"...Now, when you have a mixture of propane and n-butane, you don't have two boiling points, you have only one and it will lie between the two temperature given above, depending on the ratio of the two components of the liquid. Specifically, it depends on the molar weighted fraction."
Sorry, I still don't see it except as a special case, ie eutectic. This defines the composition of the gas ratio's in a two part mixture at any given temperture/pressure and can be calculated. In a closed system, NONE of the components actually boils, soo, I do understand what you are trying to say.
But, this contradicts what you stated earlier:
"The vapour pressure vs temperature of butane is a smooth curve, 0C just happens to be 1013mB. Similar smooth curve for propane, just higher pressure at any given temperature. For any blend of propane/butane, the partial pressure of each component is the same smooth curve of the pure companent just reduced by its mole weighted fraction. The net result is that the ratio of the two partial pressures is almost independent of temperature, meaning that the ratio of propane/butane in the gas mixture is almost independent of temperature.
The ratio of propane/butane in the gas state depends on the ratio of propane/butane in the liquid state (vapour pressure), but NOT on temperature.
So, if the GAS is say 50% propane at 10C, it will still be 50% propane at -10C"
Temperature and pressure are HIGHLY and DIRECTLY related, not independent of eachother. To say that the composition of the gas in the canister is the same at 10C and -10C is incorrect. The pressure exerted on the can, from each component, can be calculated for each temp.
However you are correct in that the propane will burn first, but not exclusivlely, in most canisters. Above the boiling point of the n-butane, it will add a considerable portion to any flame. But they will still form a balance at any given pressure (or temperature if you prefer, nearly indistinguishable.) However, below the boiling point of n-butane, it will add only a small amount to the flame. This assumes a wide open jet, of course, as in Bob's case...not enough to burn.
" A new canister with 30% propane and 70% n-butane as liquid will have around 65% propane in the gas mixture, regardless of the temperature."
This is incorrect. At temps below the boiling point of n-butane, we might see a lot more propane. At temps above that we might see a lot less, since there WILL be a difference between raw vapour pressure at -10C and raw vapour pressure at 10C for n-butane. (Since propane is always assumed to be at or above the boiling point, I am ignoring that and calling it a one. We just add a bit of pressure to a highly compressible gas/or vapour pressure if you prefer to look at it that way.)
Again, temperature and pressure are related. The molar fraction of vapour pressure will only change slightly for propane. Burn it at those two temps, I will garantee you that in the lower temp case the propane will be the primary component and the canister will burn normally at the higher temp. The open valve, of course, reverts the system to a more STP like system, I sort of doubt there is ever any actual boiling inside the cannister, though.