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Mass Relationships in Chemical Reactions

Chapter 3 4 - 5 Lectures Dr. Ali Bumajdad

Chapter 3 Topics Stoichiometry

•

Average Atomic Masses

•The Mole •Molar Mass (M.m.) and Molecular mass (M.w.) •Percent Composition •Calculation Empirical & Molecular Formula •Balancing Chemical Equation •Stoichiometric Calculations & Limiting Reactant •Theoretical yield, actual yield and percentage yield

Dr. Ali Bumajdad

Atomic mass & Average Atomic Masses

Micro World atoms & molecules

Amu Atomic mass

Macro World

grams Molar mass

•

Atomic mass

is the mass of an atom in atomic mass units (

amu

)

By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1.008 amu 16 O = 16.00 amu

(0)

Av. At. mass = (m of Isotope 1 x its abundance) + (m of Isotope 2 x its abundance) + …….

100

Natural lithium is: 7.42%

6 Li

(6.015 amu) 92.58%

7 Li

(7.016 amu)

Average atomic mass

of lithium:

7.42 x 6.015 + 92.58 x 7.016

100 = 6.941 amu

Dr. Ali Bumajdad

Average atomic mass (6.941)

Sa Ex 3.1:

Cu vaporized in a mass spectrometer and It was found to have two isotopes one of mass 62.93 amu and abundance 69.09% and the other of mass 64.93 amu and abundance 30.91%

Find its average mass

Av. m = 63.55 amu

•

The Mole

SI unit of amount of substance

number number

•

mole :

amount of a substance that contains

6.0221367 x 10 23

objects . •

mole :

number of carbon atoms in 12 g of 12 C 1 mol =

N A

6.0221367 x 10

Avogadro’s number (

N A )

23 number

Mass of 1 mole Number of atoms in 1 mole

12.00 grams of 12 C contains

6.022 x 10 23

1.008 grams of H contains

6.022 x 10 23 Mass of 1/2 mole Number of atoms in 1/2 mole

6.00 grams of 12 C contains

3.011 x 10 23 Dr. Ali Bumajdad

Molar Mass (M.m.) and Molecular mass (M.w.)

Molar mass

eggs is the mass of 1 mole of in grams marbles atoms 1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g 1 12 C atom = 12.00 amu 1 mole lithium atoms = 6.941 g of Li For any element atomic mass in amu = molar mass in grams/mol

C •From the periodic table 12.01 amu 12.01 g mass of 1 atom of C mass of 1mol of C mass of 6.022 x 10 23 atoms of C

12.01 g C One Mole of atoms S 32 g Hg 200.59 g 55.85 g 63.55 g Cu Fe

Molecular mass

(or molecular weight) is the sum of the atomic masses (in amu) in a molecule.

SO 2 1S 2O SO 2 32.07 amu + 2 x 16.00 amu 64.07 amu 1 molecule SO 2 1 mole SO 2 = 64.07 amu = 64.07 g SO 2

For any compound Molecular mass in amu = molar mass in grams/mol

M.w. of SO

= 64.07 amu M.w. of C

H

N

O

= 194.20 amu

One Mole of molecules

Q) 1 mol of apple contains 6.022 x 10 23 apple Q) 1 mol of CH 3 OH contains 6.022 x 10 23 CH 3 OH Q) 1 mol of CH 3 OH contains 6.022 x 10 23 O atoms Q) 1 mol of CH 3 OH contains 4 x 6.022 x 10 23 H atoms Q) 1/2 mol of CH 3 OH contains (6.022 x 10 23 )/2 CH 3 OH Q) 1/2 mol of CH 3 OH contains (4 x 6.022 x 10 23 )/2 H atoms 

N = n × N A



N = no. of dozen × 12 Dr. Ali Bumajdad

m M.m.

(1)

N N A

(2)

= number of mole

M.m.

