Mass Relationships in Chemical Reactions Chapter 3 4 - 5 Lectures Dr. Ali Bumajdad.

Mass Relationships in Chemical Reactions

Chapter 34 - 5 Lectures

Dr. Ali Bumajdad

Chapter 3 Topics

Dr. Ali Bumajdad

• 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

Stoichiometry

By definition: 1 atom 12C “weighs” 12 amu

On this scale

1H = 1.008 amu

16O = 16.00 amu

•Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

AmuAtomic mass

Macro Worldgrams

Molar mass

Atomic mass & Average Atomic Masses

Natural lithium is:

7.42% 6Li (6.015 amu)

92.58% 7Li (7.016 amu)

7.42 x 6.015 + 92.58 x 7.016100

= 6.941 amu

Average atomic mass of lithium:

Dr. Ali Bumajdad

Av. At. mass = (m of Isotope 1 x its abundance) + (m of Isotope 2 x its abundance) + …….100

(0)

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 amuand abundance 30.91%

Find its average mass.

Av. m = 63.55 amu

•mole : amount of a substance that contains 6.0221367 x 1023 objects.

•mole : number of carbon atoms in 12 g of 12C1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA)

number

number

number

•The Mole SI unit of amount of substance

12.00 grams of 12C contains 6.022 x 1023

1.008 grams of H contains 6.022 x 1023

Mass of 1 mole

6.00 grams of 12C contains 3.011 x 1023

Number of atoms in 1 mole

Mass of 1/2 mole Number of atoms in 1/2 mole

Dr. Ali Bumajdad

Molar mass is the mass of 1 mole of in gramseggsshoes

marblesatoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 12C 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

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

C 12.01 amu mass of 1 atom of C12.01 g mass of 1mol of C

mass of 6.022 x 1023 atoms of C

•From the periodic table

One Mole of atoms

C S

Cu Fe

Hg

32 g

55.85 g

200.59 g

12.01 g

63.55 g

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu

2O + 2 x 16.00 amu SO2 64.07 amu

1 molecule SO2 = 64.07 amu

1 mole SO2 = 64.07 g SO2

For any compound

Molecular mass in amu = molar mass in grams/mol

M.w. of SO2 = 64.07 amu

M.w. of C8H10N4O2 = 194.20 amu

One Mole of molecules

Q) 1 mol of apple contains 6.022 x 1023 apple

Q) 1 mol of CH3OH contains 6.022 x 1023 CH3OH

Q) 1 mol of CH3OH contains 6.022 x 1023 O atoms

Q) 1 mol of CH3OH contains 4 x 6.022 x 1023 H atoms

Dr. Ali Bumajdad

Q) 1/2 mol of CH3OH contains (4 x 6.022 x 1023)/2 H atoms

Q) 1/2 mol of CH3OH contains (6.022 x 1023)/2 CH3OH

N = n × NA

N = no. of dozen × 12

M.m. = molar mass in g/mol

NA = Avogadro’s number

N = number of objects

n = number of mole

m = mass

M.m. n =

m(1)

NA n =

N(2)

No. of A atomsNo. of AaBb molecules

No. of AaBb molecules

× a

a

AaBb

× a 6.022×1023

6.022×1023 × a No. of A moles

Q) 1.0 ×103 CH4 molecules contain:

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 n and NA

Find n using:

n = 0.551 g / 39.10 g mol-1

n = 0.0141 mol

N = (0.0141) (6.022 x 1023) = 8.49 x 1021 K atoms

NA n =

N(2)

Expected Dr. Ali Bumajdad

M.m. n =

m(1)

Q) How many H atoms are in 72.5 g of C3H8O ?

Answer = 5.82 x 1024 atoms H

Dr. Ali Bumajdad

NA m of 1 atom in g =

m of 1 mol in g(3)

12 m of 1 apple in g =

m of 1 dozen in g

Because

NA m of 1 atom in g =

M.m.(3)

Sa Ex. 3.2: Calculate the mass in gram of 6 atoms of Americium (Am)

Method 1: Using Eq. 3

Method 2:

243 g 6.022 x 1023 atoms

X 6 atoms

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

Use Eq. 1 and 2

n = 0.371 mol Al

N = 2.23 × 1023 Al atoms

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

N = 1.22 × 1020 Si atoms

Sa Ex. 3.5: Calculate the number of moles and the mass in gram of a sample of Co containing 5.00 ×1020 atoms

m of 5.00 ×1020 atoms = 4.89 ×10-2g

n = 8.30 ×10-4 mol

Method 1:

Method 2:n = 8.30 ×10-4 mol

m of 5.00 ×1020 atoms = 4.89 ×10-2g

Sa Ex. 3.6: (1) Calculate the molar mass of C10H6O3. (2) A sample of 1.56 ×10-2g of C10H6O3,

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 CaCO3. (2) A sample of 4.86 moles of CaCO3,

what is the mass of the CO32- ions present?

