CHAPTER 2. Stoichiometry and Bacterial Energetics
2. Stoichiometry and Bacterial Energetics Mass balance: the important concept in the engineering design of system for biological treatment Determine the amount of chemicals that must be supplied to satisfy the energy,nutrient,and environmental needs of the microorganisms. The amounts of end products generated can be estimated. Chemicals : i) oxygen (e - acceptor) ii) N, P (nutrients) iii) lime, sulfuric acid (ph maintenance)
2. Stoichiometry and Bacterial Energetics Balanced chemical equations are based upon the concept of stoichiometry The stoichiometry for microbial reactions is complicated because: st : microbial reactions often involve oxidation and reduction of more than one species. 2 nd : the microorganism have two roles as catalysts for the reaction as products of the reaction 3 rd : microorganisms carry out most chemical reactions in order to capture some of the energy released for cell synthesis and for maintaining cellular activity. reaction energetics as well as balancing for elements, electrons, charge should be considered.
2. An example stoichiometric equation Porges,Jasewicz,Hoover 956 - Wastewater: C8H2O3N 2 + 3 O2 C5H 7O2N + NH 3 + 3CO2 + H 2O Casein bacterial cell Formula weight: 84 96 3 7 32 8 = 280 = 280 [2.] * Some of the carbon in casein is fully oxidized d to CO2 (e - donner substrate) t However, the rest of the carbon in casein is incorporated into the newly synthesized biomass (carbon source) ** C 5 H 7 O 2 N represents bacterial cells. This is generally satisfactory for most practical purposes However, phosphorus normally represents about 2% of the bacterial organic dry weight. If,000kg/d of casein were consumed, 0.02x (3/84) x,000kg casein = 2kg/d of P should be present in the wastewater. *** Complex protein-containing mixture in casein is represented through empirical C 8 H 2 O 3 N 2
2. Could we predict the stoichiometry of equation 2.? Three things needed to do that:.empirical formula for cells 2.Framework for describing how the electron-donor substrate is partitioned between energy generation and synthesis 3.Means to relate the proportion of the electron-donor substrate that is used to synthesize new biomass to the energy gained from catabolism and the energy needed for anabolism.
2.2 Empirical formula for microbial cells Calculate the empirical formula C n H O a b N c C H O N 2n + 0.5a -.5c - b a - 3c + ( ) O2 nco 2 + cnh 3 + H O [2.2] 2 2 n a b c 2 COD / Weight = (2n + 0.5a -.5c - b)6 2n + a +6b +4c [2.3] T=%C/2+%H+%O/6+%N/4 [2.4] n=%c/2t %C/2T, a=%h/t %H/T, b=%o/6t %O/6T, c=%n/4t * n, a, b, c is divided by T to get the ratio of n : a : b : c as well as to make the sum of ( n+a+b+c ) equal to.0 COD : Calculated oxygen demand for full oxidation of the cellular carbon yg per unit weight of cells (g O 2 / g cell). Normally COD`=CODcr
2.2 Empirical formula for microbial cells Example: %C= 48.9% %H=5.2% %O=24.8% %N=9.46% T=48.9/2+5.2+24.8/6+9.46/4=.50 n=48.9/(2*.5) =0.354 a=5.2/.5=0.452 b=24.8/(6*.5)=0.35 c=9.46/(4*.5)=0.0588 In order to normalize to c= for N, divide by 0.0588, the following empirical formula for the cells is obtained : C 6.0 H 7.7 O 2.3 N The COD to organic weight ratio is then equal to (2X6.0 + 0.5X7.7.5 2.3) x 6 / (2x6 + 7.7 + 6x2.3 +4) =.48g COD /g cells Rounding fractional coefficients (e.g., 2.3 for O) to whole numbers introduce Rounding fractional coefficients (e.g., 2.3 for O) to whole numbers introduce errors and should not be done. **.48g of oxygen is needed for the complete oxidation of g of cells
2.3 Substrate partitioning and cellular yield 0 f e 0 f s : portion of electrons of an e - donor substrate to provide energy : portion of electrons of an e- donor substrate which are converted into microbial cells 0 0 f e + f s =
