Chapter 3 - Chemical Calculations

8/10/2019 Chapter 3 - Chemical Calculations

1/40

Stoichiometry

Atomic Mass

The Mole concept

Molar Mass Percent Composition of Compounds

Determination of Formula of Compounds

Writing and Balancing Chemical Equations Interpreting balance equations, and

Reaction Stoichiometry and Calculations


2/40

Atomic Masses

Absolute massesof atoms cannot be obtained

too small to measure the mass directly;

Relative atomic massesare used insteadmasses

relative to a chosen standard or reference. Carbon-12 is used as atomic mass referenceit is

assigned an atomic mass of 12 uexactly;

Other atoms are assigned masses relative to that of

carbon-12;

Relative atomic massesare determined using mass

spectrometer;


3/40

A Schematic Diagram of Mass

Spectrophotometer


4/40

Isotope Mass of CO2


5/40

Mass Spectrum of Chlorine


6/40

Atomic Mass Spectrum of Mercury


7/40

Calculation of Relative Atomic Masses

Example-1:

An atomic mass spectrum gives atomic mass ratio of

oxygen atom to carbon-12 as 1.3329:1. If the atomic

mass of carbon-12 is exactly 12 u, what is the atomic

mass of oxygen?

Atomic mass of oxygen = 1.3329 x 12 u

= 15.995 u


8/40

Calculation of Average Atomic Masses

Example-2:

Chlorine is composed of two stable naturally occurring

isotopes: chlorine-35 (75.76%; 34.9689 u) and

chlorine-37 (24.24%; 36.9659 u). What is the average

atomic mass of copper?

Atomic mass of chlorine

= (0.7576 x 34.9689 u) + (0.2424 x 36.9659 u)

= 35.45 u(as given in the periodic table)


9/40

Calculation of Average Atomic Masses

Example-3:

Copper is composed of two naturally occurring

isotopes: copper-63 (69.09%; 62.93 u) andcopper-65 (30.91%; 64.93 u). What is the

average atomic mass of copper?

Atomic mass of copper= (0.6909 x 62.93 u) + (0.3091 x 64.93 u)

= 63.55 u(as given in the periodic table)


10/40

Exercise #1: Relative Atomic Mass

A mass spectrometer computed the atomic mass

ratio of fluorine to carbon-12 as 1.5832-to-1. If the

atomic mass of carbon-12 is 12 u(exactly), whatis the atomic mass of fluorine in u?

(Answer: 18.998 u)


11/40

Exercise #2: Average Atomic Mass

Natural boron is composed of two isotopes:

19.78% boron-10 (atomic mass = 10.0129 amu)

and 80.22% boron-11 (atomic mass = 11.0093amu). What is the average atomic mass of

naturally occurring boron?

(Answer: 10.81 u)


12/40

Molar Quantity

The Mole:

A quantity that contains the Avogadros

number of items;

Avogadros number = 6.022 x 1023

12.01 g of carbon contains the Avogadros

number of carbon atoms.1 mole of carbon = 12.01 g

1 carbon atom = 12.01 u(or amu)


13/40

Gram-Atomic Mass

Mass of 1 carbon-12 atom = 12 u(exactly);

Mass of 1 mole of carbon-12 = 12 g;

Mass of 1 oxygen atom = 16.00 u

Mass of 1 mole of oxygen = 16.00 g

Gram-atomic mass= mass (in grams) of 1 mole ofan elementthat is, the mass (in grams) thatcontains the Avogadros number of atoms of that

element.

gram-atomic massisthe molar massof an elementin grams.


14/40

Atomic Mass & Gram-Atomic Mass

Examples:

Element Atomic mass Gram-atomic mass

Carbon 12.01 u 12.01 g/molOxygen 16.00 u 16.00 g/mol

Aluminum 26.98 u 26.98 g/mol

Silicon 28.09 u 28.09 g/molGold 197.0 u 197.0 g/mol


15/40

Molecular Mass and Molar Mass

Molecular mass= the mass of a molecule inu;

Molar mass = the mass of one mole of an element

or a compound, expressed in grams.

