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Home , Mixture, Mole fraction, Volume fraction

GP-CPC-01

UNITS – BASIC IDEAS – COMPOSITION 11-06-2020

Prof.G.Prabhakar Chem Engg, SVU GP-CPC-01 UNITS – CONVERSION (1) ➢ A two term system is followed. A base unit is chosen and the number of base units that represent the quantity is added ahead of the base unit. Number Base unit Eg : 2 kg, 4 meters , 60 seconds

➢ Manipulations Possible :

• If the nature & base unit are the same, direct addition / subtraction is permitted 2 m + 4 m = 6m ; 5 kg – 2.5 kg = 2.5 kg

• If the nature is the same but the base unit is different , say, 1 m + 10 c m both m and the cm are length units but do not represent identical quantity, Equivalence considered 2 options are available. 1 m is equivalent to 100 cm So, 100 cm + 10 cm = 110 cm 0.01 m is equivalent to 1 cm 1 m + 10 (0.01) m = 1. 1 m

• If the nature of the quantity is different, addition / subtraction is NOT possible.

Factors used to check equivalence are known as Conversion Factors. GP-CPC-01 UNITS – CONVERSION (2)

• For multiplication / division, there are no such restrictions. They give rise to a set called derived units

Even if there is divergence in the nature, multiplication / division can be carried out. Eg : Velocity ( length divided by time ) Mass flow rate (Mass divided by time) Mass Flux ( Mass divided by area (Length 2) – time). Force (Mass * Acceleration = Mass * Length / time 2)

In derived units, each unit is to be individually converted to suit the requirement = 500 kg / m3 . What is its equivalence in g / cm3 1 kg = 1000 g ; 1 m = 100 cm 500 kg * 1000 g * m3 = 500 * 10-3 g / cm3 m3 1 kg (100)3 cm3 GP-CPC-01 Vander waals equation ( P + a/V2) ( V-b) = RT

Units of b : Since it is being subtracted from V, b will have the same units as V Units of a : Since a/V2 is being added to P, this group shall have the units of P. Further a is divided by V2 and V is not available in the units of the group, units of a will be P – V2

Equations that have equal units on either side are labelled dimensionally consistent Mass Transfer Flux = Coefficient * Driving Force ; Coefficient will carry the units of flux divided by driving force = Rate Constant * n Units of Rate constant = Rate / Concentration n

Certain equations developed from experimental data are called empirical equations. Constants present in such empirical equations will have dimensions and they must be strictly adhered to.

Proper representation of quantities / conditions , number followed by base unit, is a good habit. It helps to check the correctness of the calculational approach. Sometimes , they even suggest the direction of calculation GP-CPC-01

Concept of ‘’ ➢ Certain events in chemical processing happen in direct proportion to the number of molecules of the system. Eg : Molecular , Chemical Reaction… It is convenient and purposeful to use a dimension in such cases. MOLE

➢ Avagadro’s number defines a mole of a substance , that contains 6.022 *1023 units ( molecules) of that substance. Further, it is related to mass through molecular weight by a simple relation 1 Mole = Mass / Molecular weight

➢ Mole, thus, can be considered as a secondary unit of mass The units of mole are dependent on the units of mass Mass in grams - g mol or mol Mass in pounds( lb) - lb mol Mass in kilograms - kilogram mole or kilo mole

➢ Conversion of one to another is possible with the conversion factors used for mass 1 kg = 2.205 lb = 1000 grams 1 k mol = 2. 205 lb mol = 1000 g mol GP-CPC-01 COMPOSITION ➢ All steps in industrial processing involve of substances

➢ All the members ( components) of a are in the same - A homogeneous mixture or . If they remain in different phases - A heterogeneous mixture.

