 4 moles of water?when we break down 4 moles of water?

Multiply everything by 2Multiply everything by 2

2[2[2H2H22OO(l)(l) → 2H→ 2H2(g)2(g) + O + O2(g)2(g)]]

4H4H22OO(l)(l) → 4H→ 4H2(g)2(g) + 2O + 2O2(g)2(g)

What about a number like 5.8 moles?What about a number like 5.8 moles?

Make it easier by dividing everything by 2 Make it easier by dividing everything by 2 so we can start with 1 mole of waterso we can start with 1 mole of water

[2H[2H22OO(l)(l) → 2H→ 2H2(g)2(g) + O + O2(g)2(g)]/2]/2

HH22OO(l)(l) → H→ H2(g)2(g) + ½ O + ½ O2(g)2(g)

Now simply multiply everything by 5.8Now simply multiply everything by 5.8

5.8[5.8[HH22OO(l)(l) → H→ H2(g)2(g) + ½ O + ½ O2(g)2(g)]]

5.8H5.8H22OO(l)(l) → 5.8H→ 5.8H2(g)2(g) + 2.9O + 2.9O2(g)2(g)

What if we want to find moles required when What if we want to find moles required when there are 2 reactants?there are 2 reactants?

CC33HH8(g)8(g) + 5O + 5O2(g)2(g) → 3CO→ 3CO2(g)2(g) + 4H + 4H22OO(g)(g)

Calculate the number of moles of oxygen Calculate the number of moles of oxygen required to react exactly with 4.3 moles of required to react exactly with 4.3 moles of propane, propane, CC33HH88, in the above reaction, in the above reaction

4.3 moles of 4.3 moles of CC33HH8 8 requires how many moles of requires how many moles of OO22

There is a 1:5 ratio There is a 1:5 ratio So 4.3(1) : 4.3(5)So 4.3(1) : 4.3(5)4.3 = 21.54.3 = 21.5

4.3 mol C4.3 mol C33HH8(g)8(g) + 21.5 O + 21.5 O2(g)2(g)

How many moles of products are formed?How many moles of products are formed?

9.3 Mass Calculations9.3 Mass Calculations

We can do the same sort of thing with the We can do the same sort of thing with the mass of the elements and compoundsmass of the elements and compounds

Let’s use same chemical equationLet’s use same chemical equation

CC33HH8(g)8(g) + 5O + 5O2(g)2(g) → 3CO→ 3CO2(g)2(g) + 4H + 4H22OO(g)(g)

Instead of moles we will use mass which is Instead of moles we will use mass which is more realisticmore realistic

If we have 44.1 g of propane, how much If we have 44.1 g of propane, how much oxygen will it react with?oxygen will it react with?

Let’s use what we know againLet’s use what we know again

We can convert grams of propane to We can convert grams of propane to moles of propanemoles of propane

1 mole of propane is 44.09 g of propane1 mole of propane is 44.09 g of propane

So 44.1 g x 1 mole / 44.09 g = 1 moleSo 44.1 g x 1 mole / 44.09 g = 1 mole

Know we need 5 moles of oxygen for Know we need 5 moles of oxygen for every mole of propaneevery mole of propane1 mol C1 mol C33HH88 x 5 mol O x 5 mol O2 2 / 1 mol C/ 1 mol C33HH88 = 5 mol O = 5 mol O22

Now how many grams of ONow how many grams of O2 2 are in 5 molare in 5 mol

5 mol x 32 g / 1 mol = 160 g5 mol x 32 g / 1 mol = 160 g

9.5 Mass Calculations: Comparing 9.5 Mass Calculations: Comparing Two ReactionsTwo Reactions

NaHCONaHCO3(s)3(s) + HCl + HCl(aq)(aq) → NaCl→ NaCl(aq)(aq) + H + H22OO(l)(l) + CO + CO2(g)2(g)

Reaction of baking soda and stomach acidReaction of baking soda and stomach acid

Mg(OH)Mg(OH)2(s)2(s) + 2HCl + 2HCl(aq)(aq) → 2H → 2H22OO(l)(l) + MgCl + MgCl2(aq)2(aq)

Reaction of Milk of Magnesia and stomach acidReaction of Milk of Magnesia and stomach acid

Looking at these 2 reactions, which antacid can Looking at these 2 reactions, which antacid can consume the most stomach acid, 1.0 g of consume the most stomach acid, 1.0 g of NaHCONaHCO33 or 1.0 g of Mg(OH) or 1.0 g of Mg(OH)2(s)2(s)??

