Question 1 of 7: Explosive Chemistry – Oxygen Balance of AN/FO

Question 1: Explosive Chemistry – Oxygen Balance of AN/FO (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Oxygen balance and its effect on fumes

The oxygen balance (OB) of an explosive is the percentage, by mass, of oxygen left over (positive OB) or required (negative OB) after every carbon atom in the formulation has been oxidised to CO₂ and every hydrogen atom to H₂O. It is computed from the compound's own molecular formula C_aH_bN_cO_d as

$$OB\% = \frac{1600}{MW}\left(d - 2a - \frac{b}{2}\right)$$

An explosive at exactly OB = 0% has just enough oxygen to complete combustion and produces almost pure CO₂, H₂O and N₂ – the cleanest, highest-energy detonation. A positive OB (oxygen surplus, "oxygen-rich") drives the reaction toward NO and NO₂ (the orange-brown, toxic "nitrous fumes"), because excess oxygen combines with the nitrogen instead of staying inert as N₂. A negative OB (oxygen-deficient, "fuel-rich") starves the carbon of oxygen and produces CO (odourless, highly toxic) in place of CO₂, along with unburned hydrocarbon soot. Both extremes lower the explosive's effective energy release (fume gases carry away chemical energy that should have gone into work on the rock) and both create a serious underground ventilation/ worker-safety hazard, which is why blast crews target OB as close to 0% as practical and why regulators cap fume class ratings on permitted underground explosives.

(b) Oxygen balance of 93% AN / 7% FO

Given. AN = NH₄NO₃ (MW = 14×2 + 1×4 + 16×3 = 80 g/mol); FO = CH₂ per repeat unit (MW = 12 + 1×2 = 14 g/mol); mixture 93% AN / 7% FO by mass.

Find. OB% of the mixture.

Approach. Compute each pure component's OB from its formula, then take the mass-weighted average (OB is additive on a mass basis for a physical blend).

1. Component oxygen balances. AN (a=0,b=4,d=3): $$OB_{AN}=\frac{1600}{80}\left(3-0-\frac{4}{2}\right)=\frac{1600}{80}(1)=+20.0\%$$ FO (a=1,b=2,d=0): $$OB_{FO}=\frac{1600}{14}\left(0-2-\frac{2}{2}\right)=\frac{1600}{14}(-3)=-342.9\%$$
2. Mass-weighted mixture balance. $$OB_{mix}=0.93(+20.0\%)+0.07(-342.9\%)=18.6\%-24.0\%=\boxed{-5.4\%}$$
3. Cross-check by direct O-atom balance (per 100 g mixture): 93 g AN = 1.1625 mol supplies 1.1625×3 = 3.4875 mol O; 7 g FO = 0.5 mol supplies 0.5 mol C and 1.0 mol H₂ (as CH₂, 1.0 mol H₂ equivalent). Complete combustion needs 2(0.5) + 0.5(2×0.5) … i.e. 2×C + ½×H = 2(0.5)+0.5(4.65+1.0)… carried through gives an O deficit of 0.3375 mol = 5.4 g per 100 g, i.e. OB = −5.4%, matching Step 2.

(c) Stoichiometric (oxygen-balanced) AN/FO composition

Approach. Set the mass-weighted OB to zero and solve for the AN mass fraction, or equivalently balance the whole-number combustion equation directly.

1. Solve OB_mix = 0 for AN fraction x: $$x(20.0)+(1-x)(-342.9)=0 \;\Rightarrow\; x=\frac{342.9}{362.9}=0.9449$$ so the balanced blend is 94.49% AN / 5.51% FO by mass.
2. Confirm via the balanced reaction $$3\,NH_4NO_3 + CH_2 \rightarrow 3N_2+7H_2O+CO_2$$ Mass of 3 AN = 3×80 = 240 g; mass of 1 FO = 14 g; total = 254 g. $$\%AN=\frac{240}{254}=94.49\%,\quad \%FO=\frac{14}{254}=5.51\%$$ identical to Step 1, confirming the balanced-equation route and the OB=0 algebraic route agree exactly.

This is the industry-standard "94.5/5.5" ANFO recipe; the exam's 93/7 mix (part b) is slightly fuel-heavy of stoichiometric.

(d) What controls AN/FO fume production

Beyond the bulk oxygen balance itself (part a), fume output is governed by: mixing quality – poorly prilled or segregated AN/FO leaves fuel-rich or oxidiser-rich pockets even when the bulk ratio is correct; water contamination – AN is hygroscopic, and wet or desensitised product detonates incompletely, sharply increasing CO and NO; confinement and coupling – poor confinement (large air gaps, inadequate stemming) lets the reaction quench before completion; charge diameter relative to critical diameter – sub-critical diameters propagate at low, unstable velocity of detonation (VOD) with incomplete reaction; presence of sulphide minerals in the surrounding rock (e.g. pyrite) can react with AN/FO byproducts to generate additional toxic gas independent of the explosive's own OB; and priming/ initiation energy – an undersized primer fails to drive the full column to steady-state VOD along its length.