Reading Materials: Chapter 4 Describing Physical Quantities LECTURE 9 Chapter 4 CHEM ENG 1007

Objectives At the end of these lectures you should be able to: Explain in your own words the meaning of gram-mole, lb-mole and kilogram-mole (Lecture 7) Calculate two of the quantities mass (or mass flow rate), volume (or volumetric flow rate), and moles (or molars flow rate) from a knowledge of the third quantity for any species of known density and molecular weight. (Lecture 8) Transform a material from one measure of concentration to another including mass/volume, moles/volume, ppm, ppb and molarity (Lecture 8) Calculate the average molecular weight of the mixture Convert the composition of a mixture from mass fraction (or mass percent) to mole fraction (or mole percent) and vice versa. Chapter 4 CHEM ENG 1007

§4.2.3 Mixture Concentration In the calculation of the mass fraction and mole fraction, the same units for mass and moles must be used in the numerator as in the denominator so that the calculated fraction is without units. Summary: Chapter 4 CHEM ENG 1007

Learning Check Two compounds, one with a high molecular weight and one with a low molecular weight, each comprise 50 mole% of the same mixture. Which has the greater mass fraction in the mixture Solution: Answer: the stream that has higher molecular weight will have greater mass fraction. Chapter 4 CHEM ENG 1007

Learning Check A solution of salt dissolved in water is diluted with additional water. With each of the following variables, indicate whether the dilution process will cause the value of the variable to increase, decrease, or stay the same. Support your answer Solution: Chapter 4 CHEM ENG 1007

Average/Mean Molecular Weight From mole fraction:   From mass fraction: Chapter 4 CHEM ENG 1007

Illustration 7 Calculate the average molecular weight of air (1) From its approximate molar composition of 79% N2, 21% O2 Solution: Chapter 4 CHEM ENG 1007

Illustration 7 Calculate the average molecular weight of air (2) From its approximate mass composition of 76.7 wt% N2, 23.3 wt% O2 Solution: Chapter 4 CHEM ENG 1007

Illustration 7 (3) Assume we have 100 mol of air mixture, what is this amount in grams? From (1) molar composition of 79% N2, 21% O2 Solution: Chapter 4 CHEM ENG 1007

Illustration 7 (3) Assume we have 100 mol of air mixture, what is this amount in grams? From (1) molar composition of 79% N2, 21% O2 Solution: (alternative method) Chapter 4 CHEM ENG 1007

System Properties Quantities necessary to describe the state or condition of a system. Extensive properties: depend on the size of system eg. weight, force, energy, flow rate Intensive properties: are independent of the mass or size of system, eg. temperature, pressure, density, viscosity, concentration, mass fraction, mole fraction, etc. Chapter 4 CHEM ENG 1007

Learning Check A solution of NaOH in water flows in a stream, and the mass flow rate of the stream suddenly increased. For each of the following properties of the stream, indicate whether the increase of the flow rate will cause the property to increase, decrease, or remain the same each case, explain your answer. Solution: Chapter 4 CHEM ENG 1007

Learning Check A solution of NaOH in water flows in a stream, and the mass flow rate of the stream suddenly increased. For each of the following properties of the stream, indicate whether the increase of the flow rate will cause the property to increase, decrease, or remain the same each case, explain your answer. Solution: Chapter 4 CHEM ENG 1007

Example 4.5 From the acid-neutralization problem, the volumetric flow rate of the HCl solution coming from our manufacturing process is 11,600 L/hr, and the average molarity of HCl in that stream is 0.014 M. How many mol of HCl are in 88 m3 of the solution? V = 88 m3; cHCl = 0.014 M = 0.014 mol/L; n = ? mol Chapter 4 CHEM ENG 1007

Example 4.5 b) How many mole of HCl are flowing from the process per minute when the volumetric flow rate of the solution is 11,600 L/hr? Chapter 4 CHEM ENG 1007

Example 4.5 What is the mass fraction of HCl in the solution? V and c were given; xHCl=? Chapter 4 CHEM ENG 1007

Illustration 8 A 2.25 L gas cylinder holds a mixture of hydrocarbons gases: methane, ethane and propane. The mass of the gas mixture in the cylinder is measured at 2.0 g. The mixture is at 25oC & 1 atm. Individual masses are given in the below table. Constituent mi (g of i) CH4 1.0 C2H6 0.6 C3H8 0.4 Total 2.0   Chapter 4 CHEM ENG 1007

Illustration 8 a) Mass fraction of methane? Same procedure for ethane and propane. Constituent (g of i/g total) CH4 0.5 (i.e., 1/2) C2H6 0.3 (i.e., 0.6/2) C3H8 0.2 (i.e., 0.4/2) Total 1.0 Chapter 4 CHEM ENG 1007

Illustration 8 b) Molecular weight of Propane? Very wrong!!! Constituent MW (g/mol) CH4 16 (=12+4) C2H6 30 (=2 x12+6x1) C3H8 44 (=12x3+8x1) Total - Average 21.84 Very wrong!!! Chapter 4 CHEM ENG 1007

Illustration 8 c) Mole fraction of ethane? Same procedure for methane and propane. Total number of moles is Chapter 4 CHEM ENG 1007

Illustration 8 Check result by summing mole fractions? Constituent yi CH4 0.682 C2H6 0.218 C3H8 0.099 Check result by summing mole fractions? Thought Problem: Why don’t the mole fractions sum to 1? Chapter 4 CHEM ENG 1007

Illustration 8 d) Average Molecular Weight? By mole fraction:   By mass fraction: Chapter 4 CHEM ENG 1007

Illustration 8 e) Mass concentration of propane: Apply same procedure for methane and ethane Constituent Ci (kg of i/m3) CH4 0.444 C2H6 0.267 C3H8 0.178 Chapter 4 CHEM ENG 1007

§4.2.3 Conversion between mole fraction and mass fraction Strategy for converting mass fractions/percentage to mole fractions/percentages Assume a basis of 100 mass units From the known mass fractions, calculate the mass of each species Using the molecular weights, convert the mass into moles for each species Compute the desired moles fractions or percentages. Chapter 4 CHEM ENG 1007

§4.2.3 Conversion between mole fraction and mass fraction Assume a basis of 100 moles From the known mole fractions, calculate the number of moles of each species Using the molecular weights, convert the moles into mass for each species Compute the desired mass fractions or percentages. Note: When performing such conversions, the best way is to tabulate your data Chapter 4 CHEM ENG 1007

Example 4.6 Air has the following approximate mole%: N2: 78.03 mole%; O2: 20.99 mole%; Ar: 0.94 mole% Solution: Assume 100 mol of Air Convert from mole to mass Chapter 4 CHEM ENG 1007

Example 4.6 Hence, the total number of grams is mT = 2186 + 672 + 38 = 2896 g Chapter 4 CHEM ENG 1007

Example 4.6 Alternatively, we can set up our work as shown in the below table Basis: 100 mol Component yi ni Mwi mi (ni.Mwi) xi N2 0.7807 78.07 28 2186 0.7548 O2 0.2099 20.99 32 672 0.2320 Ar 0.0094 0.94 39.95 38 0.0131 Total 1 100 2896 0.9999 Chapter 4 CHEM ENG 1007