Module 1 - Basic Calculations 2

7/29/2019 Module 1 - Basic Calculations 2

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Page 1 of 30 Module IIndustrial Training DivisionKamal El-Nashar Basic Calculations

BASIC CALCULATIONS

Table of Contents

1.1 Calculations of Number of Moles of a Substance 1

1.1.1 Relative Atomic Mass 1

1.1.2 Relative Formula Mass 4

1.1.3 Mole 5

1.2 Conversion of Percentage by mass to percentage by mole 8

1.3 Molarity and Normality of Chemical Solutions 12

1.3.1 Molarity 12

1.3.2 Normality 13

1.4 Stoichiometric Calculations using Chemical Equations 14

1.4.1 Type of Stoichiometric Problems 15

1.4.2 Mass Mass stoichiometry 16

1.4.3 Mass Volume Stoichiometry 17

1.4.4 Volume Volume Stoichiometry 18

1.4.5 Exercises 20

1.5 Material Balance 21

1.5.1 Material Balance 21

1.5.2 Material Balance Calculations 21

Process Block Diagrams 22

Defining a System using Block Diagram 24

Multiple Stage Multiple Reactors 25

TABLE OF CONTENTS


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BASIC CALCULATIONS

BASIC CALCULATIONS

1.1 CALCULATIONS OF NUMBER OF MOLES OF A SUBSTANCE

Before carrying out such calculations, it is important to define theterms:

1. Relative Atomic Mass2. Relative Molar Mass3. Mole

1.1.1 Relative Atomic Mass

The Relative Atomic Mass (Ar) of a chemical element gives us an ideaof how heavy it feels (the force it makes when a gravity pulls on it).

This is defined as the number of times an atom of an element isheavier than a hydrogen atom.

Relative atomic Mass of an element =

atomHydrogenoneofMass

elementanofatomoneofMass

If you look at the periodic table you can find out the relative atomicmasses of the elements. The number at the bottom of the symbol isthe Relative Atomic Mass (Ar):

Relative Atomic Mass (Atomic weight)

C612 CARBON

Relative atomic mass of carbon is 12 means that 1 carbon (C) atom is12 times heavier than 1 hydrogen (H) atom.

Note: Relative atomic mass is sometimes called atomic mass oratomic weight.


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BASIC CALCULATIONS

Given the following below, what are the Relative atomic Masses of

these elements:

H11 C6

12 O8

16 Na11

23 Mg12

24 Hydrogen Carbon Oxygen Sodium Magnesium

S

16

32 Cl

17

35 K

19

39 Ca

20

40 Cu

29

63 Sulfur Chlorine Potassium Calcium Copper

1. Oxygen? __________

2. Copper? __________

3. Hydrogen? __________

4. Calcium? __________

5. Sodium? __________

6. Sulphur? __________

7. Chlorine? __________

8. Carbon? __________

9. Potassium? __________

10. Magnesium? __________

EXERCISES


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BASIC CALCULATIONS

1.1.2 Relative Formula Mass (Molecular Weight)

Most atoms exist in molecules. You can use the Relative Atomic

Masses of elements to work out the mass of the molecule they makeup.

To work out the Relative Formula Mass, you simply add up theRelative Atomic Masses (atomic weights) of each atom.

Note: Relative Formula Mass is also known as Molecular Weight

Example: Find the Relative Formula Mass (Molecular weights) ofsulfuric acid, H2SO4.