➢ Composition indicates the relative amounts of components present in a mixture, homogeneous or heterogeneous

➢ Composition is a critical parameter in all the computations related to industrial processing Composition keeps varying in most of the processing steps – unit operations & unit processes GP-CPC-01 COMPOSITION ➢ Composition is specified in a number of ways. A clear understanding of the ideas and their inter relationship is of paramount importance. For any specification, a basis is necessary. Hence, different ways

Consider a mixture consisting of components – A, B & C

➢ Mass fraction – Mass percent Mass is the basis Mass fraction of A = Mass of A / Total mass = Mass of A / Mass of (A + B + C ) Sum of Mass fractions = 1 Mass Percent of A = 100 * Mass of A / Total mass = 100 * Mass of A / Mass of (A + B + C ) Sum of Mass Percent s = 100 Independent of temp & Pres. Usually used to express the compositions of & mixtures . Default Composition ( 2 / 3) GP-CPC-01 ➢ – Mole percent Mole is the basis Mole fraction of A = Moles of A / Total moles = Moles of A / Moles of (A + B + C ) Sum of Mole fractions = 1 Mole Percent of A = 100 * Moles of A / Total moles = 100 * Moles of A / Moles of (A + B + C ) Sum of Mole Percent s = 100 Independent of temp & Pres. Suitable for any mixture, more useful in case of mixtures

➢ fraction – Volume percent Volume is the basis Volume Percent of A = 100 * Volume of A / Total Volume = 100 * Volume of A / Volume of (A + B + C ) Sum of Volume Percents = 100 Volume is temp. and pres. Dependent, but it is present in both the numerator and the denominator. It will be affected in the same way for both

Ratio is considered to be independent of temperature & . Default for gas mixtures.

Volumetric Composition coincides with molar composition for IDEAL GP-CPC-01 Composition ( 3 / 3) ➢ Mass Ratio Mass of A / Mass of (B+C) , Mass of B / Mass of (C + A ) ➢ Mole Ratio Moles of A / Moles of (B + C ) Moles of B / Moles of (A + C) ➢ Hybrid Moles of A / Mass of ( B+ C ) Mass of B / moles of ( C +A) These are normally preferred in case of one component affected ( Eg : Gas Absorption, Humidification) These can sometimes assume a value greater than 1

➢ Density – Molar Density These give the mass or mole contained per unit volume Their reciprocals – and , give volume of unit mass or mole These are defined to suit specific requirements of the case

➢ Concentration units used for Liquid Mixtures Molarity , M = Number of g mol contained in a Liter of solution Normality , N = Number of gram equivalents present in a Litre of solution , m = Number of gram moles of solute present per 1000 grams

It is to be reiterated that a good grasp of composition specification will be of great help in all process calculations, mass transfer, reaction, everywhere… GP-CPC-01

• Acceleration of a body is given as 1 m/ s2. Express the value in ft / min2

1 m = 3.8 ft , 1 min = 60 sec

Acceleration = 1 m * 3.8 ft * 60 * 60 s2 = 13680 ft s2 m min2 min2 • Thermal conductivity of a steel is given as 14.7 BTU / hr – ft – OF. Express the value in cal / sec-cm-OC

1 BTU = 252 cal 1 hr = 3600 sec 1 ft = 30.48 cm 1.8 OF = 1 OC

Thermal conductivity = 14.7 BTU * 252 cal * 1 hr * 1 ft * 1.8 OF = 1.85 Cal hr – ft – OF 1 BTU 3600 sec 30.48 cm 1 OC sec-cm-OC

• Convert 1 .987 cal / g – OC to BTU / lb - OF

1 BTU = 252 cal 1 lb = 453.6 g 1 OC = (5 / 9 ) 1 OF

1 .987 cal * 1 BTU * 453.6 g * 1 OC = 1.987 BTU g – OC 252 cal 1 lb 1.8 OF lb - OF GP-CPC-01 ➢ Air has a composition of 79 % and 21 % , by volume. Express the composition on the basis of mass

Volumetric composition is taken as molar composition for ideal gases Required : Molecular weight of nitrogen (28) , molecular weight of oxygen (32) Basis : 1 mol of Air Mass = Moles * Molecular weight

