iv stoichiometry. stoichiometry the relationship (mole ratio) between elements in a compound and...

IV Stoichiometry

Stoichiometry

• The relationship (mole ratio) between elements in a compound and between elements and compounds in a reaction

e.g. H2O

means: 1 molecule H2O contains 2 atom H and 1 atom

O

and 1 mole H2O contains 2 moles H atoms and 1 mole O atoms

Chemical Compounds

• Combination of elements– A compound is made up of specific

elements in a specific ratio - Law of Constant Composition

• Chemical Formula– Written representation of a chemical

compound. So, a specific compound has a specific formula

Terms

• Formula Unit– Involves the lowest subscript which

describes the ratio.

– e.g. H2O; CH; CH2; CH4

– May or may not actually exist

• Formula Weight– The mass of the formula unit

• Empirical Formula– Written representation of the formula unit

Terms continued

• Molecule– Integral multiple of the formula unit (integer

may be 1) that actually exists

– e.g. C2H2; C2H4; C2H6

• Molecular Weight (Molar Mass)– Mass of the molecule

• Molecular Formula– Written representation of the molecule

• Mass to mole conversions

• Stoichiometry is in mole ratios

• Most measurements are made in grams

• So, you need to be able to get from grams to moles and moles to grams

• The atomic weight listed on the periodic table is listed without units. Why?

• Units depend on what you want. • If you are looking on the atomic scale,

atoms or molecules, units are amu.• If you are working on the macroscopic

scale, moles of material, units are in grams.

Convert 34.0 grams NH3 to moles.

1) Determine M of NH3.M = AW N + 3(AW H)

= 14.0 + 3(1.01)= 17.0

2) Determine the moles of NH3

34.0gNH3

1molNH3

17.0gNH3

2.00 molNH3

How many molecules in 32.0 g of oxygen gas.

A) 2.0

B) 1.0

C) 0.5

D) 6.02 x 1023

E) 1.20 x 1024

How many molecules in 32.0 g of oxygen gas.

Oxygen occurs as O2 gas.

AW of O = 16.00, so O2 = 32.00

32.0gO2

1molO2

32.00g

6.022x1023 molecules

1mol

6.02 x1023 moleculesO2

• % Composition (% weight/ weight; %w/w)

%composition grams of element

grams of compoundX100

1. Calculate the % composition form the Molecular Formula (or the Empirical Formula)

What is the % composition by weight of C2H4O2? C = 12.01; H = 1.01; O = 16.00

Step 1: Find the molecular weight of the compound.

MW = 2(AW C) + 4(AW H) + 2(AW O)

= 2(12.01) + 4(1.01) + 2(16.00)

= 60.06 g/ mol

Step 2: Find the % of each element.

%CgramsC

MWcpdx100

2(AW C)

60.06x100

2(12.01)60.06

x100= 39.99%

%HgramsH

MWcpdx100

4(AW H)

60.06x100

4(1.01)

60.06x100=6.73%

%OgramsO

MWcpdx100

2(AW O)

60.06x100

2(16.00)60.06

x100=53.28%

or: %O100 %C %H

%O100 39.99 6.73=53.28%

• Find the Chemical Formula from the % composition

• Which Chemical formula will you get?– Empirical Formula, to get the molecular

formula you would need more information than just the % composition

A compound containing only carbon, hydrogen, and oxygen was found to contain 62.02 % C and 10.42% H. What is the formula of the compound?

C = 12.01 H = 1.01 O = 16.00

1) Find the amount of O

%O = 100 - %C - %H

= 100 - 62.02 - 10.42

= 27.57 % O

2) Determine the moles of each element

(assume a 100 g sample)

molCgramsC

MW C

62.02gC

12.01g/mol5.163molC

molHgramsH

MWH

10.42gH

1.01g/mol10.3168molH

molOgramsO

MWO

27.57gO

16.00g/mol1.723molO

3) Divide by the smallest number to get whole number ratio

Mol C = 5.163 mol

Mol H = 10.3168 mol

Mol O = 1.723 mol

/ 1.723 = 2.996 = 3

/ 1.723 = 5.9877 = 6

/ 1.723 = 1

So, the compound has the ratio 3C:6H:1 O

And the Empirical formula: C3H6O

A further analysis of the compound found that the molar mass was 115.99 g/mol. What is the molecular formula of the compound?