2.5 How to write an overall reaction (Energy + Synthesis ) Example) benzoate : e - donor nitrate : e - acceptor ammonium : N source for cell synthesis Assumption (on a net yield basis); fe (= 0.6) and fs (= 0.4) Energy reaction: Re = Ra - Rd Ra: /5 NO 3- + 6/5 H + + e - -> /0 N 2 + 3/5 H 2 O - Rd: /30 C 6 H 5 COO - + 3/30 H 2 O -> /5 CO 2 + /30 HCO 3 - + H + + e - Re: /30 C 6 H 5 COO - + /5 NO 3- + /5 H + -> /5 CO 2 + /0 N 2 + /30 HCO 3- + /6 H 2 O Synthesis reaction: Rs = Rc - Rd Rc: /5 CO 2 + /20 NH4 + + /20 HCO 3- + H + + e - -> /20 C 5 H 7 O 2 N + 9/20 H 2 O - Rd: /30 C 6 H 5 COO - + 3/30 H 2 O -> /5 CO 2 + /30 HCO - 3 + H + + e - Rs: /30 C 6 H 5 COO - + /20 NH 4+ + /60 HCO 3 - -> /20 C 5 H 7 O 2 N + /60 H 2 O
2.5 Overall Reactions for Biological Growth Cell-formation (Rc ) and electron acceptor half-reactions (Ra)
2.5 Overall Reactions for Biological Growth 2) Obtain the overall reaction (R) including energy and synthesis using portions of electrons, fe (= 0.6) and fs (= 0.4) fe*re: 0.02 C 6 H 5 COO - + 0.2 NO 3- + 0.2 H + -> 0.2 CO 2 + 0.06 N 2 + 0.002 HCO 3 - + 0. H 2 O fs*rs: 0.033 C 6 H 5 COO - + 0.02 NH 4+ + 0.0067 HCO 3- -> 0.02 C 5 H 7 O 2 N + 0.0067 H 2 O R : 0.0333 C - + 3- + 4+ + + 6 H 5 COO 0.2 NO 0.02 NH 0.2 H -> 0.02 C 5 H 7 O 2 N + 0.2 CO 2 + 0.06 N 2 + 0.033 HCO 3- + 0.067 H 2 O [2.34] Th ll ti f t th i f b t i th t i b t The overall reaction for net synthesis of bacteria that are using benzoate as an e- donor, nitrate as an e- acceptor, and ammonium as nitrogen source.
Transfer of energy from energy generation to cell synthesis or maintenance via an energy carrier (ATP)
2.3 Substrate partitioning and cellular yield 0 f e 0 f s : portion of electrons of an e - donor substrate to provide energy : portion of electrons of an e- donor substrate which are converted into microbial cells 0 0 f e + f s =
0 2.3 How to convert f s into mass units (g Cell produced / g COD consumed) Y = ( n e e eq 0 f s ( M c gcells / mol / mol cells )(8gCOD cells ) / e eq donor ) Y : true yield (g cell produced / g COD) M c : empirical formula weight of cells n e : number of electron equivalents in an empirical mole of cells * o 2 (32g) + 4 e- = 2 o - ** n e e - eq are required to produce one mole of cells
0 2.3 How to convert f s into mass units (g Cell produced / g COD consumed) Y = = f 0 s ( n ( e c g cells / mol cells )( mol cells / n e e eq ) (8g COD / e eq donor ) 0 f s ( M c g cells / mol cells ) e eq / mol cells )(8g COD / e eq donor ) Example : Cells : C 5 H 7 O 2 N M c : 3 g / mol cells NH 4+ : nitrogen source n e : 20 5 + + 9 CO 2 + HCO 3 + NH 4 + H + e C 5 H 7 NO 2 + H 2O 20 20 20 20 3 = 0.706 20 8 0 Y f s = f 0 s
2.3 Substrate partitioning and cellular yield The net growth rate of microbial cells: If Y dx dt n a = ds = Y ( ) bx dt dx a / dt = Y b - ds / dt a x a ds / dt x a Y n = 0, then Y b = ds / dt - ds / dt b = = m x a Y dx a : net growth rate of active organism (M/L dt 3 T) 0 [2.5] [2.6] [2.7] ds : rate of disappearance of substrate (M/L 3 T) dt b : decay rate (T - ) decay due to endogenous respiration and predation Y : net yield of microorganisms (M/M) n m : maintenance energy (M/MT) = the substrate utilization rate per unit mass of organisms
2.3 Substrate partitioning and cellular yield - ds / dt b Yn = 0, then = = x Y a m If Y n = 0, the substrate utilization rate is then just sufficient to maintain the cells. The substrate utilization rate per unit mass of organisms is termed the maintenance energy. When the substrate utilization rate is less than m, the substrate available is insufficient to satisfy the total metabolic needs of the microorganisms. This represents a form of starvation.
