Lecture 2. Moles, Density, Specific Gravity, Fraction, Pseudo-

Molecular Weight of Air, Concentration and Flow Rate

2.1 The Mole

SI system : a mole (or g mol) = 6.022 x 1023 (Avogadro’s Number)

Small pile6.02 x 1023

g mol

Medium pile453.6 x 6.02 x 1023

lb mol

Large pile1000 x 6.02 x 1023

kg mol

2.1 The Mole• atomic mass(weight) / molecular mass(weight) – mass per mole

(kg mol, kilomole, kmol)

Examples Lecture 2.1

Carbon Atomic mass = 12

1 small pile -> mass = 12 g(g mol)

2 medium pile -> mass = 453.6 x 12 g(453.6 g mol)(lb mol)

3 large pile -> mass = 1000 x 12 g(1000 kg mol)(kg mol)

Examples Lecture 2.1

(E2.2, DMH)

If a bucket hold 2.00 lb of NaOH, how many does it contain in:(a)lb moles of NaOH?(b)g moles of NaOH?

Ans. (a) 0.050 lb mol NaOH, (b) 22.7 g mol

Examples Lecture 2.1

(E2.3, DMH)

How many pounds of NaOH are in 7.50 g mol of NaOH?

Ans. 0.661 lb NaOH

Suggestion

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

2.2 Density

ρ = density = mass/volume = m/V

2.2 Density

• Packed bed of solid particles containing void spaces

2.2 Density

• Packed bed of solid particles containing void spaces

Examples Lecture 2.2

(p. 49, DMH)

Given the density of n-propyl alcohol is 0.804 g/cm3, what would be the volume of 90.0 g of the alcohol?

Ans. 112 cm3

(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. 46.2 lb

Note

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

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,

62.43 lb/ft3 at 4 oC)• Gases• Reference substance -> air or others

• To be precise, temperature is specified.

• SI system– Density of water at 4 oC is very close to 1.0000

g/cm3 -> numerical values of sp. gr. and density are equal.

oAPI or API GravityAmerican Petroleum Industry

Note : (Baume, oBe), (Twaddel, oTW)

Examples Lecture 2.3

(E2.4, DMH)

If a 70% (by weight) solution of glycerol has a specific gravity of 1.184 at 15 oC, what is the density of the solution in (a) g/cm3? (b) lbm/ft3? And (c) kg/m3?

Ans. (a) 1.184 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) 1.184 x 103 kg solution/m3

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. 8.11 ft3

Examples Lecture 2.3

(P4, DMH)

A solution in water contains 1.704 kg of HNO3/kg H2O, and the solution has a specific gravity of 1.382 at 20 oC. What is the mass of HNO3 in kg per cubic meter of solution at 20 oC?

Ans. 870 kg/m3 solution

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%

Notes

• 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

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 = 0.500

Mole fraction : H2O = 0.69, NaOH = 0.31

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 gMW air = 2900 g air / 100 g mol air = 29.0

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)

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

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

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. 1.68 x 10-3 ppm

2.7 Flow Rate

• Mass flow rate

• Volumetric flow rate

• Molar flow rate

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

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. 0.654 kg mol/hr

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