Dealing with chemical stoichiometry

Moles of

substance

A

Moles of

substance

B

Mol/mol ratio

from balanced equation

Mass (g) of

substance A

Molecules / Atoms/FU of

substance A

Mass (g) of

substance B

Molecules /

Atoms/FU ofsubstance B

MM (g/mol) of substance

MM (g/mol) of substance

Chemical

Formula

6.022x1023

Moles of atoms/ions in

a substance

Moles of atoms/ions in

a substance

6.022x1023

Chemical

Formula

Number of

atoms/ions ina substance

Number of

atoms/ions ina substance

6.022x10236.022x1023

Steward Fall 08

Not including

volumetric stoichiometry

Chemical

Formula

Chemical

Formula

Chapter 3Chapter 3Chapter 3Chapter 3

Formulas, Equations,

and Moles: II

Amounts of Reactants and Products

How can we predict the mass of products knowing the mass of a starting reactant?

We cannot go directly between grams of reactant to grams of product. We have to use moles and the balanced

chemical equation.

Mass to mass conversions – excess reagent

MgCl2(aq) + 2 AgNO3(aq)� 2 AgCl(s) + Mg(NO3)2(aq)

If we react 1.2482 g MgCl2 with excess AgNO3, how many grams of AgCl can be produced?

Always write out and balance the chemical equation first.

Steps:

•Mass MgCl2 to moles MgCl2

•Moles of MgCl2 to moles of AgCl

•Moles of AgCl to grams AgCl

1.2482g excess ????? don’t care

Mass to mass conversions – excess reagent

Steps:

•Mass MgCl2 to moles MgCl2

•Moles of MgCl2 to moles of AgCl

•Moles of AgCl to grams AgCl

AgCl mol 1

AgCl 143.32g

MgCl mol 1

AgCl mol 2

MgCl 95.211g

MgCl mol 1MgCl g2482.1

22

22

= 3.7578 g AgCl

Mass to mass conversions

66

If we have 14.69 grams of HCl, how much NaOH would we need to completely react with HCl (with none left over)?

HCl(aq) + NaOH(aq) � H2O(l) + NaCl(aq)14.69g ????? don’t care

Steps:

•Mass HCl to moles HCl

•Moles of HCl to moles of NaOH

•Moles of NaOH to grams NaOH

Steps:

•Mass HCl to moles HCl

•Moles of HCl to moles of NaOH

•Moles of NaOH to grams NaOH

NaOH mol 1

NaOH 39.9971g

HCl mol 1

NaOH mol 1

HCl 36.4606g

HCl mol 1HCl g .6914

= 16.11g NaOH

Grams to Grams

Notice that it doesn’t matter if you are going from reactants to products or

reactants to other reactants, you still use the same steps

Limiting Reagents

What are the total number of bikes I can make? What limits the number I

can make?

Limiting Reagents

2H2 + O2 ���� 2H2O

What is the limiting reactant?

1010

Limiting Reagents

1111

2H2 + O2 ���� 2H2O

What is the limiting reactant?

Limiting Reagents

1212

2H2 + O2 ���� 2H2O

What is the limiting reactant?

Limiting Reagent Calculations

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To determine the limiting reagent, calculate the amount of products for each reactant. Two

calculations are needed.

2H2 + O2 ���� 2H2O

OH mol 4 H mol 2

OH mol 2H mol 42

2

22=

OH mol 6 O mol 1

OH mol 2O mol 32

2

22=

Limiting Reagent

(Produces the smaller

amount)

Two

calculations

CaCl2 + Na2CO3 � CaCO3 + 2NaCl

If I mix together 1.0g of each reactant (CaCl2 andNa2CO3), how much CaCO3 will I produce?