= molar mass in g/mol

= mass

N A

= Avogadro’s number

= number of objects

A a B b No. of A a B b molecules

× a a 

No. of A atoms

× a 

6.022

×10 23 No. of A a B b molecules 6.022

×10 23

× a  Q) 1.0 ×10 3 CH 4 molecules contain:

No. of A moles

1) How many H atoms.

2) How many moles of H atoms.

Q) How many atoms are in 0.551 g of potassium (K) ?

I need

and

N A (2)

N N A n

Find

using:

M.m.

0.551 g / 39.10 g mol -1 (1) n

0.0141 mol N

(0.0141) (6.022 x 10 23 ) = 8.49 x 10 21 Expected

K atoms

Dr. Ali Bumajdad

Q) How many H atoms are in 72.5 g of C 3 H 8 O ?

Answer = 5.82 x 10 24 atoms H

Dr. Ali Bumajdad

Because

m of 1 apple in g

m of 1 dozen in g 12



m of 1 atom in g

m of 1 mol in g N A m of 1 atom in g

M.m.

N A

(3) (3)

Sa Ex. 3.2: Calculate the mass in gram of

atoms of Americium (Am) Method 1: Using Eq. 3 Method 2:

243 g X 6.022 x 10 6 atoms 23 atoms

Sa Ex. 3.3: Calculate the number of moles of atoms in 10.0 g sample of Al.

and the number Use Eq. 1 and 2

= 0.371 mol Al

= 2.23 × 10 23 Al atoms

Sa Ex. 3.4: How many Si atoms in 5.68 mg of silicon computer chip?

N = 1.22 × 10 20 Si atoms

Sa Ex. 3.5: Calculate the number of moles gram and the mass in of a sample of Co containing 5.00 ×10 20 atoms Method 1:

of 5.00 ×10 20 atoms = 4.89 ×10 -2 g

= 8.30 ×10 -4 mol Method 2:

= 8.30 ×10 -4 mol

of 5.00 ×10 20 atoms = 4.89 ×10 -2 g

Sa Ex. 3.6: (1) Calculate the molar mass of C 10 H 6 O 3 . (2) A sample of 1.56 ×10 -2 g of C 10 H 6 O 3 , how many moles does this sample represent

(1)

M.m.=

174.1g

(2) No. of moles = 8.96 × 10

-5

mol

Sa Ex. 3.7: (1) Calculate the molar mass of CaCO 3 . (2) A sample of 4.86 moles of CaCO 3 , what is the mass of the CO 3 2 ions present?

M.m. = 100.09 g/mol mass of the CO3 2 = 292 g

Sa Ex. 3.8: Bees release Isopentyl acetate (C 7 H 14 O 2 ) when sting. The amount release is about 1  g (1) (2)

How many molecules How many atoms

of C 7 H 14 O 2 of

Carbon

are released are present (3)

How many atoms

are present

(1) no. of molecules

=

5 ×10

molecules (2) no. of

C

atoms = 4 × 10

(3) no. of atoms = 1 × 10

C atoms atoms

We should use Eq.1 and Eq. 2 and …

m M.m.

(1)

N N A

(2) Number of H atoms = 1.03 ×10 24 H atoms

We should use Eq.1

m M.m.

(1)

Mass of Zn =

23.3 gram

We should use Eq.1 and Eq. 2

m M.m.

(1)

N N A

(2)

Number of S atoms =

3.06 × 10

S atoms

•

Percent Composition

A) If I know M.F.

Suppose I have A

molecule C 2 H 6

= O (

M.m.