Examples:Molecular Mass Molar Mass

N2 28.02 u 28.02 g/mol

H2O 18.02 u 18.02 g/molC8H18 114.22 u 114.22 g/mol


16/40

Calculating Molar Mass

Calculating the molar mass of sucrose, C12H22O11:

(12 x 12.01 g) + (22 x 1.008 g) + (11 x 16.00 g)

= 342.3 g/mole

Molar mass of ammonium hydrogen phosphate,

(NH4)2HPO4:

(2 x 14.01 g) + (9 x 1.008 g) + (1 x 30.97 g) + (4 x 16.00 g)

= 132.06 g/mole


17/40

Percent Composition of a Compound

Composition of aluminum sulfate, Al2(SO4)3:

Molar mass of Al2(SO4)3=

(2 x 26.98 g) + (3 x 32.06 g) + (12 x 16.00 g) = 342.14 g/mole

Mass percent of Al = (53.96 g/342.14 g) x 100% = 15.77%

Mass percent of S = (96.18 g/342.14 g) x 100% = 28.11%

Mass percent of O = (192.0 g/342.14 g) x 100% = 56.12%


18/40

Formula of Compounds

Empirical Formula

A chemical formula that represents a simple whole

number ratio of the number of moles of elements in thecompound. Examples: MgO, Cu2S, CH2O, etc.

Molecular Formula

A formula that shows the actual number of atoms of

each type in a molecule.

Examples: C4H10, C6H6, C6H12O6.


19/40

Empirical Formula-1

Empirical formula from composition:

Example: A compound containing carbon, hydrogen, and oxygen hasthe following composition (by mass percent): 68.12% C, 13.73%H, and 18.15% O, Determine its empirical formula.

Solution:

Use mass percent to calculate mole and mole ratio of C:H:O

Mole of C = 68.12 g x (1 mol C/12.01 g) = 5.672 mol C

Mole of H = 13.73 g x (1 mol H/1.008 g) = 13.62 mol H

Mole of O = 18.15 g x (1 mol O/16.00 g) = 1.134 mol O

Divide all moles by mole of O (smallest value) to get simple ratio:5.672 mol C/1.134 mol O = 5; 13.62 mol H/1.134 mol O = 12, and

1.134 mol O/1.134 mol O = 1;

Mole ratio: 5C:12H:1O Empirical formula = C5H12O


20/40

Empirical Formula-2

Empirical formula from mass of elements in a sample of compound

Example: When 1.96 g of phosphorus is burned, 4.49 g of a

phosphorus oxide is obtained. Calculate the empirical formula of

the phosphorus oxide.

Solution:

Calculate moles of P and O in sample and obtain a simple mole ratio;

Mole of P = 1.96 g P x (1 mol/30.97 g) = 0.0633 mol P;

Mole of O = (4.49 g1.96 g) x (1 mol/16.00 g) = 0.158 mol O;

Divide by mole of P (smaller value) to get a simple mole ratio:0.0633 mol P/0.0633 = 1 mol P; 0.158 mol O/0.0633 = 2.5 mol O

Mole ratio: 1 mol P to2.5 mol O, OR 2 mol P to5 mol O

Empirical formula = P2O5


21/40

Empirical Formula-3

Empirical formula from data of combustion reaction:

Example: A compound is composed of carbon, hydrogen, and oxygen. When 2.32g of this compound is burned in excess of oxygen, it produces 5.28 g of CO2gasand 2.16 g of water. Calculate the composition (in mass percent) of thecompound and determine its empirical formula.

Solution:

Find mass of C, H, and O in the sample and then calculate their masspercent:

Mass of C = 5.28 g CO2x (12.01 g C/44.01 g CO2) = 1.44 g

Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1%

Mass of H = 2.16 g H2O x (2 x 1.008 g/18.02 g H2O) = 0.24 g

Mass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4%Mass of O = 2.32 g sample1.44 g C0.24 g H = 0.64 g

Mass % of O = 10062.1% C10.4% H = 27.5%

Derive empirical formula from these mass percent composition

(next slide)


22/40

Empirical Formula-3

Empirical formula from data of combustion (continued):

Calculate mole and simple mole ratio from calculated mass of each

element:

Mole of C = 1.44 g C x (1 mol/12.01 g) = 0.12 molMole of H = 0.242 g x (1 mol/1.008 g) = 0.24 mol

Mole of O = 0.64 g x (1 mol/16.00 g) = 0.04 mol

Divide all moles by mole of O (smallest mole) to obtain a simple ratio:

0.12 mol C/0.04 = 3 mol C; 0.24 mol H/0.04 = 6 mol H;

0.04 mol O/0.04 = 1 mol O

Simple molar ratio: 3 mol C : 6 mol H : 1 mol O

Empirical formula: C3H6O


23/40

Molecular Formula

Molecular formula is derived from empirical formula and molecularmass, which is obtained independently

Empirical formula = CxHyOz;molecular formula = (CxHyOz)n,

where n = (molecular mass/empirical formula mass)

Example:

A compound has an empirical formula C3H6O and its molecularformula is 116.2 u. What is the molecular formula?