M.m. = 100.09 g/mol

mass of the CO32- = 292 g

Sa Ex. 3.8: Bees release Isopentyl acetate (C7H14O2) when sting. The amount release is about 1 g (1) How many molecules of C7H14O2 are released (2) How many atoms of Carbon are present (3) How many atoms are present

(1) no. of molecules= 5×1015 molecules(2) no. of C atoms = 4× 1016 C atoms(3) no. of atoms = 1× 1017 atoms

M.m. n =

m(1)

NA n =

N(2)

We should use Eq.1 and Eq. 2 and …

Number of H atoms = 1.03 ×1024 H atoms

Mass of Zn = 23.3 gram

M.m. n =

m(1)

We should use Eq.1

We should use Eq.1 and Eq. 2

NA n =

N(2)

Number of S atoms = 3.06 × 1023 S atoms

M.m. n =

m(1)

Ligh

t

Ligh

t

Hea

vy

Hea

vy

•Percent Composition

%C =(12.01 g) x 2

46.07 gx 100% = 52.14%

%H =(1.008 g) x 6

46.07 gx 100% = 13.13%

%O =(16.00 g) x 1

46.07 gx 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

Dr. Ali Bumajdad

A) If I know M.F.

C2H6O

Suppose I have AaBb molecule

(4)% A =

(M.m.A) × a

M.m. AaBb

x 100%

Suppose I have AaBb molecule

Dr. Ali Bumajdad

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

(5)% A =

m A

mA +mB

x 100%

CyHx

m of C =24 g m of H = 8g

%C =24 g

32 gx 100% = 75 %

%H =8 g

32 gx 100% = 25 %

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

1) Find the moles of H and C from mass of H2O and CO2

2) Find mass of O (or any other element) using:

Mass of O = mtotal – (mH+mC)

3) Find the moles of O from mass of O

4) Divide by the smallest mole value

A) Empirical formula from combustion of known amount

•Calculation Empirical & Molecular Formula

Combust 11.5 g ethanol

13.5 g H2O

0.5 mol C

1.5 mol H

0.25 mol O

Empirical formula C0.5H1.5O0.25

Divide by smallest subscript (0.25)

Empirical formula C2H6O

g CO2 mol CO2 mol C

g H2O mol H2O mol H

g O mol O

22.0 g CO2

Dr. Ali Bumajdad

Q) 0.1156g f compound contains C, H and N only. The H2OAbsorber increase by 0.1676g and the CO2 absorber increase by 0.1638g what is the empirical formal of the compound?

The E.F. is CH5N

Dr. Ali Bumajdad

C) Molecular Formula from element masses or %mass and M.m.M.F.

M.F. = E.F. ×M.m. E.F.

M.m. M.F.(6)

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:

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

Dr. Ali Bumajdad

M.F. = N2O4

M.m. = 90.02 g/mol

Sa. Ex. 3.11: Compounds consists of 71.65% Cl, 24.27%CAnd 4.07% H and its molar mass = 98.96 g/mola) Determine its E.F.b) Determine its M.F.

(Assume that we have 100 grams of compound)

a)E.F. = CH2Clb)M.F. = C2H4Cl2

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.

E.F. = P2O5

M.F. = P4O10

Dr. Ali Bumajdad

1) Find the moles of elements (for %mass assume you have 100 g)

2) Divide by the smallest mole value

Balancing Chemical Equation

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

1 moles Mg + 1/2 mole O2 makes 1 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgONOT

2 grams Mg + 1 gram O2 makes 2 g MgO

Reactant Product

Coefficent

Balancing Chemical EquationsC2H6 + O2 CO2 + H2O

1.Start by balancing those elements that appear in only one reactant and one product.

start with C or H but not O

C2H6 + O2 2CO2 + 3H2O

2. Balance those elements that appear in two or more reactants or products.

C2H6 + O2 2CO2 + 3H2O72

2C2H6 + 7O2 4CO2 + 6H2O

Dr. Ali Bumajdad

Q) Balance:

1) 1C2H5OH (L) + 3O2 (g) 2CO2 (g) + 3H2O (g)

2) Na2CO3 (s) + 2HCl (aq) 2NaCl (s) + H2O (L) + CO2 (g)

3) 1C6H6 (L) + 7.5 O2 (g) 6 CO2 (g) + 3H2O (g) 2 15 12 6

Sa Ex. 3.14: balance

(NH4)2Cr2O7 (s) Cr2O3 (s) + N2(g) + 4 H2O (g)

Sa Ex. 3.15: balance

2NH3 (g) + 5/2O2 (g) 2NO (g) + 3H2O (g) 4 5 4 6 Dr. Ali Bumajdad

4Al + 3O2 2Al2O3

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

Calculation involving masses

Stoichiometric Calculations & Limiting Reactant

( Coefficient provide moles ratios not exact amount)

Dr. Ali Bumajdad

Q) How many moles of C atoms are needed to combined with 4.87 mol Cl to form C2Cl6

1) Balanced the equation

2C + 3Cl2 C2Cl6

2) Find moles of one reactant or product

2C + 3Cl2 C2Cl6

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

2 mol of C 3 mol Cl2 X 2.435 mol of Cl2

= 1.62 mol CDr. Ali Bumajdad

Q) 209 g of methanol burns in air according to the equation:

Find mass of water formed.

