1. Lecture 2. Moles, Density, Specific Gravity, Fraction, Pseudo-Molecular Weight of Air, Concentration and Flow Rate

2. 2.1 The Mole
SI system : a mole (or g mol) = x 1023 (Avogadro’s Number)
Small pile
6.02 x 1023
g mol
Medium pile
453.6 x 6.02 x 1023
lb mol
Large pile
1000 x 6.02 x 1023
kg mol

3. 2.1 The Mole (kg mol, kilomole, kmol)
atomic mass(weight) / molecular mass(weight) – mass per mole
(kg mol, kilomole, kmol)

4. Examples Lecture 2.1 Carbon Atomic mass = 12
small pile > mass = 12 g
(g mol)
medium pile -> mass = x 12 g
(453.6 g mol)
(lb mol)
large pile -> mass = 1000 x 12 g
(1000 kg mol)
(kg mol)

5. Examples Lecture 2.1 (E2.2, DMH)
If a bucket hold 2.00 lb of NaOH, how many does it contain in:
lb moles of NaOH?
g moles of NaOH?
Ans. (a) lb mol NaOH, (b) 22.7 g mol

6. Examples Lecture 2.1
(E2.3, DMH) How many pounds of NaOH are in 7.50 g mol of NaOH? Ans lb NaOH

7. Suggestion
Convert mass to the SI system first, then make the calculation.

8. 2.2 Density
ρ = density = mass/volume = m/V

9. 2.2 Density
Packed bed of solid particles containing void spaces

10. 2.2 Density
Packed bed of solid particles containing void spaces

11. Examples Lecture 2.2
(p. 49, DMH) Given the density of n-propyl alcohol is g/cm3, what would be the volume of 90.0 g of the alcohol? Ans. 112 cm3

12. (P2, DMH)
An empty 10 gal tank weighs 4.5 lb. What is the total weight of the tank plus the water when it is filled with 5 gal of water?
Ans lb

13. Note
Density of liquids and solids does not change significantly at ordinary conditions with pressure, but does change with temperature.

14. 2.3 Specific Gravity
Ratio of two densities – that of the substance of interest, A, to that of a reference substance
Liquids and solids
Reference substance -> water (1.000 g/cm3, 1000 kg/m3, lb/ft3 at 4 oC)
Gases
Reference substance -> air or others

15. To be precise, temperature is specified.
SI system
Density of water at 4 oC is very close to g/cm3 -> numerical values of sp. gr. and density are equal.

16. oAPI or API Gravity Note : (Baume, oBe), (Twaddel, oTW)
American Petroleum Industry
Note : (Baume, oBe), (Twaddel, oTW)

17. Examples Lecture 2.3
(E2.4, DMH) If a 70% (by weight) solution of glycerol has a specific gravity of at 15 oC, what is the density of the solution in (a) g/cm3? (b) lbm/ft3? And (c) kg/m3? Ans. (a) g solution/cm3 (b) (1.184 lb solution/ft3)/(1 lb water/ft3) x (62.4 lb water/ft3) = 73.9 lb solution/ft3 (c) x 103 kg solution/m3

18. Examples Lecture 2.3
(P2, DMH) The specific gravity of steel is 7.9. What is the volume in cubic feet of a steel ingot weighing 4000 lb? Ans ft3

19. Examples Lecture 2.3
(P4, DMH) A solution in water contains kg of HNO3/kg H2O, and the solution has a specific gravity of at 20 oC. What is the mass of HNO3 in kg per cubic meter of solution at 20 oC? Ans. 870 kg/m3 solution

20. 2.4 Fraction and Percent Mass fraction of A Weight fraction of A
Mole fraction of A
Percent of A = (Fraction of A) x 100%

21. Notes Sum of fraction = 1 Sum of percent = 100%
Mass fraction (or percent) = Weight fraction (or percent)
Sum of fraction = 1
Sum of percent = 100%
Common practice in industry
Composition of liquids and solids
generally -> mass (weight) percent or fraction
Composition of gases
generally -> mole percent or fraction

22. Examples Lecture 2.4
(E2.6, DMH) An industrial-strength drain cleaner contains 5.00 kg of water and 5.00 kg of NaOH. What are the mass (weight) fractions and mole fractions of each component in the drain cleaner container? Ans. Mass (weight) fraction : H2O = 0.500, NaOH = Mole fraction : H2O = 0.69, NaOH = 0.31

23. 2.5 Pseudo-Molecular Weight of Air
Assumption : Air = O2 (21 mol%) + N2 (79 mol%)
MW O2 = 32 N2 = 28.2
Pseudo-Molecular Weight of Air
Basis : 100 g mol Air
O2 21 (g mol) x 32 (g/g mol) = 672 g
N2 79 (g mol) x 28.2 (g/g mol) = 2228 g
Sum = 2900 g
MW air = 2900 g air / 100 g mol air = 29.0

24. 2.6 Concentration Mass per unit volume -> mostly for gases
g of solute/L, lb of solute/ft3, …
Moles per unit volume
g mol of solute/L, lb mol of solute/ft3, …
Molarity – g mol/L
Molality – mole solute/kg solvent
Normality – equivalents/L
ppm (parts per million), ppb (parts per billion)
Solids and liquids : mass (weight) fraction – mg/kg
Gases : mole fraction (Remember!)
ppmv (parts per million by volume)
ppbv (parts per billion by volume)

25. Examples Lecture 2.5 and 2.6
(E2.8, DMH) The current OSHA 8-hour limit for HCN in air is 10.0 ppm. A lethal dose of HCN in air is (from the Merck Index) 300 mg/kg of air at room temperature. (a) How many mg HCN/kg air is 10.0 ppm? (b) What fraction of the lethal dose is 10.0 ppm? Ans. (a) 9.32 mg HCN/kg air, (b) 0.031

26. Examples Lecture 2.5 and 2.6
(E2.9, DMH) A solution of HNO3 in water has a specific gravity of 1.10 at 25 oC. The concentration of the HNO3 is 15 g/L of solution. What is the (a) mole fraction of HNO3 in the solution? (b) ppm of HNO3 in the solution? Ans. (a) HNO3 = 3.90 x 10-5, H2O=1.00 (b) H2O = 13,640 ppm

27. Examples Lecture 2.5 and 2.6
(P3, DMH) The danger point in breathing sulfur dioxide for humans is 2620 μg/m3. How many ppm is this value? Ans x 10-3 ppm

28. 2.7 Flow Rate
Mass flow rate
Volumetric flow rate
Molar flow rate

29. Examples Lecture 2.7
(P1, DMH) Forty gal/min of a hydrocarbon fuel having a specific gravity of 0.91 flow into a tank truck with a load limit of 40,000 lb of fuel. How long will it take to fill the tank in the truck? Ans. 132 min

30. Examples Lecture 2.7
(P2, DMH) Pure chlorine enters a process. By measurement it is found that 2.4 kg of chlorine pass into the process every 3.4 minutes. Calculate the molar flow rate of the chlorine in kg mol/hr. Ans kg mol/hr

31. Problems Chapter 2
2.10, 2.13, 2.16, 2.19, 2.22, 2.34, 2.37, 2.40, 2.43, 2.46, 2.52