Solution:

Empirical formula mass = (2 x 12.01 u) + (6 x 1.008 u) + 16.00 u

= 58.1 u

Molecular formula = (C3H6O)n; where n = (116.2 u/58.1 u) = 2

Incorrect molecular formula = (C3H6O)2;

Correct molecular formula = C6H12O2


24/40

Exercise #3: Determination of Formulas

A 2.00-gram sample of phosphorus is completelyreacted with oxygen gas, which yields 4.58 g of

product that is composed of only phosphorus and

oxygen. In separate analyses, the compound is

found to have molar mass of about 284 g/mol.

(a) Determine the empirical and molecular

formulas of the compound. (b) Write an equation

for the reaction of phosphorus with oxygen gas.

(Answer: P2O5; P4O10; 4P + 5 O2P4O10)


25/40

Chemical Equation #1

Description of reaction:

Iron reacts with oxygen gas and forms solid

iron(III) oxide:

Identity: reactants = iron (Fe) and oxygen gas

(O2); product = iron(III) oxide

Chemical equation: Fe(s)+ O2(g)

Fe2O3(s) Balanced equation:

4Fe(s)+ 3 O2(g)2Fe2O3(s)


26/40



Phosphorus reacts with oxygen gas to form

solid tetraphosphorus decoxide.

Equation: P(s)+ O2(g)P4O10(s)

Balanced eqn.: 4P(s)+ 5 O2(g)P4O10(s)


27/40



Propane gas (C3H8) is burned in air (excess of oxygen)

to form carbon dioxide gas and water vapor;

Identity: reactants = C3H8(g)and O2(g);products = CO2(g)and H2O(g);

Equation: C3H8(g)+ O2(g)CO2(g)+ H2O(g);

Balanced equation:

C3H8(g)+ 5 O2(g)3CO2(g)+ 4H2O(g)


28/40

Chemical Equation 4

Description of reaction:Ammonia gas (NH3) reacts with oxygen gas to

form nitrogen monoxide gas and water vapor;

Equation: NH3(g)+ O2(g)NO(g)+ H2O(g);

Balancing the equation:

2NH3(g)+5/2O2(g)2NO(g)+ 3H2O(g);

Multiply throughout by 2 to get rid of the fraction:

4NH3(g)+ 5 O2(g)4NO(g)+ 6H2O(g);


29/40

Balancing Chemical Equations

Rules for balancing equations:

1. Use smallest integer coefficients in front of each reactants andproducts as necessary; coefficient 1 need not be indicated;

2. The formula of the substances in the equation MUST NOT be

changed.

Helpful steps in balancing equations:

1. Begin with the compound that contains the most atoms or types ofatoms.

2. Balance elements that appear only once on each side of the arrow.

3. Next balance elements that appear more than once on either side.

4. Balance free elements last.

5. Finally, check that smallest whole number coefficients are used.


30/40

Stoichiometry

Stoichiometry = the quantitative relationships betweenone reactant to another, or between a reactant andproducts in a chemical reaction.

Interpreting balanced equations:

Example: C3H8(g)+ 5 O2(g) 3CO2(g)+ 4H2O(g);

The equation implies that:

1 C3H8molecule reacts with 5 O2molecules to produce

3 CO2molecules and 4 H2O molecules; OR1 mole of C3H8reacts with 5 moles of O2to produce 3

moles of CO2and 4 moles of H2O.


31/40

Stoichiometric Calculations

Mole-to-mole relationship: Example: In the following reaction, if 6.0 moles of octane, C8H18, is

completely combusted in excess of oxygen gas, how many moles ofCO2and H2O, respectively, will be formed? How many moles of O2does it consumed?

