• Please visit and share your knowledge at our sister communities:
  • If you have not, please join our official Homebrewing Facebook Group!

    Homebrewing Facebook Group

How much malt

Homebrew Talk

Help Support Homebrew Talk:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

Brutus Brewer

Well-Known Member
Joined
Jul 12, 2006
Messages
468
Reaction score
22
I would like to formulate my own recipe for a red ale and am unsure how much malt extract to use. From most of the recipe's I've read I figure about 7 pounds. Does this sound right?
 
LME has 36 gravity points/lb./gal
DME has 40 gravity points/lb./gal

That means that 1 lb of LME in 1 gallon of water will give you an OG of 1.036. So if you want a 5 gallon batch of beer to have an OG of 1.036 you would need 5 lbs of LME.

To find out what your OG would be with any given amount of extract, use this formula. Take the amount of extract in lbs, multiply it by its gravity points (36 for LME, 40 for DME) and divide that by the total volume of the batch. Lets use your figure of 7 lbs of LME. 7 lbs x 36 gravity points = 252 ÷ 5 gallons = OG of 1.050.

John
 
Newbie here... how do you get a OG of 1.050 by dividing 252 by 5... I get 50.4.

Remember I am a newbie :)

Nevermind... did a search...

Thanks
 
So, you realized what I was saying right? The 1 in the 1.050 is a given. 1 is the specific gravity of plain water at 60˚F, and any sugars in the water will raise the SG above one, and that the 36 gravity points for a lb of LME in a gallon of water means .036.

John
 
johnsma22 said:
So, you realized what I was saying right? The 1 in the 1.050 is a given. 1 is the specific gravity of plain water at 60˚F, and any sugars in the water will raise the SG above one, and that the 36 gravity points for a lb of LME in a gallon of water means .036.

John
I was still scatching my head. Thanks! That makes sense.
 
Back
Top