Molecular formula =

(empirical formula)(# of formula units)

You can find the # of formula units from

MW/ FW

Formula weight = weight of empirical formula: C3H6O

FW = 3(12.01) + 6(1.01) + 16.00 = 58.048

# formula units =

So the molecular formula contains 2 formula units. Multiply the subscripts by 2

2(C3H6O) = C6H12O2

115.99g

58.048g1.9982

A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound?

U = 238.0 O = 15.9994

A) U2O5

B) U3O8

C) UO3

D) UO2

E) U3O

A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound?

U = 238.0 O = 15.9994

1. Determine the amount of O

100 - 84.80 = 15.20

2. Determine moles and mole ratio

molU84.80gU

238.0g/mol0.3563molU

molO15.20gO

15.9994 g/mol0.9500 molO

/ 0.3563 = 1

/ 0.3563 = 2.666

0.666 = 2/3 so O is 2 2/3 = 8/3 U1O8/3

Clear denominator, multiply by 3 U3O8

3) Determine an unknown element (X) from the % composition

A compound XO2 is 78.8% X. What element is X? O = 16.0

A) Ni

B) Co

C) P

D) Sn

3) Determine and unknown element (X) from the % composition

A compound XO2 is 78.8% X. What element is X? O = 16.0

% O = 100 - 78.8 = 21.2

21.2gO

1molO

16.0g

1.325 molO

1.325 molO

1molX

2 molO

0.6625 molX

Since Compound is 78.8% X, 0.6625 mol X = 78.8 g X

78.8g X

0.6625molX

? g X

1molX

so :?118.9 Go to Periodic Table

118.9 = Sn

Reactions

• Chemical Reaction is represented by a Chemical Equation

• General Form:aA + bB cC + dD

A/B are ?

Reactants

C/D are ?

Products

a,b,c,d are ? Stoichiometric coefficients

• Law of Conservation of Mass says?– No mass lost or gained

• Total mass of reactants = total mass of products

• Elements are re - arranged not changed

• This allows us to “Balance” Equations

• The number and kinds of atoms in the reactants have to show up as the same number and kind of atoms in the product

Ca + H2O Ca(OH)2 + H2

1 Ca + H2O 1 Ca(OH)2 + H2

1Ca + 2H2O 1 Ca(OH)2 + H2

1 Ca + 2 H2O 1Ca(OH)2 + 1 H2

Ca + 2 H2O Ca(OH)2 + H2

C3H7OH + O2 CO2 + H2O

C3H7OH + O2 3 CO2 + H2O

C3H7OH + O2 3 CO2 + 4 H2O

C3H7OH + 9/2 O2 3 CO2 + 4 H2O

2 C3H7OH + 9 O2 6 CO2 + 8 H2O

SiF4 + H2O HF + SiO2

SiF4 + H2O 4 HF + SiO2

SiF4 + 2 H2O 4 HF + SiO2

When the reaction:

C2H8N2 + N2O4 N2 + H2O + CO2

Is balanced using the smallest whole numbers, what is the coefficient of N2?

• 1

• 2

• 3

• 4

• 5

When the reaction:

C2H8N2 + N2O4 N2 + H2O + CO2

Is balanced using the smallest whole numbers, what is the coefficient of N2?

C2H8N2 + N2O4 N2 + H2O + CO2242 3

• Hydrates

• A compound (solid) that contains intact water molecules as part of the compound.

• The water can be removed by heating to leave an anhydrous residue (solid).

• Water can then be re - added to the anhydrate to yield the original hydrate

CaSO4 • 2H2OCalcium sulfate dihydrateEach mole of compound contains:1 mole calcium sulfate and 2 mole water

or1 mole Ca 1 mole S6 mol O4 mol H

CaSO4 • 2H2O(s) CaSO4(s) + 2 H2O(g)

172 136 + 2(18)

CaSO4(s) CaSO4 • 2H2O(s)

heat

Water vapor

17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left?

A) 13.4

B) 13.44

C) 3.56

D) 15.0

E) 15.03

17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left?

17.0gCaSO4 2H2O

1molCaSO4 2H2O

172.0g

1molCaSO4

1molCaSO4 2H2O

136.0gCaSO4

1molCaSO4

13.44g

13.4 g

What happened to the difference (17.0 - 13.4) = 3.6 grams of material?