Component Moles Mol.Wt Mass Mass %

Nitrogen 0.79 28 22.12 100 *22.12/28.84 = 76.7

Oxygen 0.21 32 6.72 100 * 6.72/28.84 = 23.3

28.84

• 28.84 obtained in this calculation is the mass of 1 mol of air. For mixtures, such values are called Average Molecular Weight. They are useful to convert moles to mass or vice versa of mixtures. Average Molecular Weight = Σ xi * wi ( mole fraction of a component multiplied by its molecular weight, summed over all the components)

• Atmospheric nitrogen is a mixture of nitrogen, argon, …. Its molecular weight is usually taken as 28.2 GP-CPC-01 ➢ An has 25 g of salt per g . What is the mole fraction of the salt? Use Molecular weight of salt = 40 Molecular weight of water = 18

Data provided is mass of salt / mass of water . Mole per total moles is required. Moles = Mass / Mol.wt Component Mass Mol.Wt Moles Mole fraction Salt 25 g 40 25 / 40 = 0.625 0.625/0.68 = 0.919 Water 1 g 18 1 / 18 = 0.055 0.055/0.68 = 0.081 0.680

26 g ( 25 g salt + 1 g water ) are equivalent to 0.68 g moles. Average Mol.Wt = 26 / 0.68 = 38.23

➢ An aqueous solution has 25 g of salt per 100 g of solution. What is the mole fraction of the salt? Use Molecular weight of salt = 40 Molecular weight of water = 18

1) Data provided is mass of salt / Total mass of solution. Mole per total moles is required. Moles = Mass / Mol.wt Component Mass Mol.wt Moles Mole fraction Salt 25 40 25/40 = 0.625 0.625 / 4.792 = 0.13 Water 100 – 25 = 75 18 75/18 = 4.167 4.167 / 4.792 = 0.87 4.792 Average Mol. Wt = 100 / 4.792 = 20.87 Molecular weight is mass of 1 mole of a substance GP-CPC-01 O ➢ A commercial acid at 20 C, has 36.8 mole percent H2SO4 and the balance water. Express the composition in Molarity, Molality and Normality. Take density of the solution as 1.14 g /cc Basis : 1 mol of solution Component Moles Mol.Wt Mass, in g H2SO4 0.368 98 36.064 H2O 0.632 18 11.376 TOTAL 47.44 1 kmol of solution has a mass of 47.44 Its volume is 47.44 g = = 41.61 cc ( 0.04161 L) 1.14 g/cc

Molarity = No. of g mol / L = 0.368 moles / 0.04161 L = 8.84 M

Equivalent Weight of H2SO4 = 98 / 2 = 49 , No,of Equivalents = 36.064 / 49 = 0.736 Normality = 0.736 equivalents / 0.04161 L = 17.688 N

Molality = 0.368 gmol H2SO4 * 1000 = 32.35 gmol H2SO4 = 32.35 m 11.376 g water 1000 1000 g water ( solvent) Basis of Calculation : The quantity with respect to which, all values are reported. Sometimes specified in the question statement or arbitrarily taken as 1 or 100 ( Convenience) GP-CPC-01

➢ It is desired to prepare 5 L of 2.8 m ( molal ) solution of sodium chloride. What mass of sodium chloride needs to be taken Density, ρ = 1 + 0.0044 w, where ρ is in g / cc and w is the weight percent

Basis of Calculation : 2.8 molal NaCl solution - 5 L Molecular weight of NaCl = 58.5 2.8 mol NaCl = 2.8 * 58.5 = 163.8 g NaCl 1000 g water ( Solvent)

= 163.8 g NaCl * 100 = 14 % ( weight percent) 163.8 + 1000 g solution

Density = 1 + 0.0044 * 14.07 = 1.0616 g / CC 5 L ( 5000 CC ) of solution are required. Equivalent to 5000 cc * 1.0616 g/cc = 5308 g NaCl required = 5308 * 0.14 = 743.12 g GP-CPC-01

➢ Estimate the available nitrogen in a 95 % pure, commercial ammonium sulfate

Ammonium Sulfate – (NH4)2 SO4

100 kg of 95 % commercial ammonium sulfate i.e 95 kg of (NH4)2 SO4.