A ) × a

x 100%

(4)

% C = (12.01 g) x 2 46.07 g x 100% = 52.14% % H = (1.008 g) x 6 46.07 g x 100% = 13.13% % O = (16.00 g) x 1 46.07 g x 100% = 34.73% 52.14% + 13.13% + 34.73% = 100.0%

Dr. Ali Bumajdad

B) If I do not know M.F. but I know the masses

Suppose I have A a B b molecule

m A +m B

x 100%

(5)

C y H x

of C =24 g

of H = 8g % C = % H = 24 g 32 g 8 g 32 g x 100% = 75 % x 100% = 25 %

Dr. Ali Bumajdad

% H = % P = % O = = 3.086% = 31.61% = 65.31%

•

Calculation Empirical & Molecular Formula A) Empirical formula from combustion of known amount 1) Find the moles of H and C from mass of H 2 O and C O 2 2) Find mass of O (or any other element) using: Mass of O = m total – (m H +m C ) 3) Find the moles of O from mass of O 4) Divide by the smallest mole value

Combust 11.5 g ethanol

22.0 g CO 2 13.5 g H 2 O

g CO 2 g H 2 O mol CO mol H g O mol O 2 O 2 mol C mol H 0.5

mol C

1.5

mol H

0.25

mol O Empirical formula C 0.5

H 1.5

O 0.25

Divide by smallest subscript (0.25) Empirical formula

C 2 H 6 O Dr. Ali Bumajdad

Q) 0.1156g f compound contains C, H and N only. The H 2 O Absorber increase by 0.1676g and the CO 2 absorber increase by 0.1638g what is the empirical formal of the compound?

The E.F. is CH

N

Dr. Ali Bumajdad

B) Empirical Formula from element masses or %mass 1) Find the moles of elements (for %mass assume you have 100 g) 2) Divide by the smallest mole value C) Molecular Formula from element masses or %mass and

M.m.

M.F.

1) Find the moles of elements (for %mass assume you have 100 g) 2) Divide by the smallest mole value. Now you know E.F.

3) Use: (6) M.F.

= E.F.

M.m.

M.F.

M.m.

E.F.

Dr. Ali Bumajdad

M.F. = N

O

M.m. = 90.02 g/mol

Sa. Ex. 3.11: Compounds consists of 71.65% Cl, 24.27%C And 4.07% H and its molar mass = 98.96 g/mol a) Determine its E.F.

b) Determine its M.F.

(Assume that we have 100 grams of compound)

a) E.F. = CH

Cl b) M.F. = C

H

Cl

Dr. Ali Bumajdad

Sa. Ex 3.12: White powder contain 43.64% P and 56.36% O. The compound molar mass = 283.88 g/mol.

a) What is E.F.

b) What is M.F.

1) Find the moles of elements (for %mass assume you have 100 g) 2) Divide by the smallest mole value E.F. = P 2 O 5 M.F. = P 4 O 10 Dr. Ali Bumajdad

Balancing Chemical Equation

Reactant Product 2 Mg + O 2 2 MgO

Coefficent

2 atoms Mg + 1 molecule O 2 makes 2 formula units MgO 2 moles Mg + 1 mole O 2 makes 2 moles MgO 1 moles Mg + 1/2 mole O 2 makes 1 moles MgO

48.6 grams Mg + 32.0 grams O 2 makes 80.6 g MgO NOT

2 grams Mg + 1 gram O 2 makes 2 g MgO

Balancing Chemical Equations

C 2 H 6 + O 2 CO 2 + H 2 O 1.Start by balancing those elements that appear in only

one

reactant and

one

product. start with C or H but

not

O C 2 H 6 + O 2 2 CO 2 + 3 H 2 O 2. Balance those elements that appear in

two or more

reactants or products. C 2 H 6 + O 2 2 2 CO 2 + 3 H 2 O 2 C 2 H 6 + 7 O 2 4 CO 2 + 6 H 2 O

Dr. Ali Bumajdad

Q) Balance: 1) 1 C 2 H 5 OH (L) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O (g) 2) Na 2 CO 3 (s) + 2 HCl (aq) 2 NaCl (s) + H 2 O (L) + CO 2 (g) 3) 1 C 6 H 6 2 (L) + 7.5