Solution:

Relative Formula Mass = Number of atoms x Relative Atomic Mass

Hydrogen = 2 atoms; Relative atomic mass = 1

Sulfur = 1 atom; Relative atomic mass = 32

Oxygen = 4 atoms; Relative atomic mass = 16

Relative Formula Mass = (2 x 1) + (1 x 32) + (4 x 16) = 98

Or

(Remember that Relative Formula Mass is the same as Molecular

weightand Relative Atomic Mass is the same as atomic weights)

Molecular Weights = Number of atoms x Atomic weights

Hydrogen = 2 atoms; atomic weight = 1

Sulfur = 1 atom; atomic weight = 32

Oxygen = 4 atoms; atomic weight = 16


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BASIC CALCULATIONS

Molecular weight = (2 x 1) + (1 x 32) + (4 x 16) = 98

What are the Molecular weights of these molecules:

1. NaCl? __________

2. H2O? __________

3. HCl? __________

4. Cu O? __________

5. MgCl2? __________

6. H2S? __________

7. CaCl2? __________

8. K2O? __________

9. Noah? __________

10. CO2? __________

EXERCISES


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BASIC CALCULATIONS

1.1.3 Mole

A mole is the quantity of a substance whose weight equals themolecular weight of that substance. In SI units, the term gmole(grammole) is used as a unit for the number of moles.

Number of moles =SubstanceofweightMolecular

SubstanceofMass

Example:Calculate the number of moles of 80 grams of methane, CH4.

Solution:Molecular weight of CH4 = (1 x 12)CARBON + (4 x 1) HYDROGEN = 16

Number of moles =16

80= 5 gmoles


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BASIC CALCULATIONS

1. Calculate the number of moles in 120 grams of ethane, C2H6.

EXERCISES


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BASIC CALCULATIONS

2. Calculate the mass of 2 gmoles of water, H2O.

EXERCISES


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BASIC CALCULATIONS

1.2 CONVERSION OF PERCENTAGE BY MASS (% WT) TOPERCENTAGE BY MOLE (% G. moles)

Most materials occur as mixtures of various components. Forexample, natural gas is a mixture of methane, ethane, propane, butaneand etc. The composition of mixtures can be expressed as percent bymass or percent by mole and it is always useful to convert one form toanother.

Example 1:

Natural Gas has the average composition of

Component % massMethane, CH4 72Ethane, C2H6 12

Propane, C3H8 10Butane, C4H10 6

Calculate the composition of natural gas in mole percent.

Solution:

Basis: 100 grams of Natural Gas (means that 72 % mass of methane =72 grams)

Component % mass MolecularWeight

Number ofMoles

Molepercent

Methane,

CH4

72 1616

72= 4.50

5.23

4.50x 100 =

86.0

Ethane,C2H6

12 3030

12= 0.40

5.23

0.40x 100 =

7.6

Propane,C3H8

10 4444

10= 0.23

5.23

0.23x 100 =

4.4

Butane,C4H10

6 5858

6= 0.10

5.23

0.10x 100 =

2.0

Total 100 5.23 100


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BASIC CALCULATIONS


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BASIC CALCULATIONS

Example 2:

Convert the following gas analysis from mole percent to mass percent.

Component % moleMethane, CH4 82Ethane, C2H6 10Nitrogen, N2 3.5

Carbon Dioxide, CO2 4.5

Solution:

Consider 100 gmoles of gas sample

Component %mole

MolecularWeight

Mass Mass percent

Methane,CH4

82 16 82 x 16 = 13121908

1312x 100 = 68.80

Ethane,C2H6

10 30 10 x 30 = 3001908

300x 100 = 15.70

Nitrogen,N2

3.5 28 3.5 x 28 = 98190898 x 100 = 5.10

CarbonDioxide,

CO2

4.5 44 4.5 x 44 = 1981908

198x 100 = 10.40

Total 100 1908 100.00


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BASIC CALCULATIONS

1. Air consists of nitrogen and oxygen, their mole percentage are 79

for nitrogen and 21 for oxygen. Express their mass percent in an airsample.

E XERCISES


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BASIC CALCULATIONS

2. A gas has the following compositions

Component % massMethane, CH4 64Ethane, C2H6 15

Propane, C3H8 11Butane, C4H10 10

Express the gas composition in mole percent

EXERCISES


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BASIC CALCULATIONS

1.3 MOLARITY AND NORMALITY OF CHEMICAL SOLUTIONS

Concentration of solutions can be expressed through some

quantitative methods. In the basic course we dealt with:

a. Percent by mass (% mass, % weight)b. Parts per million (ppm)c. Mass per volume (kgs / lit)

In this lesson, we will deal with other methods of expressing theconcentration of solutions. These methods are:

1. Molarity (M)2. Normality (N)

1.3.1 Molarity (M)

Molarity or molar concentration is the number of moles of solute perlitre of solution.