Mass of nitrogen = 2 * 14 = 28 Formula weight of (NH4)2 SO4 = (14 +4) * 2 + 32 = 4 * 16 = 132 Available Nitrogen = 95 kg * 28 /132 = 0.2015 kg/kg Theoretical available = 28/132 = 0.2121 kg /kg 100 Kg 1 kg of ammonium sulfate can provide 0.2414 kg of nitrogen to the . But, because of 5 % impurities, it offers 0.2293 kg/kg of nitrogen as 95 % pure salt

➢ Obtain the mass based composition of MgSO4.7 H20

Molecular Weight of MgSO4 = 24.32 +32 + 64 = 120.32 Molecular weight of 7H2O = 7 * 18 = 126

Mass percent of MgSO4 = 120.32 *100 = 48.85 Mass percent of water = 126 * 100 = 51.15 120.32 + 126 120.32 + 126

➢ Available nitrogen content of a commercial urea is 45 %. How pure is this urea?

Urea - NH2CONH2 Formula weight = 60 Nitrogen in urea = 28 Theoretical availability of nitrogen in urea = 28 /60 = 0.467 Actual availability = 45 / 100 = 0.45 Purity of commercial urea = 0.45 / 0.467 = 0.9636 or 96.36 % GP-CPC-01 ➢ A liquid mixture of components A, B & C is found to contain 25 mol % of B and 1.5 moles of C per mole of B. The molecular weights of A, B & C are 56, 58 & 72 , respectively. What is the molecular weight of the mixture? Basis : 1 mol of mixture Moles of B = Mole fraction of B = 0.25 Correspondingly, Moles of C = 1.5 * 0.25 = 0.375 Moles of A = 1 – ( 0.25 + 0.375 ) = 0.375 Average molecular weight of the mixture = (0.375 * 56 ) + (0.25* 58) + (0.375 * 72) = 62.5

➢ A gas mixture containing A, B and C analyses 40 mole % A, 18.75 mass % B and 20 mol % C. If the molecular weights of A and C are 40 and 50 respectively, what is the molecular weight of B ? Basis : 1 mol of mixture mole fraction of A = 0.4 mole fraction of C = 0.2 mole fraction of B = 1 – (0.4 + 0.2) = 0.4 Equiv. mass = 0.4 * 40 = 16 Equiv. Mass = 0.2 * 50 = 10 Since mass fraction of B = 0.1875 , mass fraction of ( A + C ) = 1 – 0.1875 = 0.8125 0.8125 mass fraction corresponds to 26 (= 16 + 10) Total mass = 26 / 0.8125 = 32 0.4 * 40 + 0.2 * 50 + 0.4 * MB = 32 MB = 15 GP-CPC-01

➢ A liquid mixture of components A, B & C is found to contain 10 kg of A, 25 mass % of B and 1.5 moles of C per mole of B. Molecular weights of A, B & C are 56, 58 & 72 , respectively. What is the molecular weight of the mixture?

Let X be the mass of mixture

This mixture contains 10 kg of A, 0.25 * X Kg of B and (0.25 * X / 58) * 1.5 * 72 kg of C ( 0.4655 *X kg of C) Now, 0.25 * X = 0.25 10 + 0.25 * X + 0.4655 * X

0.25 * X = 2.5 + 0.179 * X X = 35.21 Kg A 10 Kg 10 / 56 = 0.1786 kmol B 0.25 * 35.21 = 8.8025 kg 8.8025 / 58 = 0.1518 kmol C 0.4655 * 35.21 = 16.39 kg 16.39/ 72 = 0.2276 kmol Total = 0.558 kmol

Mass of 0.558 kmol is 35.21 Kg Mass of 1 mol = 35.21 / 0.558 = 63.1 = Average Molecular weight GP-CPC-01

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