CH3OH + O2 CO2 + H2O

m H2O =n ×M.m.=13.04×18.016 = 235 g H2O

1) Balanced the equation

2) Find moles of one reactant or product

Dr. Ali Bumajdad

2CH3OH + 3O2 2CO2 + 4H2O

2CH3OH + 3O2 2CO2 + 4H2O

209 g

32.04 g/moln = = 6.52 mol

4) Convert moles of required substance into masses

M.m. n =

m (1)

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

2 mol of CH3OH 4 mol H2O6.52 mol of CH3OH X = 13.04 mol H2O

Q) How many grams of Ca must react with 83.0 g of Cl2 to form CaCl2

1) Balanced the equationCa + Cl2 CaCl2

2) Find moles of one reactant or product

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

83.0 g?

Ca + Cl2 CaCl2

83.0 g

70.9 g/moln = = 1.17 mol

4) Convert moles of required substance into masses

m Ca =n ×M.m.= 1.17×40.08 = 46.9 g CaM.m. n =

m (1)

= 1.17 mol Ca

1 mol of Ca 1 mol Cl2 X 1.17 mol of Cl2

Sa. Ex. 3.16: LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L)What is the mass of CO2 require to react with 1.00 Kg of LiOH?

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

m CO2 =n ×M.m.= 20.88×44.0 = 920 g CO2

M.m. n =

m (1)

2LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L)

1.00 × 103 g ?

1.00 × 103 g

23.95 g/moln = = 41.75 mol (expected)

2LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L)

2 mol of LiOH 1 mol CO2

41.75 mol of LiOH X = 20.88 mol CO2

Dr. Ali Bumajdad

6 green used up

6 red left over

Limiting Reactant

Dr. Ali Bumajdad

(the reactant the completely consumed in a chemical reaction)

2 mol 1mol 2mol

1 mol 1/2mol 1mol

2.1 mol 1mol 2mol

1.9 mol 1mol 1.9mollimiting

limiting

Limiting reactant determine the amount of product

2H2 + O2 2H2Oe.g.

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

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

Calculation involving a limiting reactant

Q) In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.1) Balanced the equation

2) Find moles of the two reactants

2Al + Fe2O3 Al2O3 + 2Fe

Dr. Ali Bumajdad

3) Identify the limiting reactant (How?)

4.60

2= 2.3

3.76

1= 3.76

limiting

2Al + Fe2O3 Al2O3 + 2Fe124g ?

124 g

26.98 g/moln = = 4.60 mol

601g

601 g

159.69 g/moln = = 3.76 mol

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

4) Convert moles of required substance into masses

m Al2O3 =n ×M.m.= 2.30×101.96 = 235 g Al2O3M.m. n =

m (1)

= 2.30 mol Al2O3

2 mol of Al 1 mol AL2O3

4.60 mol of Al X

Sa.Ex. 3.18: 18.1 g of NH3 and 90.4g of CuO reacted NH3 + CuO N2 + Cu + H2O

1) Which is the limiting reactant?2) How many grams of N2 will be formed?

2NH3 + 3CuO N2 + 3Cu + 3H2O1) Balanced the equation

2) Find moles of the two reactants

Dr. Ali Bumajdad

3) Identify the limiting reactant (How?)

1.06

2= 0.53

1.14

3= 0.38

limiting

limiting18.1g ?

18.1 g

17.03 g/moln = = 1.06 mol

90.4g

90.4 g

79.55 g/moln = = 1.14 mol

2NH3 + 3CuO N2 + 3Cu + 3H2O

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

= 0.380 mol N2

4) Convert moles of required substance into masses

m N2 =n ×M.m.= 0.380×28.0 = 10.6 g N2M.m. n =

m (1)

3 mol of CuO 1 mol N2

1.14 mol of Al X

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 Yieldx 100

(7)

Sa.Ex. 3.19: 8.60 kg of H2 and 68.5kg of CO reacted H2 + CO CH3OH

1) Theoretical yield?2) % yield if the actual yield = 3.57 × 104 g CH3OH

1) Balanced the equation

2) Find moles of the two reactants

Dr. Ali Bumajdad

3) Identify the limiting reactant (How?)

4.27×103

2= 2135

2.44×103

1= 2440

limiting

limiting

2H2 + CO CH3OH

8.60 ×103g ?

8.60 ×103g

2.016 g/moln = = 4.27×103 mol

68.5 × 103g

68.5 × 103g28.02 g/mol

n = = 2.44×103 mol

2H2 + CO CH3OH

3) Use the coefficient ratio between the limiting reactant

and the required substance to find out the number of mole

of the required substance

= 2135 mol CH3OH

4) Convert moles of required substance into masses

m CH3OH =n ×M.m.= 2135×32.04 = 6.86×104 g CH3OH

M.m. n =

m (1)

2 mol of H2 1 mol CH3OH4.27×103 mol of H2 X

% yield = 6.86×104 g

3.57 × 104 g= 52.0%× 100

Dr. Ali Bumajdad