Molarity (M) =

SolutionofLitre

SoluteofMoles

or it can be written as

M =SolutionofLitrexMassMolarRelative

SoluteofMass

or it can be written as

M =SolutionofLitrexweightMolecular

SoluteofWeight


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BASIC CALCULATIONS

Example:

Calculate the molarity of an aqueous solution of sodium chloride

containing 284 g of NaCl in 2.20 lit of solution.

Solution:

Weight = 284 grams

Molecular weight = (1 x 23)Na + (1 x 35.5)Cl = 58.5

Moles of NaCl =NaClofweightMolecular

NaClofWeight=

58.5

284= 4.85 gmoles

Molarity (M) =SolutionofLitre

SoluteofMoles=

Litre2.20

gmoles4.85= 2.205 M

1.3.2 Normality (N)

Normality (N) is the number of equivalents of solute per litre of solution.

Equivalent:

One equivalent is the amount of acid that can give one mole ofhydrogen (H+) ions.

One equivalent is the amount of a base that can give one mole ofhydroxide (OH

-) ions.

Example 1:

One equivalent of H2SO4 =usedIonsHydrogenofMoles

SOHofweightMolecular 42

=2

gmoles16)x(432)x(11)x(2 OxygenSulfurHydrogen

= 49 gequivalents


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BASIC CALCULATIONS

Example 2:

One equivalent of Ca(OH)2 = usedIonsHydroxideofMoles

Ca(OH)ofweightMolecular 2

=2

gmoles1)x(216)x(240)x(1 HydrogenOxygenCalcium

= 74 gequivalents

Example 3:

One equivalent of Na2SO4 =replacedbetoNaofMoles

SONaofweightMolecular 42

=2

gmoles16)x(432)x(123)x(2 OxygenSulfurSodium

= 71 gequivalents

Calculation of Normality

Example:

Calculate the normality of an aqueous solution of phosphoric acidcontaining 275 g of H3PO4 in 1.20 L solution in reactions that replaceall three hydrogen ions.

Solution:

Molecular weight of H3PO4 = (3 x 1)H + (1 x 31)P + (4 x 16)O

= 98 gmoles

One equivalent of H3PO4 =3

98= 32.7 gequivalents

Number of equivalents of H3PO4 in solution = g32.7

g274

= 8.4


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BASIC CALCULATIONS

Normality =solutionoflitre

soluteofsEquivalent

=1.28.4

= 7.00 N

1.4 STOICHIOMETRIC CALCULATIONS USING CHEMICALEQUATIONS

Stoichiometry deals with calculation about the masses (sometimesvolumes) of reactants and products involved in a chemical reaction.

There are a number of methods available for solving stoichiometryproblems. The method we consider the best is the mole method.Three (3) basic steps are involved.

Step 1: Calculate moles of the known Substances

Step 2: Using balanced equation, calculate moles of unknownsubstances.

Step 3: Calculate quantities (mass or volume) of unknowns.

1.4.1 Type of Stoichiometry Problems

There are 3 types of stoichiometry problems:

3. Mass Mass (weight weight)

4. Mass Volume or Volume Mass

5. Volume - Volume


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BASIC CALCULATIONS

1.4.2 Mass Mass Stoichiometry

In this type, quantities are given or asked for in a mass units

Example:

Calculate the number of grams of oxygen required to burn 72 g of C2H6to CO2 and H2O.

C2H6 + O2 CO2 + H2O

Solution:

First write the balanced equation:

2C2H6 + 7O2 4CO2 + 6H2O

Step 1: Calculate moles of C2H6

Moles of C2H6 =HC 1)x(612)x(2

g72

= 2.4 moles

Step 2: Calculate the moles of oxygen needed.