Lost as water vapor

2. 15.00 grams of the hydrate

Na2SO4 • XH2O(s) was heated to remove the water. After heating, 7.95 grams of material remained. What is the formula of the hydrate: (find the value of X)

A) Na2SO4 • H2O

B) Na2SO4 • 2H2O

C) Na2SO4 • 3H2O

D) Na2SO4 • 5H2O

E) Na2SO4 • 7H2O

Na2SO4 • XH2O(s) Na2SO4(s) + XH2O 15.00 7.95 15.00-7.95

7.05Find moles of both products and compare

molNa 2SO4 (7.95g) 1mol

142g

0.0559 mol

molH2O(7.05g) 1mol

18.0g

0.391mol

/ 0.0559 = 1

/ 0.0559 = 6.99 = 7

So, X = 7, formula = Na2SO4 • 7H2O

Limiting Reactant

• Limiting Reactant is that element or compound that determines the amount of product that you get.

• It is the reactant that is used up

You are the owner of a bike shop. A shipment came in with 183 frames, 150 seats, 252 pedals, 131 brake assemblies. How many bikes can you sell? (enter the number)

Each bike needs, 1 frame, 1 seat, 2 pedals, 1 brake.

Given: 183 frames

150 seats

252 pedals

131 brake assemblies

Pedals will allow you to make only 126 bikes so that is the limiting reactant

NH3 + O2 N2 + H2O

Balance

NH3 + O2 N2 + H2O

NH3 + O2 N2 + H2O

2 33/2

4 63 2

4 63 2 NH3 + O2 N2 + H2O

Remember: Stoichiometry is mole ratios

How many moles of N2 can be formed from 4 mol NH3 and 4 mol O2?

1. Determine the LR.

4 molNH3

3molO2

4 molNH3

3molO2 required

Mol O2 given > mol O2 required. So, NH3 is LR

2. Determine amount of product

4 molNH3

2 molN2

4 molNH3

2 molN2 produced

4 63 2 NH3 + O2 N2 + H2O

How many moles of N2 can be formed from 6 mol NH3 and 4 mol O2?

1. Determine the LR.

6 molNH3

3molO2

4 molNH3

4.5 molO2 required

Mol O2 given < mol O2 required. So, O2 is LR

2. Determine amount of product

4 molO2

2 molN2

3molO2

2.67 molN2 produced

4 63 2 NH3 + O2 N2 + H2O

How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2?

A) 5B) 4C) 3.75D) 2.67E) 2.5

4 63 2 NH3 + O2 N2 + H2O

How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2?

1. Determine the LR.

5 molNH3

3molO2

4 molNH3

3.75 molO2 required

Mol O2 given > mol O2 required. So, NH3 is LR

2. Determine amount of product

5 molNH3

2 molN2

4 molNH3

2.5 molN2 produced

If 10.0 grams each of NH3 and O2 are reacted, how many grams of water and N2 are formed?

1. Find moles of each reactant.

2. Determine the Limiting Reactant Determine mole of O2 needed

Compare to what was given: 0.442 mol required > 0.312 mol given

So, O2 = LR

4 63 2 NH3 + O2 N2 + H2O

10.0gNH3

1molNH3

17.0g

0.589 molNH3

10.0gO2

1molO2

32.0g

0.312 molO2

0.589molNH3

3molO2

4 mol NH3

0.442 molO2 required

3. Determine the amount of product based on the LR

total mass of products = 17.0 g

What happened to conservation of mass?

3.0 g un-reacted NH3

0.312molO2

6 molH2O

3 mol O2

18.0 gH2O

1 mol H2O

11.2 gH2O

0.312 molO2

2 molN2

3 mol O2

28.0 gN2

1 mol N2

5.8gN2

10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed?

Cr = 52.0 S = 32.0

A) 20.9 g

B) 20.0 g

C) 19.2 g

D) 3.12 g

E) 0.80 g

10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed?

Cr = 52.0 S = 32.0

1. Write and balance the equation:

2Cr + 3S --> Cr2S3

2. Determine moles of each

10.0gCr

1molCr

52.0g

0.192 molCr

10.0gS

1molS

32.0g

0.313molS

3. Determine LR

4. Determine the amount of product based on the LR

0.192 molCr

3molS

2 mol Cr

0.288 molS required

0.288 molS required < 0.313 molS given

so : Cr LR

0.192 mol Cr

1molCr2S3

2 mol Cr

200.0gCr2S3

1molCr2S3

19.2gCr2S3

• Determination of an unknown element

• A metal oxide has the formula XO3 and reacts with H2 to form free metal X and H2O. If 15.99 grams XO3 yields 6.00 grams H2O, what element is X?

