Units for Chemical Engineers

1

The Mole What is a mole?

For our purposes we say that a mole is a certain amount of material corresponding to a specified number of molecules, atoms, electrons, or any other specified types of particles.

In the SI systems a mole is composed of 6.022 x 1023 molecules. This number is known as Avogadro’s Number named in honour of the early 19th century Italian chemist.

However, for convenience in calculations and for clarity, chemical engineers will make use of other specifications for moles such as pound mole, or larger units such as the kg mol (kilomole, kmol, comprised of 1000 moles), and so on.

You will find that such nonconforming definitions of the amount of material will help avoid excess details in many calculations.

To keep the units straight, we will use the designation of mol or g mol for the SI mole.

What would a ton mole of molecules consist of?

One important calculation you should become skilled at is to convert the number of moles to mass and the mass to moles.

To do this you make use of the molecular weight – the mass per mole: �� =

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘šđ?‘œđ?‘™đ?‘’

Thus, the calculations you carry out are: đ?‘” đ?‘šđ?‘œđ?‘™ =

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘–đ?‘› đ?‘” đ?‘€đ?‘Š

đ?‘™đ?‘? đ?‘šđ?‘œđ?‘™ =

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘–đ?‘› đ?‘™đ?‘? đ?‘€đ?‘Š

and if we rearrange them: đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘–đ?‘› đ?‘” = đ?‘€đ?‘Š Ă— đ?‘”đ?‘šđ?‘œđ?‘™ đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘–đ?‘› đ?‘™đ?‘? = đ?‘€đ?‘Š Ă— đ?‘™đ?‘?đ?‘šđ?‘œđ?‘™ Therefore, from the definition of the molecular weight, you can calculate the mass knowing the number of moles or the number of moles knowing the mass.

Exercise 1: Use of Molecular Weights to Convert Mass to Moles If a bucket holds 2.00 lb of NaOH: a. How many pound moles of NaOH does it contain? b. How many gram moles of NaOH does it contain?

Professor Mo Zandi

2 Exercise 2: Use of Molecular Weights to Convert Moles to Mass How many pounds of NaOH are in 7.50 g mol of NaOH?

Values of the molecular weights are built up from the values of atomic weights based on a scale of the relative masses of the elements. The atomic weight of an element is the mass of an atom based on the scale that assigns a mass of exactly 12 to the carbon isotope 12C. The value 12 is selected in this case because an atom of carbon 12 contains 6 protons and 6 neutrons for a total molecular weight of 12.

A compound is composed of more than one atom, and the molecular weight of the compound is nothing more than the sum of the weights of atoms of which it is composed.

Molecular weights for mixtures You can calculate average molecular weights for mixtures of constant composition even though they are not chemically bonded if their compositions are known accurately. The next example shows how to calculate the fictitious quantity called the average molecular weight of air. Of course, for a material such as fuel oil or coal where composition may not be exactly known, you cannot determine an exact molecular weight, although you might estimate an approximate average molecular weight, which is good enough for most engineering calculations.

Example: Average Molecular Weight of Air Calculate the average molecular weight of air, assuming that air is 21% O2 and 79% N2.

Solution Because the composition of air is given in mole percent, a basis of 1 g mol is chosen. The MW of the N2 is not actually 28.0 but 28.2 because the value of the MW of the pseudo 79% N 2 is actually a combination of 78.084% N2 and 0.934% Ar. The masses of the O2 and pseudo N2 are

Therefore, the total mass of 1 g mol of air is equal to 29.0 g, which is called the average molecular weight of air. (Because we chose 1 g mol of air as the basis, the total mass calculated directly provides the average molecular weight of 29.0.)

Professor Mo Zandi

3 Exercise 3: Calculation of Average Molecular Weight Since the discovery of superconductivity almost 100 years ago, scientists and engineers have speculated about how it can be used to improve the use of energy. Until recently most applications were not economically viable because the niobium alloys used had to be cooled below 23 K by liquid He. However, in 1987 superconductivity in Y-BaCu-O material was achieved at 90 K, a situation that permits the use of inexpensive liquid N 2 cooling.

What is the molecular weight of the cell of a superconductor material shown in the figure below?

(The figure represents one cell of a larger structure.)

Mole and mass fraction Mole fraction is simply the number of moles of a particular substance in a mixture or solution divided by the total number of moles present in the mixture or solution. This definition holds for gases, liquids, and solids.

Similarly, the mass fraction is nothing more than the mass or weight of the substance divided by the total mass of all of the substances present in the mixture or solution. Although mass fraction is the correct term, by custom ordinary engineering usage frequently employs the term weight fraction.

These concepts can be expressed as: đ?‘šđ?‘œđ?‘™đ?‘’ đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ??´ =

đ?‘šđ?‘œđ?‘™đ?‘’đ?‘ đ?‘œđ?‘“ đ??´ đ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘šđ?‘œđ?‘™đ?‘’đ?‘

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘“đ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ??´ =

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘œđ?‘“ đ??´ đ?‘Ąđ?‘œđ?‘Ąđ?‘Žđ?‘™ đ?‘šđ?‘Žđ?‘ đ?‘

and

Mole percent and weight percent are the respective fractions times 100. Be sure to learn how to convert from mass fraction to mole fraction and vice versa without thinking, because you will have to do so quite often in the future. Unless otherwise specified, when a percentage or fraction is given for a gas, it is assumed that it refers to a mole percentage or a mole fraction. When a percentage or fraction is given for a liquid or a solid, it is assumed that it refers to a weight percentage or a mass fraction.