O 2 15 (g) 6 CO 2 12 (g) + 3 H 2 O (g) 6 Sa Ex. 3.14: balance (NH 4 ) 2 Cr 2 O 7 (s) Cr 2 O 3 (s) + N 2 (g) + 4 H 2 O (g) Sa Ex. 3.15: balance 2 NH 3 (g) + 5/2 O 2 (g) 2 NO (g) + 3 H 2 O (g) 4 5 4 6

Dr. Ali Bumajdad

4Al + 3O 2 2Al 2 O 3

Stoichiometric Calculations & Limiting Reactant

Calculation involving masses

( Coefficient provide moles ratios not exact amount) 1. Balanced the equation 2. Find moles of one reactant or product 3. Use coefficient ratio to calculate the moles of the required substance 4. Convert moles of required substance into masses Dr. Ali Bumajdad

Q) How many moles of C atoms are needed to combined with 4.87 mol

to form C 2 Cl 6

1) Balanced the equation

2C + 3Cl 2 C 2 Cl 6

2) Find moles of one reactant or product

2C + 3Cl 2 C 2 Cl 6 ?

2.435 mol

3) Use coefficient ratio to calculate the moles of the required substance

2 mol of C 3 mol Cl 2 X 2.435 mol of Cl 2

= 1.62 mol C Dr. Ali Bumajdad

Q) 209 g of methanol burns in air according to the equation: CH 3 OH + O 2 CO 2 + H 2 O Find mass of water formed.

1) Balanced the equation

2 CH 3 OH + 3 O 2 2 CO 2 + 4

2) Find moles of one reactant or product

H 2 O 2 CH 3 OH + 3 O 2 2 CO 2 + 4 H 2 O 209 g

= = 6.52 mol 32.04 g/mol

3) Use coefficient ratio to calculate the moles of the required substance

2 mol of CH 3 OH 4 mol H 2 O 6.52 mol of CH 3 OH X

= 13.04 mol H 2 O

4) Convert moles of required substance into masses

(1)

H 2 O =

M.m.=

13.04

×18.016

235 g H 2 O

M.m.

Dr. Ali Bumajdad

Q) How many grams of Ca must react with 83.0 g of Cl 2 to form CaCl 2

1) Balanced the equation

Ca + Cl 2 CaCl 2

2) Find moles of one reactant or product

Ca + Cl 2 83.0 g CaCl 2 83.0 g

= = 1.17 mol 70.9 g/mol

3) Use coefficient ratio to calculate the moles of the required substance

1 mol of Ca 1 mol Cl 2 X 1.17 mol of Cl 2

= 1.17 mol Ca 4) Convert moles of required substance into masses

(1)

M.m.

Ca =

M.m.=

1.17

×40.08

46.9 g Ca

Sa. Ex. 3.16: LiOH(s) + CO 2 (g) Li 2 CO 3 (s) + H 2 O(L) What is the mass of CO 2

1) Balanced the equation

require to react with 1.00 Kg of LiOH? 2LiOH(s) + CO 2 (g) Li 2 CO

2) Find moles of one reactant or product

3 (s) + H 2 O(L) 2LiOH(s) + CO 2 1.00 × 10 3 g ?

(g) Li 2 CO 3 (s) + H 2 O(L) 1.00 × 103 g

= = 41.75 mol (expected) 23.95 g/mol

3) Use coefficient ratio to calculate the moles of the required substance

2 mol of LiOH 1 mol CO 2 41.75 mol of LiOH X

= 20.88 mol CO 2 4) Convert moles of required substance into masses

(1)

CO 2 =

M.m.=

20.88

×44.0

920 g CO 2

M.m.

Dr. Ali Bumajdad

Limiting Reactant

(the

reactant

the

completely consumed

in a chemical reaction) 6 green used up 6 red left over

Dr. Ali Bumajdad

e.g. 2H 2 + O 2 2H 2 2 mol 1mol 2mol O 1 mol 1/2mol 1mol 2.1 mol 1mol 2mol limiting 1.9 mol 1mol 1.9mol

limiting  Limiting reactant determine the amount of product

Dr. Ali Bumajdad

•

Why limiting reactant is important?

