principles of chemistry i chem 1211 chapter 3 dr. augustine ofori agyeman assistant professor of...

PRINCIPLES OF CHEMISTRY I

CHEM 1211

CHAPTER 3

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 3

STOICHIOMETRY(CHEMICAL CALCULATIONS)

STOICHIOMETRY

- The area of study involved with calculation of the quantities of substances consumed or produced in a chemical reaction

- Chemical reactions are represented by chemical equations

- Reactants are substances that are consumed

- Products are substances that are formed

CHEMICAL EQUATIONS

- Reactants are written on the left side of a chemical equation and products on the right side

- An arrow pointing towards the products, is used to separate the reactants from the products

- The plus sign (+) is used to separate different reactants or different products

- Chemical equations must be consistent with experimental facts

(reactants and products in a reaction that actually takes place)

- Chemical equations must be balanced (equal numbers of atoms of each kind on both sides)

(Daltons atomic theory)

CHEMICAL EQUATIONS

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

States of reactants and products

Physical states of reactants and products are represented by:(g): gas

(l): liquid(s): solid

(aq): aqueous or water solution

CHEMICAL EQUATIONS

- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)

- The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation

C2H5OH(l) + O2(g) → 2CO2(g) + H2O(g)

2 C atoms 2 C atoms

Place the coefficient 2 in front of CO2 to balance C atoms

BALANCING CHEMICAL EQUATIONS

C2H5OH(l) + O2(g) → 2CO2(g) + 3H2O(g)

(5+1)=6 H atoms 3(1x2)=6 H atoms

Place 3 in front of H2O to balance H atoms

- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)

- The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation

BALANCING CHEMICAL EQUATIONS

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

1+(3x2)=7 O atoms (2x2)+3=7 O atoms

Place 3 in front of O2 to balance O atoms

- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)

- The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation

BALANCING CHEMICAL EQUATIONS

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

2 C atoms(5+1)=6 H atoms

1+(3x2)=7 O atoms

2 C atoms(3x2)=6 H atoms

(2x2)+3=7 O atoms

- Check to make sure equation is balanced- When the coefficient is 1, it is not written

- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)

- The coefficients in a chemical equation are the smallest set of whole numbers that balance the equation

BALANCING CHEMICAL EQUATIONS

Balance the following chemical equations

Fe(s) + O2(g) → Fe2O3(s)

C12H22O11(s) + O2(g) → CO2(g) + H2O(g)

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

BALANCING CHEMICAL EQUATIONS

TYPES OF CHEMICAL REACTIONS

Five Types of Chemical Reactions

- Combination reaction

- Decomposition reaction

- Single-replacement reaction

- Double-replacement reaction

- Combustion reaction

COMBINATION REACTION

- Addition or synthesis reaction- Two or more reactants produce a single product

X + Y → XY

N2(g) + 3H2(g) → 2NH3(g)

2Mg(s) + O2(g) → 2MgO(s)

SO3(g) + H2O(l) → H2SO4(aq)

DECOMPOSITION REACTION

- Two or more products are formed from a single reactant

XY → X + Y

2H2O(l) → 2H2(g) + O2(g)

BaCO3(s) → BaO(s) + CO2(g)

2NaN3(s) → 2Na(s) + 3N2(g)

SINGLE-REPLACEMENT REACTION

- Substitution or Displacement reaction- An atom or molecule replaces another atom or molecule

A + BY → B + AY

Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq)

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Cl2(g) + 2NaBr(aq) → 2NaCl(aq) + Br2(g)

- Metal replaces metal and nonmetal replaces nonmetal- Cation replaces cation and anion replaces anion

DOUPLE-REPLACEMENT REACTION

- Exchange or metathesis (transpose) reaction- Parts of two compounds switch places to form two new compounds

AX + BY → AY + BX

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

CaCl2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaCl(aq)

COMBUSTION REACTION

- Reaction between a substance and oxygen (air) accompanied by the production of heat and light

- A common synonym for combustion is burn

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + heat

2CH3OH + 3O2(g) → 2CO2(g) + 4H2O(g) + heat

2Mg(s) + O2(g) → 2MgO(s) + heat

Hydrocarbons are the most common type of compounds that undergo combustion producing CO2 and H2O

ACIDS

- Ionize in aqueous solutions to form hydrogen (H+) ions- Proton (H+) donors

- H+ and H3O+ are used interchangeably

Ionize- Dissolving in solution (water) to form ions

ExamplesHCl(aq) → H+(aq) + Cl-(aq)