Professor Mo Zandi

4 Exercise 4: Conversion between Mass (Weight) Fraction and Mole Fraction An industrial-strength drain cleaner contains 5.00 kg of water and 5.00 kg of NaOH. What are the mass (weight) fraction and mole fraction of each component in the drain cleaner?

Self-Assessment Test 1. Indicate whether the following statements are true or false: a. The pound mole is composed of 2.73 Ă— 1026 molecules. b. The kilogram mole is composed of 6.023 Ă— 1026 molecules. c. Molecular weight is the mass of a compound or element per mole.

2. What is the molecular weight of acetic acid (CH3COOH)?

Density In ancient times, counterfeit gold objects were identified by comparing the ratio of the weight to the volume of water displaced by the object, which is a way to measure the density of the material of the object, to that of an object known to be made of gold. A striking example of quick thinking by an engineer who made use of the concept of density was reported by P. K. N. Paniker in the June 15, 1970, issue of Chemical Engineering: The bottom outlet nozzle of a full lube-oil storage tank kept at a temperature of about 80°C suddenly sprang a gushing leak as the nozzle flange became loose. Because of the high temperature of the oil, it was impossible for anyone to go near the tank and repair the leak to prevent further loss.

After a moment of anxiety, we noticed that the engineer in charge rushed to his office to summon fire department personnel and instruct them to run a hose from the nearest fire hydrant to the top of the storage tank. Within minutes, what gushed out from the leak was hot water instead of valuable oil.

Some time later, as the entering cold water lowered the oil temperature, it was possible to make repairs.

Density (we use the Greek symbol Ď ) is the ratio of mass per unit volume such as kg/m3 or lb/ft3: đ?œŒ = đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś =

đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘š = đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’ đ?‘Ł

Density has both a numerical value and units. Densities for liquids and solids do not change significantly at ordinary conditions with pressure, but they can change significantly with temperature for certain compounds if the temperature change is large enough, as shown in the figure below. Note that between 0°C and 70°C, the density of water is relatively constant at 1.0 g/cm 3. On the other hand, for the same temperature range, the density of NH 3 changes by approximately 30%. Usually we will ignore the effect of temperature on liquid density unless the density of the material is especially sensitive to temperature or the change in the temperature is particularly large.

Professor Mo Zandi

5

Densities of liquid H2O and NH3 as a function of temperature.

Flowrate In the process industries, process streams are normally delivered to or removed from a process in pipes. The flow rate of a process stream is the rate at which material is transported through a carrying pipe. In some books authors usually use an overlay dot (for example áš ) to denote a rate except for the volumetric flowrate. In this course we usually use F for mass flowrate and V for the volumetric flowrate unless the unit of the quantity suggests otherwise. The mass flow rate (F) of a process stream is the mass (m) transported through a pipe per unit time (t): đ??š = đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘“đ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = áš =

đ?‘š đ?‘Ą

The molar flowrate (áš…) of a process stream is the moles (n) of a substance transported through a pipe per unit time: áš… = molar flowrate =

đ?‘› đ?‘Ą

The volumetric flowrate (V) of a process stream is the volume (V) transported through a line per unit time: đ?‘‰ = đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ =

đ?‘Ł đ?‘Ą

For a known density we can write: đ??š = đ?‘‘đ?‘’đ?‘›đ?‘ đ?‘–đ?‘Ąđ?‘Ś Ă— đ?‘Łđ?‘œđ?‘™đ?‘˘đ?‘šđ?‘’đ?‘Ąđ?‘&#x;đ?‘–đ?‘? đ?‘“đ?‘™đ?‘œđ?‘¤đ?‘&#x;đ?‘Žđ?‘Ąđ?‘’ = đ?œŒ . đ?‘‰

Professor Mo Zandi

6 Exercise 5: Unit conversion and converting volumetric flowrate to mass flowrate Water (density = 62.4 lb/ft3 ) flows at an average velocity of 10 feet per minute through a 6 inch diameter pipe. What is the mass flow rate of water in kg/s?

Home Activity: Measure Your Shower Water Flowrate Your water flow rate is simply the amount or bulk of water which moves through your pipes over a minute or hour. The easiest way to get a fairly accurate measure of your water flow rate is to time yourself filling up a bucket. So for example if you fill up a 10 litre bucket in 1.5 minutes, then your flow rate will be: 10/1.5 = 6.66 Litres per minute For this activity, measure and determine your home/accommodation shower or bath cold water flowrate in SI units (kg/h) and complete the activity page on Blackboard.