From the balanced equation: 2C2H6 + 7O2 4CO2 + 6H2O

2 moles of C2H6 needs 7 moles of oxygen. Therefore,

Moles of oxygen needed = 2.4 moles C2H6 x62

2

HCmoles2

Omoles7

= 8.4 moles O2

Step 3: Calculate the oxygen needed

Grams of oxygen needed = Moles of O2 x Molecular weight of O2

= 8.4 x 32 = 268.8 grams


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BASIC CALCULATIONS

1.4.3 Mass Volume Stoichiometry

In these types of problems, either the known or unknown is a gas.

Example:

Calculate the volume in litres of O2 measured at 0C and 1 ATM whichcould be obtained by heating 28 g of KNO3.

Solution:

First write the balanced equation.

2 KNO3 2 KNO2 + O2

Molecular weight of KNO3 = (1 x 39)K + (1 x 14)N + (3 x 16)O = 101

Step 1: Calculate moles of KNO3

Moles of KNO3 =101

grams28= 0.277 moles

Step 2: Calculate the moles of oxygen produced.

From the balanced equation: 2 KNO3 2 KNO2 + O2

2 moles of KNO3 produced 1 mole of oxygen. Therefore,

Moles of oxygen produced = 0.277 moles KNO3 x3

2

KNOmoles2

Omoles1

= 0.1386 moles O2

Step 3: Calculate the volume of oxygen produced

At 0C and 1 ATM (STP), 1 mole of O2 occupies 22.4 L

Volume of oxygen produced = 0.1386 Moles of O2 x2Omole1

L22.4

= 3.1 L O2

at STP


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BASIC CALCULATIONS

1.4.4 Volume Volume Stoichiometry

This is based on Gay Lussacs law of combining volumes which

states that At the same temperature and pressure, whenever gasesreact or form, they do so in the ratio of small whole numbers ofvolume.

Example:

CH4 (g) + 2 CO2 (g) CO2 (g) + 2 H2O

All compounds are in the gaseous state and at the same temperature

and pressure.

1 volume of CH4 reacts with 2 volumes of O2 to produce 1 volume ofCO2 and 2 volumes H2O vapor. The ratios are 1:2:1:2 forCH4:O2:CO2:H2O.

in solving volume volume stoichiometry problems, steps 1 and 3 arenot necessary; only step 2 is required.

Example:

Calculate the volume of O2 required and volumes of CO2 and H2Oproduced from the complete combustion of 1.5 L of C2H6. All Volumesare being measured at 400C and 1 ATM.

Solution:

Balanced equation:

2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (g)

All substances are gases measured at same temperature andpressure.


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BASIC CALCULATIONS

Volume of O2 needed = 1.5 L C2H6 x62

2

HCofL2

OofL7

= 5.25 L of O2 (g)

Volume of CO2 produced = 1.5 L C2H6 x62

2

HCofL2

COofL4

= 3.0 L of CO2 (g)

Volume of H2O produced = 1.5 L C2H6 x62

2

HCofL2

OHofL6

= 4.5 L of H2O

(g)


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BASIC CALCULATIONS

1. Calculate Number of moles in 71 grams of HCl.

2. Mass of 3 gmoles of NaOH.

3. A solution of 10 mass % NaCl. Find mole percent of saltin solution.

4. Mole percent of CaSO4 in hard water = 2.5 % mole.Find mass percent of CaSO4 in hard water.

5. Calculate normality (N) of 12.1 grams of H2SO4 in 750mL of solution in reactions that replace both hydrogenions.

6. Calculate the normality of 14.1 grams of Na2SO4 in 625mL of solution in reactions that replaces both sodiumions.

7. Find the molarity of a solution containing 300 grams ofCaCl2 in 500 mL solution.

8. Calculate the number of grams of Zinc chloride that canbe prepared from 340 grams of zinc.

Zn(s) + 2 HCl(Aq) ZnCl2 (Aq) + H2 (g)

