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I generally use rum extract to make a "non-alcoholic" version of my eggnog recipe (which calls for rum and bourbon). However, the rum extract in my cupboard still technically has some alcohol in it, and therefore might not always be an acceptable substitute, depending on who I'm cooking for.

What is the approximate difference in alcohol concentration when using an extract to mimic flavor, compared to the original spirit? I have always heard that the alcohol in (e.g.) rum extract is "trivial" or "insignificant," and certainly don't disagree -- but I am hoping for a more quantitative measure.

  • You might be better to cook down some rum, probably light rum. Most of the alcohol will cook off. The alcohol in the extract is not insignificant, it's 70 proof, but in reality it will cook out of whatever you are making... – farmersteve Dec 14 '17 at 2:08
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As an example, McCormick Rum Extract has 35 percent alcohol, and lists the other ingredients as water, run, and natural flavor, in that order. Depending on the volume of your other ingredients, the alcohol content may become insignificant. Even imitation extracts, rum and others, contain alcohol. Bakery emulsions are an alcohol-free alternative, a water-based flavor concentrate, such as those marketed by LorAnn Oils and Flavors.

  • All extracts from McCormicks are about 35% alcohol. Most of alcohol will evaporate during cooking. I was shocked at how much alcohol is in Almond, vanilla, rum whatever. You can just buy it in the store at 35% (70 proof) anyone can in the USA... weird... – farmersteve Dec 14 '17 at 2:03

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