2.3 Substrate partitioning and cellular yield Y n is less than Y, because some of the electrons originally present in the substrate must be consumed for energy of maintenance. When considering net yield, the portion of e - used for synthesis is f s rather than f 0 s0 : the portion of e - for energy production is f e rather than f 0 e 0 0 f e + f s = f e + f s = f < s f 0 s 0 f e > f e
2.4 Energy Reactions Oxidation Reduction reaction Microorganisms energy for growth and maintenance electron donor : cf. Chemoorganotrophs : organic matter (most common) chemolithotrophs : inorganic compound ( NH 3, S 2- ) electron acceptor : - oxygen under aerobic conditions - nitrate, sulfate, carbon dioxide under anaerobic conditions Fermentation : organic matter for both the electron donor and acceptor
2.4 Energy Reactions Aeorbic oxidation C 6 H 2 O 6 + 6 O 2 6 CO 2 + H 2 O Denitrification Free Energy + 5C6H2O6 + 24NO3 + 24H 30CO2 + 42H 2O + 2N2 Sulfate reduction kj 2,880 2,720 2 + 2C6H2O6 + 6SO4 + 9H 2CO2 + 2H 2O + 3H 2S + 3HS 492 Methanogenesis C C 6 6 H H 2 2 O O 6 6 3CO 2 2CO 2 + 3CH Ethanol Fermentation 4 + 2CH CH 3 2 OH 428 244 / mol glu cose [2.8] [2.9] [2.0] [2.] [2.2] Microorganisms would like to obtain as much energy from a reaction as possible Preference for electron acceptors : oxygen > nitrate > sulfate >carbon dioxide > fermentation E.coli use several ldifferent electron acceptor (facultative) ti Methanogens use acetate or carbon dioxide (strict anaerobes)
2.4 Energy Reactions Aerobic organisms need to send less electrons from their donor substrate to oxygen (the largest energy = -2,880 kj) than anaerobic organisms in order to generate the energy required to synthesize a given amount of new biomass. Aerobic organisms : f e 0 small, f s 0 large : Y is larger than anaerobic microorganisms : grows faster than anaerobic microorganisms
2.4 Energy Reactions Constructing energy reaction Oxidation half reaction for glucose C 24 6 H 2 O 6 + H + 2O CO 2 + H + e 4 4 [2.3] Reduction half-reaction for nitrate 5 6 + NO + H + e N2 + 5 0 3 H 5 3 2O [2.4] Overall balanced reaction + 7 C H2O6 + NO3 + H CO2 + H 2O + 24 5 5 4 20 0 6 N 2 [2.5] Eq. 2.9 29results ife Eq. 25i 2.5 is multiplied li dby the least common denominator of f20 + 5 C H2O6 + 24 NO3 + 24 H 30 CO2 + 42 H2O+ 2N2 6 2,720 [2.9]
2.4 Energy Reactions How to construct reduction half reaction. Only one element is allowed to have its oxidation state changed in a half reaction. Example : alanine (CH 3 CHNH 2 COOH) : Carbon (2 e - equivalent CH ;7e - ;4e - - 3, CHNH 2, COOH; e ) : The other elements (H; +, O; -2, N; -3 )must retain their oxidation state the same on both sides of equation. 2. In half-reactions, for organic compounds, the oxidized form is always CO 2 or, in some cases, HCO 3 -or CO -2 3 depending on solution environment like ph.