9. How many moles of KCLO3 could be prepared from 24.7L of Chlorine gas at STP?

3 Cl2 (s) + 6 KOH(Aq) KclO3 (Aq) + 3 H2 (g)

10. Assuming STP, how many litres of oxygen are neededto produce 19.8 L of SO3 according to this balanceequation?

2 SO2 (g) + O2 (g) 2 SO3 (g)

EXERCISES


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BASIC CALCULATIONS

1.5 MATERIAL BALANCE

A mass and energy balance is a conceptual tool for studying the way inwhich a process is put together.

Applications

Test different design concepts Conduct energy audits of existing and proposed facilities Determine utility requirements for a range of plant operating

conditions

Calculate raw materials requirements for different product mixes Conduct process change impact studies

Benefits

Optimise process design A mass and energy balance produces a graphical consensus

document Optimise energy use

1.5.1 Material Balancing

Generally speaking, material balancing is a method of calculating theamounts of substances, called reactants, that must be put intoprocess in order to produce the desired amounts of products. Ingeneral the term reactants refers to the starting materials in achemical reaction.

1.5.2 Material Balance Calculations in a Steady State System

In a steady state system the weight or mass of the material at the startof the system is the same as that remaining at the end of the system.A short way of saying this is that the Input equals the Output. Thefollowing diagram.

Mass input Process Mass output

Mass or weight at start of a process = Mass or weight at the finish ofthe process.


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BASIC CALCULATIONS

The process may well change the chemical composition of the startingmaterial but the relative masses of the process products will still beequal to the masses of the starting materials.

In a steady state process the masses in a process will be constant, thismeans that the masses of the start material will be in BALANCE withmasses at the end. Calculations using the start material masses willallow the expected masses of the products to be found. If the finalmasses are different from the start masses then the process must beleaking or producing unexpected products that are not being weighed.

PROCESS BLOCK DIAGRAMS

Squares or rectangles are used to indicate a part of the process. Thisis a simple way to show where in the process reactions are occurring.

Mass input Mass output

Example: Calculation is sugar dissolved into a Tank of water.

In the tank is 4m3 of water and 200 kilograms of sugar. Thecalculation is to find the total mass of sugar solution and thepercentage mass of the sugar in the solution.

To understand the calculation we make a block diagram of the systemsimilar to the following: The arrows indicate the direction of theprocess.

Sugar200 kg Sugar solution

4200kgWater4000kg

This above diagram shows the actions of putting water into a tank andadding sugar. Notice that the start materials sugar and water are onthe left and the reactor tank is central with the products on the right.This is the usual format to be used for block diagrams. Arrows

indicates the input and output into and from a reactor.

Process Reactor

Tank ofsugar

solution


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BASIC CALCULATIONS

The calculation is simplified by the diagram as 200 kilograms of sugaris weighed into 4 cubic meters of water. One cubic meter of water hasa mass of 1000 kilograms. The system is in steady state so the

input must equal the output. The total mass is 4 times 1000 plus the200 kilograms of sugar, which equals 4200 kilograms.

Mass of water = 4 m3

3m1

kg1000= 4000 kg

Mass of sugar = 200 kg

Mass of sugar solution = 4000 kg + 200 kg = 4200 kg

The percentage sugar in the sugar solution will be the mass of sugardivided by the total mass of sugar and water.

% sugar = waterofmasssugarofmass

sugarofmass

100

% sugar = 4000200

200

100 = 4.7 %


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BASIC CALCULATIONS

Defining a System using Block Diagram

A continuous process may contain many reaction vessels each havinga different input and output value. The total input at the start must stillequal the total output at the finish. Each reaction vessel may bedescribed as a square or rectangle with input stated to the left andoutput on the right. Arrows indicate the direction of the process withthe relative input flow rates and output flow rates.