You must do 2 calculations – one for each reactant

Remember…mass to mass calculations

3

3

3

2

3

2

22 CaCO g902.0CaCO mol 1

CaCO 100.087g

CaCl mol 1

CaCO mol 1

CaCl 110.99g

CaCl mol 1CaCl g00.1=

3

3

3

32

3

32

3232 CaCO g944.0CaCO mol 1

CaCO 100.087g

CONa mol 1

CaCO mol 1

CONa 105.99g

CONa mol 1CONa g00.1=

1.00g 1.00g ????? don’t care

The answer is the smaller of the two amounts.

Now that we know the limiting reagent, we can also calculate how much excess reagent is left over

CaCl2 + Na2CO3 � CaCO3 + 2NaCl

1.00g 1.00g still don’t care0.902g

limiting excess

Do the same mass to mass calculations you did before.

32

32

32

2

32

2

22 CONa 0.955g CONa mol 1

CONa 105.99g

CaCl mol 1

CONa mol 1

CaCl 110.99g

CaCl mol 1CaCl g 1.00=

Since the reaction only used 0.955g of the 1.00g we added…

323232 CONa 0.05 g045.0CONa g955.0CONa g 00.1 ==−

How much actually reacts?

Percent Yield

The amounts calculated using stoichiometric calculations are called theoretical yields. This is the amount of

products that would be formed if the reaction is 100%

complete.

In the real world, most reactions do not go to 100%

completion. To calculate the percent yield, we use a

simple formula:

Percent yield =

Actual Yield(from experiment)

Theoretical yield(from calculations)

X 100

Yields of Chemical Reactions

So, if we perform a reaction between MgCl2and AgNO

3

(like we calculated earlier) and get 2.12 grams of AgCl...what would be our percent yield?

Percent yield =3.123 g

3.7578 gX 100

= 83.11%

Concentration of Solutions: Molarity

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Concentration: amount of solute present in a given amount of solvent or solution

M Moles of solute

Liters of solution=

Notice that in molarity (M), the liters are for the entire

solution, not just the solvent.

Molarity

Na+

Na+

Cl-

Cl-Na+ Cl-

NaCl

Dealing with chemical stoichiometry

Moles of

substance

A

Moles of

substance

B

Mol/mol ratio

from balanced equation

Mass (g) of

substance A

Molecules / Atoms of

substance A

Mass (g) of substance B

Molecules /

Atoms ofsubstance B

MM (g/mol) of substance

MM (g/mol) of substance

Chemical

Formula

6.022x1023

Moles of atoms/ions in

a substance

Moles of atoms/ions in

a substance

6.022x1023

Chemical

Formula

Number of atoms/ions in

a substance

Number of atoms/ions in

a substance

6.022x10236.022x1023

Steward Fall 08

Volume (L) of substance A

Volume (L) of substance B

Con. (mol/L) of substance cA

Conc. (mol/L) of substance B

Concentration of Solutions

• What is the concentration of a 1 L solution made by dissolving 5.0 g NaCl in water?

• 500 mL solution?

• How much NaCl (in g) do we need to make 1 L of a 0.125 M solution?

Concentration of Solutions – Mad Libs

• 1) Water-soluble ionic compound

• 2) Number greater than 100

• 3) Number less than 15

• 4) Number between 50 and 100

• 5) Any water-soluble compound

• 6) Number greater than 100

Concentration of Solutions - Mad Libs

• How many grams of (1) _______ are required to prepare a (2) _____ mL solution whose concentration is (3) _____?

• What is the concentration of a solution made by adding (4) ______ grams of (5) ______ to _______ mL of water?

Dilution

• Dilution: making a less concentrated solution (by adding water); number of moles of solute stays the same, amount of solvent increases

• The number of moles of solute stays the same but the ratio changes (lower molarity)

– moles = M * V

– Mconc * Vconc = Mdil * Vdil

– M1V1 = M2V2

• Example: What is the final concentration of a 100.0 mL solution of 3.00 M HCl if the final volume is 400.0 mL?

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Dilution

• How would you prepare 500.0 mL of a 0.500 M solution of H2SO4 starting with concentrated (18 M) solution?