HNO3(aq) → H+(aq) + NO3-(aq)

Sum of charges on each side of equation must be equal

Strong Acids

- Transfer 100% (or very nearly 100%) of their protons to H2O in aqueous solution

- Completely or nearly completely ionize in aqueous solution

- Strong electrolytes

ExamplesHCl, HNO3, H2SO4, HBr, H3PO4

ACIDS

ACIDS

Weak Acids

- Transfer only a small percentage (< 5%) of theirprotons to H2O in aqueous solution

ExamplesOrganic acids: acetic acid, citric acid

BASES

- Proton (H+) acceptors- Produce hydroxide ion (OH-) when dissolved in water

Examples NaOH → Na+(aq) + OH-(aq)

Ca(OH)2 → Ca2+(aq) + 2OH-(aq)

Strong Bases

- Completely or nearly completely ionize in aqueous solution

-Strong electrolytes

Examples- Hydroxides of Groups 1A and 2A are strong bases

LiOH, CsOH, Ba(OH)2, Ca(OH)2

most common in lab: NaOH and KOH

BASES

Weak Bases

- produce small amounts of OH- ions in aqueous solution

Examplesmethylamine, cocaine, morphine

most common: NH3

- Small amounts of NH4+ and OH- ions are

produced in aqueous solution

- The name aqueous ammonia is preferred over ammonium hydroxide

BASES

ACID-BASE REACTION

- Also referred to as Neutralization Reaction

- Occurs when solutions of an acid and a base are mixed

- Products are salt and water when the base is a metal hydroxide

ExampleHCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)

A cation from a base combines with an anion from an acid to form a salt

ACID-BASE REACTION

Gas Formation

- Carbonates and bicarbonates react with acids to form CO2 gas

HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2CO3(aq) 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2CO3(aq)

- H2CO3 (carbonic acid) is unstable and decomposes to produce CO2

H2CO3(aq) → H2O(l) + CO2(g)

- Hydrogen sulfide (H2S) is produced when Na2S reacts with an acid 2HCl(aq) + Na2S(aq) → 2NaCl(aq) + H2S(g)

- Also called redox reactions- Involve transfer of electrons from one species to another

Oxidation - loss of electronsReduction - gain of electrons

Ionic solid sodium chloride (Na+ and Cl- ions) formed from solidsodium and chlorine gas

2Na(s) + Cl2(g) → 2NaCl(s)

The oxidation (rusting) of iron by reaction with moist air4Fe(s) + 3O2(g) → 2Fe2O3(s)

OXIDATION-REDUCTION REACTION

- There is no transfer of electrons from one reactant to another reactant

BaCO3(s) → BaO(s) + CO2(g)

- Double-replacement reactions

- Most reactions we have already come across

NONREDOX REACTION

OXIDATION NUMBER (STATE)

The concept of oxidation number - provides a way to keep track of electrons in redox reactions

- not necessarily ionic charges

Conventionally - actual charges on ions are written as n+ or n-

- oxidation numbers are written as +n or -n

Oxidation - increase in oxidation number (loss of electrons)Reduction - decrease in oxidation number (gain of electrons)

OXIDATION NUMBERS

1. Oxidation number of uncombined elements = 0 Na(s), O2(g), H2(g), Hg(l)

2. Oxidation number of a monatomic ion = chargeNa+ = +1, Cl- = -1, Ca2+ = +2, Al3+ = +3

3. Oxygen is usually assigned -2H2O, CO2, SO2, SO3

Exceptions: H2O2 (oxygen = -1) OF2 (oxygen = +2)

4. Hydrogen is usually assigned +1 (-1 when bonded to metals)+1: HCl, NH3, H2O-1: CaH2, NaH

5. Halogens are usually assigned -1 (F, Cl, Br, I) Exceptions: when Cl, Br, and I are bonded to oxygen

Cl2O: Cl O Cl

6. The sum of oxidation numbers for - neutral compound = 0- polyatomic ion = chargeH2O = 0, CO3

2- = -2, NH4+ = +1

+1 -2 +1

OXIDATION NUMBERS

CO2

The oxidation state of oxygen is -2 CO2 has no charge

The sum of oxidation states of carbon and oxygen = 01 carbon atom and 2 oxygen atoms