A) Nd

B) Ti

C) S

D) Mo

E) H

• Determination of an unknown element

• A metal oxide has the formula XO3 and reacts with H2 to form free metal X and H2O. If 15.99 grams XO3 yields 6.00 grams H2O, what element is X?

XO3 + H2 --> X + H2O3 3

Find moles XO3 from stoichiometry and moles of H2O

6.00gH2O

1molH2O

18.0g

1molXO3

3molH2O

0.111 molXO3

Determine MW XO3

15.99g XO3

0.111mol

? g

1mol

?143.9

AW X = 143.9 - 3(16.0) = 95.9 = MO

Theoretical Yield

• Theoretical Yield– Maximum amount of product that can

obtained if all the reactant is converted to product

• Actual Yield– Actual amount of product obtained

• % Yield

%Yield ActualYield

TheoreticalYieldx100

%Yield AY

TYx100

One hundred grams of potassium chlorate was heated. What is the final state of affairs?

K = 39.1 Cl = 35.45 O = 16.00

One hundred grams of potassium chlorate was heated. What is the final state of affairs?

K = 39.1 Cl = 35.45 O = 16.00

1. Determine the formula for potassium chlorate

A. KClO B. KClO2 C. KClO3 D. KClO4

One hundred grams of potassium chlorate was heated. What is the final state of affairs?

K = 39.1 Cl = 35.45 O = 16.00

1. Determine the formula for potassium chlorate

A. KClO B. KClO2 C. KClO3 D. KClO4

2. Write and Balance the reaction

3. Determine Theoretical Yields of Products

KClO3 --> KCl + O2 2 2 3

KClO3 --> KCl + O2 2 2 3

100. grams Convert to moles-mole ratio-convert to g

100.g KClO3 1molKClO3

122.5g

3molO2

2 molKClO3

32.0gO2

1molO2

39.2gO2

100.g KClO3 1molKClO3

122.5g

2 molKCl

2 molKClO3

74.5 g KCl

1molKCl

60.8g KCl

Or: Conservation of mass: 100.0 g - 39.2 g = 60.8 g KCl

Part B. If the reaction only produced 50.37 g KCl, 1) What is the %Yield and 2) how many grams of O2 2) were produced.

1) %Yield AY

TYx100

%Yield 50.37

60.8x100 =82.8%

2) gO2 %Yield

100x TY

gO2 82.8%

100x 39.2= 32.4 gO2

Consecutive Reactions

• Consecutive Reactions– Sequence of reactions (steps) that are

required to reach desired products

FeS2 + O2 --> Fe2O3 + SO2

SO2 + O2 --> SO3

SO3 + H2O --> H2SO4

FeS2 + O2 --> Fe2O3 + SO2

SO2 + O2 --> SO3

SO3 + H2O --> H2SO4

4 11 2 8

2 2

4 11 2 8

8 8

4(

4

8(

8 8 8

4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4

Now you can have any kind of problem

4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4

How many moles of sulfuric acid can be made from 5 mol FeS2 and 17 mol O2?

Have LR problem.

1) Determine the LR

5 molFeS2 15 molO2

4 molFeS2

18.75 molO2 required

Mol O2 given < mol O2 required so O2 LR

2) Determine the amount of product

17 molO2 8 molH2SO4

15 molO2

9.1molH2SO4

Combustion Analysis

• Combustion analysis– Reaction of a compound that contains

carbon, hydrogen and sometimes oxygen burned in air (O2) to produce CO2 and H2O

• Can be used to determine empirical formula and % composition

A 0.3000 gram sample containing only carbon, hydrogen and oxygen was burned in air to produce 0.440 grams CO2 and 0.180 grams H2O. Determine the empirical formula of the sample.

CxHyOz + O2 --> CO2 + H2Ox

y

2?

Determine moles of C and H from moles of CO2 and H2O

molC0.440gCO2 1molCO2

44.0g

1molC

1molCO2

0.01molC

molH0.180gH2O 1molH2O

18.0g

2 molH

1molH2O

0.02molH

Determine moles of O from g of O in sampleg O = g sample - g C - g Hg O = 0.3000 -(0.01 mol x 12.0 g/mol) - (0.02 mol x 1.01g/mol)g O = 0.16 g mol O = 0.16g/ 16.0 g/mol = 0.01 mol

Mol C = 0.01

Mol H = 0.02

Mol O = 0.01

/0.01 = 1

/0.01 = 2

/0.01 = 1

Empirical formula = CH2O

• A) impossible

• B) Very hard

• C) able to be done

• D) un understandable

• E) hard but workable

How will the exam on Thursday be?