2. Mass Balance (Stoichiommetry) Calculations

Introduction

Basic mass balance calculations go hand in hand with balancing chemical reactions. Just as the ability to balance chemical reactions is a crucial, basic skill for any process chemical industry, so is the ability to perform these calculations. The principle of conservation of mass is again the foundation.

If an ore consumes sulfuric acid, we can determine how much per tonne of ore based on the composition of the ore and the relevant chemical reactions. If we need oxygen to be fed into an autoclave to treat a concentrate, we can calculate the flow rate required, based on the flow rate of concentrate, its composition and the pertinent chemical reactions. If we want to know how much copper metal we can produce from an ore, that depends on the amount of each copper mineral and the degree of reaction for each under specified conditions. These and myriad other mass balance questions can be addressed by straightforward calculations. The problem is that there is no single procedure available, because there is such a variety of types of calculations. Rather, we need to develop the basic principles and seek to apply them systematically. A degree of interpretation is needed to address each problem; to translate the information into a starting point for the calculations. 

First, amounts of matter can be measured in three basic ways: moles, mass and volume. In a chemical reaction like (in acid solution):

Cu₂S _(s) + O_{2 (g)} + 2 H⁺ _(aq) = 2 Cu⁺² _(aq) + SO₄^2- _(aq) + H₂O _(l)          (114) 

1 mole of Cu₂S produces 2 moles of Cu⁺² and uses up two moles of H⁺. But, we do not buy or sell things in units of moles. Mostly we use mass and sometimes volume. How do we relate the mass of Cu⁺² formed to the mass of Cu₂S available? This requires the atomic weights (denoted AW) of the elements in g/mol, from which we can determine the formula weights of compounds or ions. Tables of atomic weights are available in materials engineering and chemistry texts. 

Element 
Atomic weight (g/mol) 
Cu 
63.546 
S 
32.064 

Formula weight for Cu₂S = 2 x 63.546 + 32.064 = 159.156 g/mol. For every 1 kg of Cu₂S we have:  

1000 g Cu₂S x  = 6.28314 mole Cu₂S          [1]

To determine the amount of copper we can extract from 1000 g of Cu₂S in kg we have to relate the moles of Cu⁺² to the moles of Cu₂S, i.e. 1 mol Cu₂S = 2 mol Cu⁺². This yields: 

6.28314 mole Cu₂S x   x = 798.5 g Cu^{+2}> = 0.7985 kg    [2]

This illustrates the point that at heart of all such calculations is a progression of conversions:

mass → moles → … → moles → mass          [3]

HOWEVER, there are a number of things that must be borne in mind about this generalization if it is to be used correctly, i.e.

• The calculation is begun on the left, with one of mass or moles units.
• There might be no moles conversions involved, i.e. mass to mass only. 
• There might be one conversion to/from moles required. 
  
• There might be more than one moles to moles conversions required. 
  
• The calculation sequence ends on the right, with one of moles or mass units. 

The following examples will serve to illustrate some of the points.

2.1 Examples 

1. An ore contains 0.5% copper, in the form of Cu₂S. Ore will be processed at a rate of 500,000 tonnes per year (t/y). What is the maximum annual copper production rate in t/y?

Note: 0.5% copper means 0.5% by weight, unless otherwise specified. Providing the copper content and the mineral form inevitably causes confusion. Students often assume that the question indicates that the ore contains 0.5% Cu₂S. That is not what it says. It says, 0.5% Cu, i.e. 5 kg Cu/tonne of ore, and that Cu is present as Cu₂S. (In terms of Cu₂S we have 0.626% Cu₂S, which is equivalent to 0.5% Cu.) 

The maximum copper production means that 100% of the Cu₂S will be leached. Where do we start the calculations; mass or moles? Our data is in units of mass – 50 kg Cu/t ore and 500,000 t ore/y. On that basis we expect to start at the mass end of the sequence [3]. Where do we end? We need Cu production in t/y, i.e. mass units. Do we need to convert from mass to moles? We have mass Cu in ore and need to relate that to mass Cu produced. For both we have the same chemical species – Cu metal. Hence no moles conversions are needed. It is this sort of analysis that must always be done:

• Does the data suggest mass or moles units for starting the calculations?
• Does the outcome require mass or moles units at the end of the calculations?
• Are conversions to/from moles required?

x x   = 2500 t Cu/y           [4]

Note: Set up the calculation including the units, and include the chemical species in the units. Make sure the units cancel to get the correct units at the end. This is essential if mistakes are to be avoided.