1(x) + 2(-2) = 0x = +4

CO2

x -2 for each oxygen

OXIDATION NUMBERS

CH4

x +1

1(x) + 4(+1) = 0x = -4

OXIDATION NUMBERS

NO3-

x -2

1(x) + 3(-2) = -1x = +5

OXIDATION NUMBERS

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

+1-4 +4 +10 -2 -2

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

-4

+1 +1

+48e- loss

-2 -20

8e- gain

OXIDATION NUMBERS

Carbonlosses

8 electrons

EachOxygen

atom gains 2 electrons

Oxidation

Loss of electronsIncrease in oxidation number

Reducing Agent (electron donor)

Reduction

Gain of electronsDecrease in oxidation number

Oxidizing Agent (electron acceptor)

MnemonicOIL RIG

Oxidation Involves Loss; Reduction Involves Gain

Redox reactions are characterized by transfer of electrons

OXIDATION NUMBERS

• Oxidation Number Method

• Half Reaction Method

- Acidic Solutions- Acidic Solutions

- Basic Solutions- Basic Solutions

BALANCING REDOX EQUATIONS

MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)

OXIDATION NUMBER METHOD

Step 1: Balance the net ionic equation for all atoms other than H and O

Step 2: Assign oxidation numbers to all atoms

+7 -1 +2 0

MnO4-(aq) + Br-(aq) → Mn2+(aq) + Br2(aq)

MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)

-2

Step 3: Determine which atoms have changed oxidation numbers and by how much

Step 4: Multiply net gain and net loss of electrons by suitable factors to balance

+7 5 e- gain

2 e- loss -1

Net gain: 5 x 2 = 10Net loss: 2 x 5 = 10

+2

0

MnO4-(aq) + 2Br-(aq) → Mn2+(aq) + Br2(aq)

OXIDATION NUMBER METHOD

Step 5: Multiply the coefficients by respective factors

Net gain: 5 x 2 = 10 Net loss: 2 x 5 = 10

2[MnO4-(aq)] + 5[2Br-(aq)] → 2Mn2+(aq) + 5Br2(aq)

2MnO4-(aq) + 10Br-(aq) → 2Mn2+(aq) + 5Br2(aq)

Step 6: Balance the equation for O by adding H2O and for H by the adding H+

2MnO4-(aq) + 10Br-(aq) + 16H+(aq)

→ 2Mn2+(aq) + 5Br2(aq) + 8H2O(l)

Net charge: (2 x -1) + (-10) + (+16) = +4 Net charge: (2 x +2) = +4

OXIDATION NUMBER METHOD

Cl-(aq) → Cl2(aq)

Cr2O72-(aq) → Cr3+(aq)

HALF REACTION METHOD

Cr2O72-(aq) + Cl-(aq) → Cr3+(aq) + Cl2(aq) (In Acid)

Oxidation half-reaction

Reduction half-reaction

+6 -2 -1 +3 0

Step 2: - Balance all the elements except hydrogen and oxygen

2Cl-(aq) → Cl2(aq)

Oxidation half-reaction

Reduction half-reaction

Cr2O72-(aq) → 2Cr3+(aq)

HALF REACTION METHOD

2Cl-(aq) → Cl2(aq)

Step 2: - Balance all the elements except hydrogen and oxygen- Balance oxygen using H2O

Oxidation half-reaction

Reduction half-reaction

Cr2O72-(aq) → 2Cr3+(aq) + 7H2O(l)

HALF REACTION METHOD

Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l)

2Cl-(aq) → Cl2(aq)

Step 2: - Balance all the elements except hydrogen and oxygen- Balance oxygen using H2O- Balance hydrogen using H+

Oxidation half-reaction

Reduction half-reaction

HALF REACTION METHOD

Step 2: - Balance all the elements except hydrogen and oxygen - Balance oxygen using H2O- Balance hydrogen using H+

- Balance charge using electrons (e-)

2Cl-(aq) → Cl2(aq) + 2e-

Oxidation half-reaction

Reduction half-reaction

Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

HALF REACTION METHOD

3 x [2Cl-(aq) → Cl2(aq) + 2e-]

Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

Step 3: Multiply both balanced half-reactions by suitable factors to equalize electron count

6Cl-(aq) → 3Cl2(aq) + 6e-

Oxidation half-reaction

Reduction half-reaction

HALF REACTION METHOD

Step 4: - Add the half-reactions and cancel identical species - Check that the elements and charges are balanced

6Cl-(aq) → 3Cl2(aq) + 6e-

Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

Cr2O72-(aq) + 14H+(aq) + 6Cl-(aq)

→ 3Cl2(aq) + 2Cr3+(aq) + 7H2O(l)