Significant Figures & Uncertainty Chemical engineers all agree that measurement should include three pieces of information: 1. The magnitude of the variable being measured 2. Its units 3. An estimate of its uncertainty (error) The last is likely to be either disassociated from the first two or ignored completely. If you have no idea of the accuracy of a number of a measurement or the error associated with a measurement is not given, a conservative approach is to imply that the last digit is known within upper and lower bounds.

Example: The flowrate of a liquid has been reported as 1.43 kg/h. This indicates a value of 1.43 Âą 0.005, meaning that the value can be deemed to be between 1.425 and 1.435. Another interpretation of 1.43 is that it means 1.43 Âą 0.01.

Significant Figures

Professor Mo Zandi

7 Significant figures can be used as an indication of the precision with which a quantity is measured or known. The significant digits of a number are the digits from the first nonzero digit on the left to either (1) the last digit (whether it is nonzero or zero) on the right if there is a decimal point or (2) the last nonzero digit of the number if there is no decimal point.

Exercise: Determine significant figures of each the following numbers: 2.50 250 250.

250.0 25.040 .025 .0250 .02504 The quantities that are known exactly (e.g. pure integers) have an infinite number of significant figures.

Be aware that some textbooks and authors do not attribute significance of the trailing zeros on the right-hand side of a decimal point. The rationale behind attributing additional significant figures to the trailing zeros is that they would not be added unless there was a reason for displaying additional accuracy.

But what should you do when you add, subtract, multiply, and divide numbers that have associated uncertainty?

Generally, several options exist in establishing the degree of certainty in a number. Besides common sense (which comes with experience), three common decision criteria are: 1- Absolute error 2- Relative error 3- Statistical analysis (it is beyond this course and won’t be discussed)

Absolute Error In this method, you have to consider two cases: a) numbers with a decimal point, and b) numbers without a decimal point.

For case (a), suppose we assume that the last significant figure in a number represents the associated uncertainty. For case (b), if a number is stated without a decimal point, we will assume that the trailing zeros do not imply any additional accuracy.

When you multiply or divide numbers, generally you should retain in your final answer the lowest number of significant figures that occur among all of the numbers involved in the calculations even though you carry along 10 or 20 digits during the calculations themselves. In other words, in long calculations, the final result should be rounded off to the correct number of significant figures.

Professor Mo Zandi

8 When quantities are combined by addition and/or subtraction, the final result cannot be more precise than any of the quantities added or subtracted. Therefore, the position of the last significant digit in the number that has the lowest degree of precision is the position of the last permissible significant digit in the result.

Please note that absolute errors can lead to gross distortions in the specified uncertainty of a number. In this case, perhaps Relative Error can often be a better way to determine significant figures in your answers.

Relative Error In this method, you must avoid increasing the precision of your answer very much over the precision in your measurements or data when presenting results of calculations. You do have to use common sense in applying the concept of relative error to scales that use both relative and absolute units.

Choosing a Basis A basis is a reference chosen by you for the calculations you plan to make in a particular problem, and a proper choice of basis often can make a problem much easier to solve than a poor choice. The basis may be a period of time such as hours, or a given mass of material, or some other convenient quantity.

To select a sound basis (which in many problems is predetermined for you but in some problems is not so clear), ask yourself the following three questions: 1. What do I have to start with (e.g., I have 100 lb of oil; I have 46 kg of fertilizer)?

2. What answer is called for (e.g., the amount of product produced per hour)?

3. What is the most convenient basis to use? (For example, suppose that the composition of a given material is known in mole percent. Then selecting 100 kg moles of the material as basis would make sense. On the other hand, if the composition of the material in terms of mass is known, then 100 kg of the material would be an appropriate basis.)

These questions and their answers will suggest suitable bases. Sometimes when several bases seem appropriate, you may find it is best to use a unit basis of 1 or 100 of something, for example, kilograms, hours, moles, or cubic feet. For liquids and solids in which a mass (weight) analysis applies, a convenient basis is often 1 or 100 lb or kg; similarly, because gas compositions are usually provided in terms of moles, 1 or 100 moles is often a good choice for a gas.

Always state the basis you have chosen for your calculations by writing it prominently on your calculation sheets or in the computer program used to solve the problem.

To sum up, be sure to state the basis of your calculations so that you will keep clearly in mind their real nature, and so that anyone checking your problem solution will be able to understand on what basis your calculations were performed.

More frequently than you probably would like, you will have to change from your original selection of a basis in solving a problem to one or more different bases in order to put together the information needed to solve the entire problem. Consider the following example.

Professor Mo Zandi

9 Example Changing Bases Considering a gas containing O2 (20%), N2 (78%), and SO2 (2%), find the composition of the gas on an SO2-free basis, meaning gas without the SO2 in it. Solution First choose a basis of 1 mol of gas (or 100 mol). Why? The composition for the gas is in mole percent. Next you should calculate the moles of each component, remove the SO2, and adjust the basis for the calculations so that the gas becomes composed of only O2 and N2 with a percent composition totaling 100%: Basis: 1.0 mol of gas

The round-off in the last column is appropriate given the original values for the mole fractions.

Self-Assessment Test 1. What are the three questions you should ask yourself when selecting a basis?

2. Why do you sometimes have to change bases during the solution of a problem?

Professor Mo Zandi