2. Assume conditions as in example 1 with the ore containing 0.5% copper, in the form of Cu2S. Ore will be processed at a rate of 500,000 tonnes per year (t/y). Assume that 80% of the Cu2S is leached. How much sulfuric acid will be required in kg per tonne of ore? Refer to reaction (114). 


Now we introduce the idea of incomplete reaction: 80% of the Cu in Cu2S will dissolve. The remainder will not. Data is in mass units. Start at the mass end again. The answer is in mass units. Conclude at the mass end. We need to relate mass of Cu to mass of H2SO4. Conversions to/from moles will be needed. Note that we need 2 mole H+/mol Cu2S, and that 2 mol H+ = 1 mol H2SO4. Also note that we have data in mass Cu/t ore and the answer we want is mass H2SO4/t ore. The “/t ore” unit will follow through the calculations from start to finish. Therefore we do not need the ore flow rate. 


5 kg Cu x  1000 g Cu x 0.8 g Cu aq x  1 mol Cu    x 1 mol Cu2S  x 1 mol H2SO4               
    t ore              kg Cu   g Cu in ore    63.546 g Cu     2 mol Cu        mol Cu2S  


 x 98.0775 g H2SO4 x 10-3 kg H2SO4 =  3.09 kg H2SO4/t ore [5] 
               mol H2SO4         g H2SO4 


The 80% extent of Cu2S dissolution is expressed as 0.8 g Cu aq, meaning 0.8 g Cu   g Cu in ore 
dissolved per 1 g Cu in the ore. Note: Stoichiommetric factors (such as 1 mol Cu2S/2 mol Cu) are set up so as to make sure the units cancel. To get mol Cu in the numerator to cancel, we use 1 mol Cu2S/mol Cu, rather than the other way round. This way there is no need to try to think through which way round the stoichiommetric factors have to go. This can prevent errors. 




3. Assume conditions as in example 1 with the ore containing 0.5% copper, in the form of Cu2S and with 80% of it leached. Ore will be processed at a rate of 500,000 tonnes per year (t/y). How much oxygen will be needed in m3/h at STP? Refer to reaction (114). 


“Oxygen” refers to molecular oxygen, i.e. O2. (You must know this.) STP refers to standard temperature and pressure (0°C, or 273.15 K and 1 atm pressure). The data again is in units of mass, as per question 1. The required answer has units of m3/h for a gas. We will assume the ideal gas law, 


PV = nRT [6] 


This relates pressure, volume and temperature to amount of gas in moles. Hence the end of calculation sequence [3] must be in moles. And since we have mass data and a moles output, mass to moles conversion is going to be required. Further, the answer we need is in volume per unit time. In order to obtain this result we need data that also involves time-1. The piece of data we have is the ore flow rate in t/y. Then, once we have an answer in moles/h of O2, we can use the ideal gas law to obtain volumetric flow rate. We start with mass of copper. In order to use reaction (114), we need to convert to moles of Cu, to moles of Cu2S, to moles of O2, as per the reaction.   


 5 kg Cu x 1000 g Cu x 0.8 g Cu aq x 500,000 t ore x 1 y    x 1 d  x 1 mol Cu    x    
t ore kg Cu   kg Cu         g Cu in ore                    y     365 d   24 h   63.546 g Cu           


5/2 mol O2 =  4.49105 x 104 mol O2/h [7] 
 2 mol Cu    


Then convert mole O2/h to volume/h using the ideal gas law: 


4.49105 x 104 mol O2 x 0.08206 L atm x 273.15 K x    1     x 10-3 m3 
         h          mol K        1 atm           L 


= 100.7 m3 O2/h [8] 




The calculations can be schematically represented as: 


mass Cu → moles Cu → moles O2 [9] 
           ¯ 
      volume O2 




4. A gold ore contains 4 g Au/t ore (as gold metal) and 0.012% copper as CuO. The ore will be leached with sodium cyanide and 95% of the CuO will be leached and will form [Cu(CN)3]2-. Gold forms [Au(CN)2]-, and 90% of it will be leached. The leach slurry contains 40% solids. The ore is processed at a rate of 2000 t/h. If the initial sodium cyanide concentration is 0.5 g/L, what will it be after the leaching process? The solution density is 1.01 g/mL. The reactions involved are: 


2CuO s + 7CN-aq + H2O l = 2[Cu(CN)3]2-aq + CNO-aq + 2OH-aq (115) 


2Au s + 1/2O2 g + H2O l + 4CN-aq = 2[Au(CN)2]-aq + 2OH-aq (116) 