Net Charge: (-2) + (+14) + (6 x -1) = +6 Net Charge: (2 x +3) = +6

HALF REACTION METHOD

2MnO4-(aq) + 3C2O4

2-(aq) + 2H2O(l) →

2MnO2(s) + 6CO32-(aq) + 4H+(aq)

MnO4-(aq) + C2O4

2-(aq) → MnO2(s) + CO32-(aq) (In Base)

Balance equation as if it were acidic

Note H+ ions and add same number of OH- ions to both sides

Cancel H+ and OH- (=H2O) with H2O on other side

BASIC SOLUTIONS

2MnO4-(aq) + 3C2O4

2-(aq) + 2H2O(l) + 4OH-(aq)

→ 2MnO2(s) + 6CO3

2-(aq) + 4H+(aq) + 4OH-(aq)

BASIC SOLUTIONS

Balance equation as if it were acidic

Note H+ ions and add same number of OH- ions to both sides

Cancel H+ and OH- (=H2O) with H2O on other side

MnO4-(aq) + C2O4

2-(aq) → MnO2(s) + CO32-(aq) (In Base)

2MnO4-(aq) + 3C2O4

2-(aq) + 2H2O(l) + 4OH-(aq)

→ 2MnO2(s) + 6CO3

2-(aq) + 4H+(aq) + 4OH-(aq)

Balance equation as if it were acidic

Note H+ ions and add same number of OH- ions to both sides

Cancel H+ and OH- (=H2O) with H2O on other side

4H2O

MnO4-(aq) + C2O4

2-(aq) → MnO2(s) + CO32-(aq) (In Base)

2

BASIC SOLUTIONS

Balance equation as if it were acidic

Note H+ ions and add same number of OH- ions to both sides

Cancel H+ and OH- (=H2O) with H2O on other side

MnO4-(aq) + C2O4

2-(aq) → MnO2(s) + CO32-(aq) (In Base)

2MnO4-(aq) + 3C2O4

2-(aq) + 4OH-(aq) →

2MnO2(s) + 6CO32-(aq) + 2H2O(l)

BASIC SOLUTIONS

THE MOLE

The amount of substance of a system, which contains as manyelementary entities as there are atoms in 12 grams of carbon-12

- abbreviated mol

1 mole (mol) = 6.02214179 x 1023 entities

- known as the Avogadro’s number (after Amedeo Avogadro)

- usually rounded to 6.022 x 1023

THE MOLE

The number of entities (or objects) can be atoms or molecules

1 mol C = 6.022 x 1023 atoms C

1 mol CO2 = 6.022 x 1023 molecules CO2

2 conversion factors can be derived from each

THE MOLE

How many atoms are there in 0.40 mole nitrogen?

= 2.4 x 1023 nitrogen atoms

How many molecules are there in 1.2 moles water?

= 7.2 x 1023 water molecules

nitrogenmol1

atomsnitrogen10x6.022xnitrogenmol0.40

23

watermol1

moleculeswater10x6.022xwatermol1.2

23

How many H atoms are there in 1.2 moles water?

= 1.4 x 1024 H atoms

molecule)water(1

atoms)H(2x

water)mol(1

molecules)water10x(6.022xwatermol1.2

23

THE MOLE

MOLAR MASS

- The mass of a substance in grams that is numerically equal tothe formula mass of that substance

- Add atomic masses to get the formula mass (in u) = molar mass (in g/mol)

- The mass, in grams, of 1 mole of the substance

MOLAR MASS

Consider the following

Sodium (Na) has an atomic mass of 22.99 uThis implies that the mass of 1 mole of Na = 22.99 g

Molar mass of Na = 22.99 g/mol

Formula mass of NaCl = 58.44 uThe mass of 1 mole of NaCl = 58.44 g

Molar mass of NaCl = 58.88 g/mol

Formula mass of CaCO3 = 100.09 uThe mass of 1 mole of CaCO3 = 100.09 g

Molar mass of CaCO3 = 100.09 g/mol

MOLAR MASS

Calculate the mass of 2.4 moles of NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)

= 85.00 g /mol NaNO3

= 204 g NaNO3

= 2.0 x 102 g NaNO3

3

333 NaNOmol1

NaNOg85.00xNaNOmol2.4NaNOg

MOLAR MASS

How many moles are present in 2.4 g NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)