What are we trying What are we trying to figure out?to figure out?

Want to know which Want to know which reacts with the most reacts with the most moles of HClmoles of HCl

So need to do a mole So need to do a mole to mole ratioto mole ratio

First must change First must change grams of each antacid grams of each antacid to molesto moles

Find the number of moles is in each massFind the number of moles is in each mass The molar mass of NaHCOThe molar mass of NaHCO33 is 84.01g so 1.00 g is 84.01g so 1.00 g

NaHCONaHCO33 x 1 mole / 84.01 g = 0.0119 mol x 1 mole / 84.01 g = 0.0119 mol

The molar mass of The molar mass of Mg(OH)Mg(OH)2 2 is 58.33 g so 1.00 g is 58.33 g so 1.00 g

Mg(OH)Mg(OH)2 2 x 1 mole / 58.33 g = 0.0171 molx 1 mole / 58.33 g = 0.0171 mol

Now we use the mole ratio for eachNow we use the mole ratio for each 1 mole NaHCO1 mole NaHCO33 / 1 mole HCl = 0.0119 mole / x / 1 mole HCl = 0.0119 mole / x

1 mole 1 mole Mg(OH)Mg(OH)22 / 2 mole HCl = 0.0171 mole / x / 2 mole HCl = 0.0171 mole / x

When we do the math we see thatWhen we do the math we see that 0.0119 mole NaHCO0.0119 mole NaHCO33 reacts with 0.0119 mole HCl reacts with 0.0119 mole HCl

0.0171 mole 0.0171 mole Mg(OH)Mg(OH)22 reacts with 0.0342 mole HCl reacts with 0.0342 mole HCl

So which 1 gram sample of antacid So which 1 gram sample of antacid consumes the most stomach acid?consumes the most stomach acid?

A 1 g sample of Milk of Magnesia will consume A 1 g sample of Milk of Magnesia will consume more acidmore acid

9.6 The Concept of Limiting 9.6 The Concept of Limiting ReactionsReactions

When you have When you have multiple ingredients multiple ingredients involved in producing involved in producing something, one will something, one will run out first and run out first and cause a limit on how cause a limit on how much product you can much product you can makemakeRemember the Remember the Sandwich making Sandwich making example!!example!!

CHCH4(g)4(g) + H + H22OO(g)(g) → 3H→ 3H2(g)2(g) + CO + CO(g)(g)

What mass of HWhat mass of H22OO(g)(g) is needed for 249 g of CH is needed for 249 g of CH4(g)4(g)? ?

(same steps we’ve been using)(same steps we’ve been using) Need to make sure equation is balancedNeed to make sure equation is balanced Need to convert grams CHNeed to convert grams CH4(g)4(g) to moles CH to moles CH4(g)4(g)

Need to compare moles CHNeed to compare moles CH4(g)4(g) to moles H to moles H22OO(g)(g)

Need to change moles HNeed to change moles H22OO(g)(g) to grams H to grams H22OO(g)(g)

Let’s do the ProblemLet’s do the ProblemIs Equation Balanced? Yeah!!Is Equation Balanced? Yeah!!