S1 or 100 kg/hr S3 or 220 kg/hr

S2 or 120 kg/hr

S1 = 100 kg/hr

S2 = 120 kg/hr

S3 = 220 kg/hr

MULTIPLE REACTORS A & B50 m3/hr

50 m3/hr

110 m3/hr

60 m3/hr60 m3/hr

Reaction Vessel

A B


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BASIC CALCULATIONS

MULTIPLE STAGE MULTIPLE REACTORS

Predictions can be made of the flow rates at different stages of asteady state system from flow rates at other reactors.

A C G

E

HB F

D I

Flow rates

A = 500 m3/hr

B = 400 m3/hr Total input at A + B = output atG + H + I = 900 m3/hr

C = 450 m3/hr

D = 450 m3/hr Total output at C + D = outputat G + H + I

E = 250 m3/hr Flow rate of C G = E

F = 250 m3/hr Flow rate of D - I = F

G = 200 m3/hr

H = 500 m3/hr and Flow rate E + F = H

I = 200 m3/hr


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BASIC CALCULATIONS

1. Calculate how much is the Feed as per given in the diagram below.

2. Feed pretreatment unit in Gas processing plants receives mixture ofhydrocarbons with impurities such as water and chlorine gas. Giventhe data and diagram below, calculate how many hydrocarbon gasesin k g s/hr are coming out of the unit.

INLETSEPARATOR

PHASE 1

GAS5 tons/hr

PHASE 1LIQUID

10 tons/hr

ASSOCIATEDGAS

20 tons/hr

WATER50 kgs/hr

CHLORIDESCRUBBER

NaOHsolution

25 kgs/hr

WASTEWATER40 kgs/hr

SEPARATORHYDROCARBON

GAS? kgs/hr

DECANTERDRUM

WATER30 kgs/hr

COALESCERDRUM

WATER15 kgs/hr

ADSORBER

WATER5 kgs/hr

HYDROCARBONLIQUID

20 tons/hr

20 kgs/hr

60 kgs/hr

40 kgs/hr30 kgs/hr

Feed =?

AB

C

D

EXERCISES


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BASIC CALCULATIONS

3. Reflux in the distillation towers provides cooling at the topmost partof the tower and it increases the % yield of high octane fuel. Giventhe data and diagram below, calculate the reflux rate in tons/hr.

4. Methyl Tertiary Butyl Ether or MTBE is used as an additive for liquidfuel. It contains 18.2 wt% O2. It is colorless and odorless. It provideseasy ignition to vehicles because of its oxygen content. Given the dataand diagram below, calculate how many tons/hr of methanol is neededto produce 26 tons/hr of MTBE. Overhead of finishing reactorconstitutes unreacted methanol and isobutylene plus side reaction ofdimethyl ether. Overhead flowrate is set at 2% of total MTBE flow.

STATICMIXER

MAINREACTOR

FINISHINGREACTOR

METHANOL? tons/hr

ISOBUTYLENE17 tons/hr

MTBE26 tons/hr

Unreacted METHANOLUnreacted ISOBUTYLENE

Dimethyl Ether2 % of total MTBE Flow

ATMOSPHERICPIPESTILL

CRUDE OIL50 tons/hr

REFLUXDRUM

ASPHALT

2 tons/hr

STRIPPER

STRIPPER

REFLUX

? Tons/hr

HIGH OCTANEFUEL

27 tons/hr

KEROSENE12 tons/hr

DIESELFUEL

5 tons/hr


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BASIC CALCULATIONS

5. Dehydrogenation is an endothermic reaction. It converts propane

into propylene with the presence of a platinum on an alumina base

catalyst. Propane is fed to the Oleflex reactor and propylene is theresulting product. Since efficiency is not 100 %, some of thepropane are still unreacted. Given the data and diagram below,calculate how much propane is fed into the Oleflex reactor.

Hint: See the definition of acatalyst.

CATALYST100 kgs/hr

SPENT

CATALYST

OLEFLEXREACTORPROPANE

? Tons/hrPROPYLENE

35 Tons/hr

Unrelated PROPANE15 Tons/hr

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