= 85.00 g /mol NaNO3

= 0.028 mol NaNO3

= 2.8 x 10-2 mol NaNO3

3

333 NaNOg85.00

NaNOmol 1xNaNOg2.4NaNOmol

PERCENTAGE COMPOSITION

- Percentage by mass contributed by individual elements in a compound

100%xcompoundofmass

element of masselement%

100%xcompoundofmassformula

element)ofatomsofumberelement)(nofmass(atomicelement%

PERCENTAGE COMPOSITION

Calculate the percentage of carbon, hydrogen, and oxygen, inethanol (C2H5OH)

% 13.13100%xu 46.07

u)(6) (1.01H%

% 73.34100%xu 46.07

u)(1) (16.00O%

% 52.14100%xu 46.07

u)(2) (12.01C%

PERCENTAGE COMPOSITION

Calculate the percent composition by mass of each elementin the following compounds

C9H8O4

(NH4)2PtCl4

C2H2F4

C8H10N4O2

Pt(NH3)2Cl2

COMBUSTION ANALYSIS

- Used for the determination of mass percentages and empiricalformula of compounds

- A combustion train is usually used for analysis of compoundscontaining only carbon, hydrogen, and oxygen

- A first compartment with CaCl2 traps H2O

- A second compartment with NaOH traps CO2

- Masses of trapped H2O and CO2 are then determined

COMBUSTION ANALYSIS

furnace

CaCl2 NaOHO2

sample

H2O trap CO2 trap

COMBUSTION ANALYSIS

Combustion of a 0.2000-g sample of a compound made up ofonly carbon, hydrogen, and oxygen yields 0.200 g H2O and0.4880 g CO2. Calculate the mass and mass percentage of

each element present in the 0.2000-g sample.

- Convert mass H2O/CO2 to moles using molar mass- Determine moles H/C from number of atoms and moles H2O/CO2

- Convert moles H/C to mass H/C using molar mass- Determine mass O by subtracting total mass H and C

from mass sample- Calculate percentages as discussed earlier

COMBUSTION ANALYSIS

OHmol0.0111g18.02

mol1xg0.200OHmol 22

Hmol0.0222OHmol1

Hmol2xOHmol0.0111Hmol

22

H0.0224Hmol1

H01.1xHmol0.0222Hmass g

g

COMBUSTION ANALYSIS

22 COmol0.01109g44.01

mol1xg0.4880COmol

Cmol0.01109COmol1

Cmol1xCOmol0.01109Cmol

22

Cg0.1332Cmol1

Cg12.01xCmol0.01109Cmass

COMBUSTION ANALYSIS

Mass O = 0.2000 g sample – (0.0224 g + 0.1332 g)

= 0.0444 g O

% .6066100%x0.2000

g 0.1332C%

% .211100%x0.2000

g 0.0224H%

% .222100%x0.2000

g 0.0444O%

EMPIRICAL FORMULA

Given mass % elements:

- Convert to g elements assuming 100.0 g sample

- Convert to mole elements using molar mass

- Calculate mole ratio (divide each by the smallest number of moles)

- Round each to the nearest integer

- Multiply through by a suitable factor if necessary( __.5 x 2 or __.33 x 3 or __ .25 x 4)

EMPIRICAL FORMULA

Determine the empirical formula for a compound that gives the following percentages upon analysis (in mass percents):71.65 % Cl 24.27 % C 4.07 % H

- Assume 100.0 g of sample and convert grams to moles

Clmol2.021Clg35.45

Clmol1xClg71.65

Cmol2.021Cg12.01

Cmol1xCg24.27

Hmol04.4 Hg 1.01

Hmol1xHg07.4

71.65 g Cl

24.27 g C

4.07 g H

EMPIRICAL FORMULA

1.0002.021

2.021:Cl

1.0002.021

2.021:C

- Calculate mol ratios

99.12.021

4.04:H

- Round to nearest integers and write empirical formula

Cl: 1, C: 1, H: 2 giving CH2Cl

MOLECULAR FORMULA

Given the molar mass:

- Determine the empirical formula

- Calculate the empirical formula mass

- Divide the given molar mass by the empirical formula mass to obtain a whole-number multiple

- Multiply all subscripts in the empirical formula by the multiple

MOLECULAR FORMULA

From previous example, calculate the molecular formula if themolar mass is known to be 98.96 g/mol

Empirical formula = ClCH2

Empirical formula mass = 35.45 + 12.01 + 2(1.01) = 49.48 g/mol

2g/mol49.48

g/mol98.96

MassFormulaEmpirical

massMolar

Molecular formula = (CH2Cl)2 = C2H4Cl2

CHEMICAL FORMULA

Subscripts represent both atomic and molar amounts

Consider Na2S2O3:

- Two atoms of sodium, two atoms of sulfur, and three atoms ofoxygen are present in one molecule of Na2S2O3

- Two moles of sodium, two moles of sulfur, and three moles ofoxygen are present in one mole of Na2S2O3

CHEMICAL FORMULA

How many moles of sodium atoms, sulfur atoms, and oxygenatoms are present in 1.8 moles of a sample of Na2S2O3?