Reaction: 2C8H18(l) + 25 O2(g) 16CO2(g) + 18H2O(g)

Calculations:

Mole CO2formed = 6.0 mol C8H18x (16 mol CO2/2mol C8H18) = 48 moles

Mole H2O formed = 6.0 mol C

8H

18x (18 mol H

2O/2mol C

8H

18) = 54 moles

Mole O2consumed = 6.0 mol C8H18x (25 mol O2/2mol C8H18) = 75 moles


32/40


Mass-to-mole-to-mole-to-mass relationship: Example-1: In the following reaction, if 690 g of octane, C8H18, is

completely combusted in excess of oxygen gas, how many grams of

CO2are formed?

Reaction: 2C8H18(l)+ 25 O2(g) 16CO2(g)+ 18H2O(g)

Calculation-1:

Moles C8H18reacted = 690 g C8H18x (1 mol/114.2 g) = 6.0 moles

Moles CO2formed = 6.0 mol C8H18x (16 mol CO2/2 mol C8H18)

= 48 moles CO2

Mass of CO2formed = 48 mol CO2x (44.01 g/mol) = 2.1 x 103g


33/40




H2O are formed?


Calculation-2:


Moles H2O formed = 6.0 mol C8H18x (18 mol H2O/2 mol C8H18)

= 54 moles CO2

Mass of H2O formed = 54 mol H2O x (18.02 g/mol) = 970 g


34/40




oxygen gas are consumed?


Calculation-3:


Moles O2consumed = 6.0 mol C8H18x (25 mol O2/2 mol C8H18)

= 75 moles O2

Mass of H2O formed = 75 mol O2x (32.00 g/mol) = 2.4 x 103g g


35/40

Stoichiometry Involving Limiting Reactant

Limiting reactant

one that got completely consumed in a

chemical reaction before the other

reactants.

Product yields depend on the amount of

limiting reactant


36/40

A Reaction Stoichiometry

Example:

In the reaction: 2Cu(s)+ S(s) Cu2S(s),

2 moles of copper are required to react completely with1 mole of sulfur, which will produce 1 mole of

copper(I) sulfide.

If a reaction is carried out using 1 mole of copper and 1

mole of sulfur, then copper will be the limiting reactant

and sulfur is in excess. Only 0.5 mole of copper(I)

sulfide is obtained.


37/40

Exercise #4: Stoichiometry Calculations

Ammonia is produced from the reaction with

hydrogen according to the following equation:

N2(g) + 3H2(g) 2NH3(g)

If 25 N2molecules are reacted with 60 H2

molecules in a sealed container, which molecules

will be completely consumed? How many NH3

molecules are formed?

(Answer: H2; 40 NH3molecules)


38/40

Exercise #5: Limiting Reactants and

Reaction Yields

Ammonia is produced in the following reaction:

N2(g) + 3H2(g) 2NH3(g)

(a) If 118 g of nitrogen gas is reacted with 31.5 g of

hydrogen gas, which reactant will be completely consumed

at the end of the reaction? (b) How many grams of the

excess reactant will remain (unreacted)? (c) How many

grams ammonia will be produced when the limiting

reactant is completely reacted and the yield is 100%?

(Answer: (a) N2; (b) 6.0 g; (c) 143.4 g of NH3)


39/40

Theoretical, Actual and Percent Yields

Theoretical yield:

yield of product calculated based on the stoichiometry

of balanced equation and amount of limiting reactant

(assuming the reaction goes to completion and the

limiting reactant is completely consumed).

Actual Yield:

Yield of product actually obtained from experiment

Percent Yield = (Actual yield/Theoretical yield) x 100%


40/40

Exercise #6: Limiting Reactant & Yields

In an ammonia production, the reactor is charged with N2

and H2gases at flow rates of 805 g and 195 g per minute,

respectively, at 227oC, and the reaction is as follows:

N2(g) + 3H2(g) 3 NH3(g)(a) What is the rate (in g/min) that ammonia is produced if

the yield is 100%? (b) If the reaction produces 915 g of

NH3per minute, calculate the percentage yield of the

reaction.

(Answer: (a) 978.5 g/min; (b) Yield = 93.5%)

Date post:	02-Jun-2018
Category:	Documents
Upload:	ahmedamer1
View:	264 times
Download:	0 times

Chapter 3 - Chemical Calculations

Documents