This question has a lot of data. 0.012% Cu as CuO means 0.12 kg Cu/t of ore. Ultimately we need to relate concentrations of Cu and Au (in mass units) in the ore to a concentration of NaCN in g/L. In addition, the concentrations for Cu and Au are given for the ore (a solid), while the concentration for NaCN is needed in solution. So we also have to relate the solids and the solution. This is given by the % solids in the slurry: 40% solids means that for 1000 kg of slurry, 400 kg is solids and 600 kg is solution, i.e. 400 kg solids/600 kg solution. Using this we can relate mass of solution to mass of ore. But we will need a volume of solution to get the required NaCN concentration in g/L. Hence we will need the solution density. Finally, relating the masses of Cu and Au to mass NaCN will require conversion to/from moles, as per the reactions above. Note that the ore mass flow rate has not entered into these considerations and will not be needed. The calculation below provides the amount of NaCN consumed by reaction with CuO. 


0.12 kg Cu x 1000 g Cu x    1 t ore     x 400 kg ore x 0.95 g Cu aq x   1 mol Cu 
         t ore         kg Cu     1000 kg ore   600 kg aq      g Cu in ore       63.546 g Cu 


x 7 mol NaCN x 49.0075 g NaCN = 0.205143 g NaCN/kg aq solution [10] 
    2 mol Cu          mol NaCN 


Next we need to convert this to g NaCN/L. This is done using the density. 


0.205143 g NaCN x 1.01 kg aq = 0.2072 NaCN/L consumed [11] 
       kg aq   L 


The same calculation can be done for gold: 


4 g Au x     1 t ore    x 400 kg ore x 0.90 g Au aq x 1 mol Au       x  4 mol NaCN 
   t ore   1000 kg ore    600 kg aq        g Au ore     196.9665 g Au    2 mol Au 


x 49.0075 g NaCN = 0.0011943 g NaCN/kg aq solution [12] 
       mol NaCN 


Convert to g NaCN/L: 


0.0011943 g NaCN x 1.01 kg aq = 0.001206 g NaCN/L [13] 
         kg aq                 L 


Note that gold consumes very little NaCN. The final NaCN concentration then is simply the initial less what was consumed: 


0.5 – 0.2072 – 0.00121 = 0.292 g/L NaCN [14] 


We can summarize the calculations as follows: 


 Concentration Cu as mass Cu    x  mass ore  
            mass ore       mass aq 


     ¯  
mass Cu → moles Cu → moles NaCN → mass NaCN [15] 
mass aq      mass aq        mass aq    mass aq         ¯ 
              x density = mass NaCN 
                      vol. solution 




 2.2 Generalized Scheme for Mass Balance Calculations 


The illustration below summarizes a general outline for conducting these basic kinds of mass balance calculations. The units of various data combinations containing mass terms indicate starting at the mass end of the calculation sequence. Combinations involving moles for data indicate that the starting point should be moles. Answer unit combinations involving mass indicate that the sequence should end at the mass end. Similarly unit combinations involving moles indicate ending at the moles end. Volumes for gases imply ending at the moles end, since volume and moles are related by PV = nRT (or more complex equations of state if necessary).  


Data may need to be manipulated to get started. For instance, in example 4 we had to combine a mass concentration x a mass-to-volume ratio for a slurry. And the final result on the mass – moles … moles – mass sequence may need to be manipulated to get the required units. In example 3 we had to convert moles of O2 to volume of O2 via the ideal gas law.  


Moles as a starting point is not common in industrial practice, although not inconceivable. For instance, for a given flow of gas into a process, one could determine the necessary input of reactants.  


The points made above are summarized below. 


To get started on the calculation sequence, what do the units for data and units for answers tell you about where to start/end on the calculation sequence? 


Does the data suggest mass or moles units for starting the calculations? 
Does the outcome require mass or moles units at the end of the calculations? 
Are conversions to/from moles required? (NO if the chemical species of concern at the start and end are the same; YES if they differ.) 


There are several possibilities for how the calculation sequence could have to be done: 


The calculation is begun on the left, with one of mass or moles units. 
There might be no moles conversions involved, i.e. mass to mass.  
There might be one conversion to/from moles required. 
There might be more than one moles to moles conversions required. 
The calculation sequence ends on the right, with one of moles or mass units. 


Set up the calculations including the units, and include the chemical species in the units. Make sure the units cancel to get the correct units at the end. This is essential if mistakes are to be avoided. 


Proceed in an orderly way. Arrange stoichiommetric factors such that the units cancel as needed. Including chemical species in the units will clearly show how this should be done. Figure 1 summarizes common data/answer relationships.