I mol Na2S2O3 contains 2 mol Na, 2 mol S, and 3 mol O

Namol3.6OSNamol1

Namol2xOSNamol1.8Namol

322322

Smol3.6OSNamol1

Smol2xOSNamol1.8Smol

322322

Omol5.4OSNamol1

Omol3xOSNamol1.8Omol

322322

CHEMICAL CALCULATIONS

Calculate the number of molecules present in 0.075 g of urea,(NH2)2CO

Given mass of urea: - Convert to moles of urea using molar mass

- Convert to molecules of urea using Avogadro’s number

= 7.5 x 1020 molecules (NH2)2CO

CO)(NHmole1

CO)NH(molecules10x6.022x

CO)(NHg60.07

CO)(NHmole1xCO)(NHg0.075

22

2223

22

2222

CHEMICAL CALCULATIONS

How many grams of carbon are present in a 0.125 g of vitamin C,C6H8O6?

Given mass of vitamin C: - Convert to moles of vitamin C using molar mass

- Convert to moles of C (1 mole C6H8O6 contains 6 moles C)- Convert moles carbon to g carbon using molar mass

= 0.0511 g carbon

Cmol1

Cg12.01x

OHCmol1

Cmol6x

OHCg176.14

OHCmol1xOHCg0.125

686686

686686

CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)

Coefficients represent both molecular and molar amounts

Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

- 1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O

- 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

- make sure the equation is balanced- calculate moles of propane from given mass and molar mass- determine moles of oxygen from mole ratio (stoichiometry)

- calculate mass of oxygen

= 349 g O2

CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)

2

2

83

2

83

8383 Omol1

Og32.00x

HCmol1

Omol5x

HCg44.11

HCmol1xHCg96.1

Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

What mass of oxygen will react with 96.1 g of propane?

- make sure the equation is balanced- calculate moles of propane from given mass and molar

mass- determine moles of CO2 from mole ratio (stoichiometry)

- calculate mass of CO2

= 288 g CO2

CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)

2

2

83

2

83

8383 COmol1

COg44.01x

HCmol1

COmol3x

HCg44.11

HCmol1xHCg96.1

Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

What mass of CO2 will be produced from 96.1 g of propane?

- Also called limiting reagent- The reactant that is completely consumed in a reaction

- The reactant(s) with leftovers is (are) known as the excess reactant(s) or excess reagent(s)

To determine the limiting reactant:- Write and balance the equation for the reaction

- Use given amount of each reactant to determine amount of desired product- The reactant that gives the smallest amount of product is the limiting

LIMITING REACTANT

Consider the following reaction for producing nitrogen gas from gaseous ammonia and solid copper(II) oxide:

LIMITING REACTANT

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)

If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant?

- Make sure the equation is balanced- Calculate moles of desired product from each reactant

LIMITING REACTANT

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)

23

2

3

33 Nmol0.530

NHmol2

Nmol1x

NH g 17.03

NHmol1xNHg18.1

22 Nmol0.380

CuOmol3

Nmol1x

CuO g 79.55

CuOmol1xCuOg90.4

CuO is limiting since it produces smaller amount of N2

PERCENT YIELD

%100xyieldltheoretica

yieldactualyieldPercent

Theoretical YieldThe calculated quantity of product formed,

assuming all of the limiting reactant is used up

Actual YieldThe amount of product actually obtained in a

reaction (always less than the theoretical yield)

PERCENT YIELD

Given actual yield:

- Determine the limiting reactant

- Calculate the theoretical yield from the limiting reactant

- Calculate percent yield

PERCENT YIELD

Calculate the percent yield of N2 from the previous example if 9.04 g of N2 is actually produced

- CuO is the limiting reactant

- Calculate the theoretical yield

2NH3(g) + 3CuO(s) → N2(g) + 3Cu(s) + 3H2O(g)

22

22 Ng10.6Nmol1

Ng28.02x

CuOmol3

Nmol1x

CuO g 79.55

CuOmol1xCuOg90.4

%85.3%100xNg10.6

Ng9.04%100x

yieldltheoretica

yieldactualyieldPercent

2

2