Because in the stoichiometric calculation I should only use the coefficient of the limiting reactant

•

How do I know that a reaction contains a limiting reactant ?

If I've been given the masses or number of moles of two reactant then I might have limiting reactant Dr. Ali Bumajdad

Calculation involving a limiting reactant

1. Balanced the equation 2. Find moles of the two reactants 3. Identify the limiting reactant (How?) 4. Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance 5. Convert moles of required substance into masses

Q) In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe Calculate the mass of Al 2 O 3

1) Balanced the equation

2Al + Fe 2 O 3 formed.

Al 2 O 3 + 2Fe

2) Find moles of the two reactants

2Al + Fe 2 O 3 124g 601g Al 2 O 3 ?

+ 2Fe limiting

= 124 g

26.98 g/mol

= 4.60 mol

= 601 g = 3.76 mol

159.69 g/mol 3) Identify the limiting reactant (How?)

4.60

= 2.3

limiting 3.76

= 3.76

Dr. Ali Bumajdad

3) Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance

2 mol of Al 1 mol AL 2 O 3 4.60 mol of Al X

= 2.30 mol Al 2 O 3 4) Convert moles of required substance into masses

m M.m.

(1)

Al 2 O 3 =

M.m.=

2.30

×101.96

235 g Al 2 O 3 Dr. Ali Bumajdad

Sa.Ex. 3.18: 18.1 g of NH 3 NH 3 and 90.4g of CuO reacted + CuO N 2 + Cu + H 2 O 1) Which is the limiting reactant?

2) How many grams of N2 will be formed?

1) Balanced the equation

2 NH 3 + 3 CuO N 2

2) Find moles of the two reactants

+ 3 Cu + 3 H 2 O 2NH 3 18.1g

+ 3CuO N 2 90.4g

limiting

= 18.1 g

17.03 g/mol

= 1.06 mol

= 90.4 g

79.55 g/mol

= 1.14 mol

3) Identify the limiting reactant (How?)

+ 3Cu + 3H 2 O 1.06

= 0.53

1.14

= 0.38

limiting

Dr. Ali Bumajdad

3) Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance

3 mol of CuO 1 mol N 2 1.14 mol of Al X

= 0.380 mol N 2 4) Convert moles of required substance into masses

m M.m.

(1)

N 2 =

M.m.=

0.380

×28.0

10.6 g N 2 Dr. Ali Bumajdad

•

Theoretical yield, actual yield and percentage yield

•

Mass of product

•

Maximum yield

•

Calculated using stiochiometry

•

Mass of product

•

Known by experiment

•

Never more that theoretical yield % Yield

Actual Yield = Theoretical Yield

(7)

x 100

Dr. Ali Bumajdad

Sa.Ex. 3.19: 8.60 kg of H 2 H 2 and 68.5kg of CO reacted + CO CH 3 OH 1) Theoretical yield?

2) % yield if the actual yield = 3.57 × 10 4

1) Balanced the equation

g CH 3 OH 2 H 2 + CO CH

2) Find moles of the two reactants

3 OH 2 H 2 limiting 8.60 ×10 3 g + CO CH 68.5

= 8.60 ×10 3 g

2.016 g/mol

= 4.27

×10 3 mol × 10 3 g

68.5 × 10 3 g =

28.02 g/mol

3 OH = 2.44

×10 3 mol

3) Identify the limiting reactant (How?)

4.27

×10 3 = 2135

limiting 2.44

×10 3 = 2440

1 Dr. Ali Bumajdad

3) Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance

2 mol of H 2 4.27

×103 mol of H 2 1 mol CH X 3 OH

= 2135 mol CH 3 OH 4) Convert moles of required substance into masses

m M.m.

(1)

CH 3 OH =

M.m.=

2135 ×32.04

3.57 × 10 4 g

% yield = × 100 =

52.0% 6.86

×10 4 g

6.86

×10 4 g CH 3 OH Dr. Ali Bumajdad