Convert 249 g CHConvert 249 g CH44 to moles CH to moles CH44

249g x 1mole/16g = 15.5 mol CH249g x 1mole/16g = 15.5 mol CH44

Change moles Change moles CHCH44 to moles to moles HH22OO 15.5 mol CH15.5 mol CH44 x 1 mol H x 1 mol H22O / 1 mol CHO / 1 mol CH44 = 15.5 mol H = 15.5 mol H22OO

As proportion 1 mol CHAs proportion 1 mol CH44/15.5 mol CH/15.5 mol CH44 = 1 mol H = 1 mol H22O/ ? mol HO/ ? mol H22OO

Change moles Change moles HH22OO to grams to grams HH22OO 15.5 mol H15.5 mol H22O x 18 g / 1 mol = 279 g HO x 18 g / 1 mol = 279 g H22OO

So in order for us to use up 249 g methane we need to So in order for us to use up 249 g methane we need to have 279 g waterhave 279 g water

But what if we didn’t have the right But what if we didn’t have the right amount of each?amount of each?

Well, 1 of them would be a limiting reactantWell, 1 of them would be a limiting reactant

To solve we will use the same steps as before To solve we will use the same steps as before with a little twistwith a little twist

Steps for Solving Limiting Reactant Steps for Solving Limiting Reactant ProblemsProblems

Write and Balance EquationWrite and Balance Equation

Convert Masses of Reactants to Convert Masses of Reactants to MolesMoles

Use Mole Ratios to Determine what is Use Mole Ratios to Determine what is LimitingLimiting

Use Amount of Limiting Reactant and Use Amount of Limiting Reactant and Mole Ratios to Compute the Number Mole Ratios to Compute the Number of Moles of Product of Moles of Product

Convert Moles to Grams (if needed)Convert Moles to Grams (if needed)

ProblemProblem

25,000 g Nitrogen gas and 5,000 g Hydrogen 25,000 g Nitrogen gas and 5,000 g Hydrogen gas react to form ammonia. How much gas react to form ammonia. How much ammonia is made when reaction stops?ammonia is made when reaction stops?

Step 1Step 1 NN22 + 3H + 3H22 → 2NH→ 2NH33

Step 2Step 2 25,000 g 25,000 g NN22 x 1 mole x 1 mole NN22 / 28 g / 28 g NN2 2 = 892 mol N= 892 mol N22

5,000 g H5,000 g H22 x 1 mole H x 1 mole H22/ 2.0 g H/ 2.0 g H22 = 2,500 mol H = 2,500 mol H22

Step 3 (just pick one)Step 3 (just pick one) 892 mol N892 mol N22 x 3 mol H x 3 mol H22 / 1 mol N / 1 mol N22 = 2,680 mol H = 2,680 mol H22

Or Or 2,500 mol H2,500 mol H22 x 1 mol N x 1 mol N22 / 3 mol H / 3 mol H22 = 833.33 mol N = 833.33 mol N22

This tells us we need 2,680 mol HThis tells us we need 2,680 mol H22 and 833.33 and 833.33 mol Nmol N2 2 to use up all the reactants, do we have to use up all the reactants, do we have these amounts? these amounts? No, we do not have enough HNo, we do not have enough H22 (only have 2,500 mol (only have 2,500 mol

and need 2,680 mol) so it will run out = limiting and need 2,680 mol) so it will run out = limiting reactantreactant

Step 4Step 4 2,500 mol H2,500 mol H22 x 2 mol NH x 2 mol NH33/3mol H/3mol H22 = 1,650 mol NH = 1,650 mol NH33

Step 5Step 5 1,650 mol NH1,650 mol NH33 x 17 g NH x 17 g NH33/ 1 mol NH/ 1 mol NH33= 28,100 g NH= 28,100 g NH33

Practice !!Practice !!

9.8 Percent Yield9.8 Percent Yield

Products stop forming when 1 reactant Products stop forming when 1 reactant runs outruns out

Theoretical Yield = how much you would Theoretical Yield = how much you would get in a perfect worldget in a perfect world

Actual Yield = what you actually getActual Yield = what you actually get

Percent Yield = comparing these twoPercent Yield = comparing these two Actual / theoretical x 100 = % yieldActual / theoretical x 100 = % yield