2.4 Energy Reactions How to construct reduction half reaction Exemple: Amino acid alanine (CH 3 CHNH 2 COOH) - Product is CO 2 step CO2 CH3CHNH2COOH - Add other species (H 2 O) - Electron on the left side - N in the reduced form (NH 3 or NH + 4 ) like in alanine on the left side step 2 CO2 + H 2 O + NH 3 + e CH 3 CHNH 2 COOH -Balance the reaction for the elements except O and H step 3 3CO2 + H 2O + NH 3 + e CH 3CHNH 2COOH -Balance the O through addition or substraction of water. step 4 3CO2 + NH 3 + e CH 3CHNH 2COOH + 4H 2O
2.4 Energy Reactions How to construct reduction half reaction -Balance the H by introducing H+. + step 5 3 CO2 + NH 3 + 2 H + e CH 3 CHNH 2 COOH + 4 H 2 O -Balance the charge on the reaction by adding sufficient i e - to the left side. + step 6 3CO2 + NH 3 + 2H + 2e CH 3CHNH 2COOH + 4H 2O -Convert the eqn. to the electron-equivalent form by dividing by the coefficient on e -. + step CO + NH 3 + H + e CH 3CHNH 2COOH + 4 2 2 7 2 H 2O 3 [2.6 ]
2.4 Energy Reactions Example 2.3 RPODUCTS FROM METHANOGENESIS OF ALANINE Anaerobic methanogenic fermentation of alanine (00-mM at ph near neutral) Reaction O 2 Reaction O 2 Sum 2 CH 3 + CO2 + H + e CH 4 + H 2O [2.8] 8 8 4 5 CH 3CHNH 2COOH + H 2O CO2 + NH 2 2 6 2 CHNH 2 + 4 + HCO 2 COOH + H 2O CH 4 + CO + 2 + NH 4 + HCO 6 8 24 2 2 3 3 + H + + e [2.20] [2.9] - 75 % methane, 25% carbon - Bicarbonate alkalinity generates (see Example 2.3. p40)
2.4 Energy Reactions Different half-reaction forms are sometimes used for different reasons - Different half-reaction forms are sometimes used for different reasons - All are technically correct as long as the eqns are written in a balanced form. - Regardless of the form selected /2 mole of alanine gives one e - equivalent. + CO + NH 3 + H + e CH 3CHNH 2COOH + H 4 2 2 3 2 2O [2.22] + 5 CO + NH 3 + HCO3 + H + e CH 3CHNH 2COOH + H 6 2 2 2 2 2 2O + 5 CO 2 + NH 3 + HCO 3 + H + e CH 3CHNH 2COOH + H 6 2 2 2 2 2 2 2O [2.23] [2.24] 2 3 + 7 NH + HCO3 + H + e CH 3CHNH 2COOH + H 4 2 2 2 3 2O And also see o-2 in Table 2.3 (p.36) [2.25]
2.5 Overall Reactions for Biological Growth Bacterial growth involves two basic reactions: - i)e Energy production, and ii) Clll Cellular synthesis Rc : half-reaction for synthesis (cell formation), e - acceptor Ra : acceptor half-reaction Rd : donor half-reaction Re : energy reaction Rs : synthesis reaction Re = Ra Rd [2.26], Rs = Rc Rd [2.27]
2.5 Overall Reactions for Biological Growth How to write an overall reaction (Energy + Synthesis ) Example) benzoate : e - donor nitrate : e - acceptor ammonium: N source Assumption; fe (= 0.6) and fs (= 0.4) Ra: /5 NO 3- + 6/5 H + + e - -> /0 N 2 + 3/5 H 2 O - Rd: /30 C 6 H 5 COO - + 3/30 H 2 O -> /5 CO 2 + /30 HCO 3 - + H + + e - Re: /30 C 6 H 5 COO - + /5 NO 3- + /5 H + -> /5 CO 2 + /0 N 2 + /30 HCO 3- + /6 H 2 O Synthesis reaction: Rs = Rc - Rd Rc: /5 CO 2 + /20 NH4 + + /20 HCO 3- + H + + e - -> /20 C 5 H 7 O 2 N + 9/20 H 2 O - Rd: /30 C 6 H 5 COO - + 3/30 H 2 O -> /5 CO 2 + /30 HCO 3 - + H + + e - Rs: /30 C 6 H 5 COO - + /20 NH 4+ + /60 HCO 3 - -> /20 C 5 H 7 O 2 N + /60 H 2 O
2.5 Overall Reactions for Biological Growth 2) Obtain the overall reaction (R) including energy and synthesis using portions of electrons, fe (= 0.6) and fs (= 0.4) fe*re: 0.02 C 6 H 5 COO - + 0.2 NO 3- + 0.2 H + -> 0.2 CO 2 + 0.06 N 2 + 0.002 HCO 3 - + 0. H 2 O fs*rs: 0.033 C 6 H 5 COO - + 0.02 NH 4+ + 0.0067 HCO 3- -> 0.02 C 5 H 7 O 2 N + 0.0067 H 2 O R : 0.0333 C - + 3- + 4+ + + 6 H 5 COO 0.2 NO 0.02 NH 0.2 H -> 0.02 C 5 H 7 O 2 N + 0.2 CO 2 + 0.06 N 2 + 0.033 HCO 3- + 0.067 H 2 O [2.34] Th ll ti f t th i f b t i th t i b t The overall reaction for net synthesis of bacteria that are using benzoate as an e- donor, nitrate as an e- acceptor, and ammonium as nitrogen source.