– Remember to add acid to water.

• How much water do you need to add to 25.0 mL of a 4.50 M solution of NaOH to make a 1.00 M NaOH solution?

Solution Stoichiometry Example• Consider the reaction:

– 3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)

• If we mix 25.0 mL of 0.200 M CaCl2 solution with 50.0 mL of 0.250 M Na3PO4 solution, what mass of Ca3(PO4)2(s) is formed?

• Notice this is a limiting reagent problem. You need to be able to work problems of this type. Be sure to practice these!

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Problem Solution:

3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)

• 0.200 M x 0.0250 L = 0.00500 mol CaCl2

• 0.250 M x 0.0500 L = 0.0125 mol Na3PO4

• Limiting Reactant:

– 0.00500 mol CaCl2 � 0.00167 mol Ca3(PO4)2

– 0.0125 mol Na3PO4 � 0.00625 mol Na3PO4

• 0.00500 mol CaCl2 x (1 mol/3 mol) x (312 g / 1 mol) = 0.520 g Ca3(PO4)2

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Titrations

• Titration: method of determining the concentration of an unknown solution by using a solution with a known concentration (standard)

– moles = M * V

• Equivalence point: point at which acid has completely reacted with/been neutralized by base (moles acid = moles base)

• End point: point at which the indicator changes color (slight change in solution’s pH)

Titrations

• Solution stoichiometry: use molarity to convert between volume and moles (instead of molar mass)

• 23.78 mL of 0.2500 M NaOH neutralized 20.00 mL of HCl. What is the concentration of HCl?

Group Work

• How many mL of 1.018 M H2SO4 are needed to neutralize 20.00 mL of 0.9989 M NaOH?

Percent Composition of Compounds

Percent by mass: percent by mass of each element in a compound

• H2O2

– 2 moles H, 2 moles O

– H = 2 mol x 1.008 g/mol H = 2.016 g H

– O = 2 mol x 16.00 g/mol O = 32.00 g O

– H2O2 = 2.016 g H + 32.00 g O = 34.02 g

– %H = 2.016 g H / 34.02 g H2O2 x 100% = 5.926%

• What is %O? Solve 2 ways…..

Percent Composition of Compounds

• What is the percent by mass of Cr, S, and O in Cr2(SO4)3?

Empirical Formulas

• Calculate the empirical formulas:

– 50% S, 50% O

– 43.64% P, 56.36% O

• Assume 100 g of the material:

– 50 g O x 1 mol / 16.0 g = 3.125 mol O

– 50 g S x 1 mol / 32.066 g = 1.559 mol S

– mol O / mol S = 3.125 / 1.559 = 2.004 or 2

– SO2

• Assume 100 g of material:

– 43.64 g P � mol P (1.4089 mol P)

– 56.36 g O � mol O (3.5225 mol O)

– Ratio of two moles (2.5 O / 1 P); need whole numbers

3232

Molecular Formulas

• We just found the empirical formula to be P2O5. This compound has a molecular mass of 141.9 g/mol.

• We know the molecular formula has a molar mass of 283.8 g/mol. How many multiples of P2O5 do we need to reach a mass of 283.8?

Practice Problem

• A compound contains 30.4 % nitrogen and 69.6 % oxygen. The molecular mass of the compound is 92 g/mol.

– What is the empirical formula of the compound?

– What is the molecular formula of the compound?

3434

Answers to Practice Problem

• Assume a sample of 100 g, which then contains 30.4 g N and 69.6 g O.

• 30.4 g N x (1 mol / 14 g) = 2.17 mol N

• 69.6 g O x (1 mol / 16.0 g) = 4.35 mol O

• ratio = 4.35 mol O / 2.17 mol N = 2.00

• Empirical formula = NO2

• Empirical mass = 46 g/mol

• Molecular mass / empirical mass = 92 / 46 = 2

• Molecular formula = (NO2)2 = N2O4

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