2.5 Overall Reactions for Biological Growth General equation for constructing stoichiometric equations for microbial synthesis and growth: R = fe * Re + fs * Rs = fe * (Ra Rd) + fs * (Rc Rd) [2.35] = fe *Ra + fs *Rc Rd ( fe + fs ) < fs +fe =.0 and Rd ( fe + fs ) = Rd > R = fe *Ra + fs *Rc Rd [2.37] [2.37] represents net consumption of reactants and production of products when the microorganisms consume one e - equivalent of electron donor (an electron-equivalent basis) [2.37] represents not Energy balance, but Mass Balance (Stoichiometric),
2.5 Overall Reactions for Biological Growth Example: benzoate ; e - donor and C source (Rd = O-3 from Table 2.3) NO3- ; e - acceptor and N source (Rc, Ra = C-2, I-7 from Table 2.4) fs = 0.55, fe = 0.45 fe*ra: 0.09 NO 3- + 0.54 H + + 0.45 e - -> 0.045 N 2 + 0.27 H 2 O fs*rc: 0.096 NO 3- +0.0982CO 2+0.5696 H + + 0.55 e - ->0.096C 5H 7O 2N + 0.26 H 2O -Rd: 0.0333 C 6 H 5 COO - + 0.4333 H 2 O -> 0.2 CO 2 + 0.0333 HCO 3- + H + + e - R: 0.0333 C 6 H 5 COO - + 0.096 NO 3- + 0.096 H + -> 0.096 C 5 H 7 O 2 N + 0.045 N 2 + 0.0333 HCO 3- +0.08 CO 2 + 0.0528 H 2 O e - = 0.45 e - + 0.55 e - 0.09 mole nitrate is converted to nitrogen gas 0.096 mole nitrate is converted into organic nitrogen of the cells
2.5 Overall Reactions for Biological Growth Cell-formation (Rc ) and electron acceptor half-reactions (Ra)
Ex. 2.4 Nitrification Stochiometry Ammonium to nitrate, Aerobic condition, Inorganic carbon for cell synthesis, fs=0., NH -N=22mg/L 3 4 22mg/L, Wastewater 000m. NH 4 : e- donor, N source, O 2 : e- acceptor, CO 2 : C source, fs=0.& fe=0.9 fe*ra: 0.225O 2 + 0.9 H + + 0.9 e - -> 0.45 H 2 O (Table 2.4) fs*rc: 0.02 CO 2 + 0.005 NH 4+ + 0.005 HCO 3- + 0. H + + 0. e - -> 0.005 C 5 H 7 O 2 N + 0.045045 H 2 O (Table 2.4) -Rd: 0.25 NH 4+ + 0.375 H 2 O -> 0.25 NO 3- +.25 H + + e - (I- in Table 2.2) R: 0.3 NH 4+ + 0.225O 2 + 0.02 CO 2 + 0.005 HCO 3- -> 0.005 C 5 H 7 O 2 N + 0.25 NO 3- +0.25 H + + 0.2 H 2 O
Ex. 2.4 Nitrification Stochiometry R: 0.3 NH 4+ + 0.225O 2 + 0.02 CO 2 + 0.005 HCO 3-0.005 C 5 H 7 O 2 N + 0.25 NO 3- +0.25 H + + 0.2 H 2 O 0.3 (4) =.82 g NH 4 -N, 0.225 (32) = 7.2 g O 2, 0.005005 (3) = 0.565 g cells, 0.25 (4) =.75g NO - 3 -N Total NH 4 -N in a wastewater of,000 m 3 = (22 mg/l)(000 m 3 )(0 3 l/m 3 )( kg/0 6 mg) = 22 kg Then, - Oxygen consumption = 22 kg(7.2 g/.82 g) = 87 kg - Cell dry weight produced = 22 kg(0.565 g/.82 g) = 6.83 kg - Effluent NO 3 N con. = 22mg/l(.75 g/.82 g) = 2 mg/l
2.5. Fermentation Reactions Fermentation: an organic compound serves as e - donor and e - acceptor Knowledge of all reactants and products in a given case is all that is required to construct a balanced equation for the reaction, and we don t need to know the intermediates along the path, as long as they do not persist. (It is because no matter how complicated are the pathways by which chemical A is transformed into chemical B, the laws of energy and mass conservation must be obeyed ) Simple Fermentation: only one reduced product (e.g., ethanol lfrom glucose)
Example 2.6 Simple Fermentation Stochiometry Example: Glucose to ethanol fermentation, ti - fs=0.22, fe= -0.22 = 0.78 - Donor: glucose - Product: ethanol, - Cell synthesis reaction from C-,Table 2.4, R = fe *Ra + fs *Rc Rd 0.78*Ra: 0.3 CO 2 + 0.78 H + + 0.78 e - -> 0.065 CH 3 CH 2 OH + 0.95 H 2 O (o-5 in Table 2.3) 0.22*Rc: 0.044 CO 2 + 0.0NH 4+ + 0.0 HCO 3- + 0.22 H + + 0.22 e - -> 0.0 C 5 H 7 O 2 N + 0.099 H 2 O -Rd: 0.047 C 6 H 2 O 6 + 0.25 H 2 O -> 0.25 CO 2 + H + + e - (o-7 in Table 2.3) R: 0.047 C 6 H 2 O 6 + 0.0NH 4+ + 0.0 HCO 3- --> 0.0 C 5 H 7 O 2 N + 0.065 CH 3 CH 2 OH + 0.076 CO 2 + 0.044 H 2 O Every equivalent of glucose (one e - ) fermented, 0.065 mole ethanol is formed CO 2 Champagne, NH 4+ is required for cell growth.
2.5 Mixed fermentation Mixed Fermentation: more than a single product and/or a single reactant i)more than a single product (mixed e - acceptors) n i = e ai equiv = n ai Ra = e airr ai, where and j= equivaj n i= e eai = ii)more than a single reactant (mixed e - donors) Rd = n e dir di, where edi = n i = j= equiv di equiv equivdjd Ra : a half reaction for the e - acceptors e eai equiv : fraction of n reduced end products (acceptors) ai : equivalence of ai produced n i= and e di =
Example 2.7 Citrate Fermentation (multiple products) Bacteroides sp. Converts mol citrate into mol formate, 2mol acetate, and mol bicarbonate. Write R e. Energy reaction from Table 2.3 i) # of equivalents e formate = 2/(2+6) = 0. e acetate = 6/(2+6) = 0.889 - formate: reduced end product, 2 e - for one mole formate (o-6) - acetate: reduced end product, 6 e - for two moles acetate (o-) - bicarbonate: not considered because it s oxidized end product, like CO 2 ii) Ra = e formate Ra formate + e acetate Ra acetate 0.*Raf: 0.0555 HCO 3- + 0. H + + 0. e - -> 0.0555 HCOO - + 0.0555 H 2 O 0.889*Raa: 0. CO 2 + 0. HCO 3- + 0.889 H + + 0.889 e - -> 0. CH 3 COO - + 0.333 H 2 O Ra: 0. CO +066HCO - + +e - ->0 0555 - +0CHCOO - 2 0.66 3 +H >0.0555 HCOO 0. 3 COO +0388HO 0.388 2
2.5 Overall Reactions for Biological Growth iii) Overall reaction, R = Ra Rd (from table 2-3) Ra: 0. CO 2 + 0.66 HCO 3- + H + + e - ->0.0555 HCOO - + 0. CH 3 COO - + 0.388 H 2 O -Rd: 0.0555(COO - )CH 2 COH(COO - )CH 2 COO - +0.444 H 2 O->0.66CO 2 +0.66HCO 3- +H + +e - R e : 0.0555 (COO - )CH 2 COH(COO - )CH 2 COO - + 0.056 H 2 O -> 0.0555 HCOO - + 0. CH 3 COO - + 0.056 CO 2 By normalization, this equation becomes : (COO - )CH 2 COH(COO - )CH 2 COO - + H 2 O HCOO - + 2 CH 3 COO - + CO 2