Day 96

The last official day of classes for the fall semester, and Ingredients of Brewing ran right up to the imaginary wire–first, an 80-minute test on material covered in the past three weeks (hop planting, hop diseases and pests, and the start of water chemistry, remember?), followed by another hour to finish up water chemistry. The calculation of brewer’s salts was the last eensy-weensy squiblet of learning squeezed from the Ingredients tube:

  1. Calculate how many milligrams per litre you will need of the ion. Gently set this number aside for a moment, with promises that you will play with it anon.
  2. Calculate the molecular weight of the brewer’s salt you are using, and the percentage of this that the desired ion represents.
  3. Retrieving the first number from where you set it aside a moment ago, divide this by the second number to give the number of milligrams of brewer’s salt needed.

Example: Using gypsum (calcium sulphate dihydride), we want to raise the concentration of calcium  in 10 hectolitres of strike water by 60 mg/L. How much gypsum do we add?

  1. Amount of calcium needed is 60 mg/L x 1000 litres = 60,000 mg
  2. Molecular weight of gypsum (CaSO4·2H2O) is 40.1 (calcium) + 32.1 (sulphur) + 64 (4 oxygen) + 4 (4 hydrogen) + 16 (2 oxygen) = 172.2. The percentage of calcium is therefore 40.1/172.2 = 23.3%
  3. Amount of calcium needed/percentage of calcium = 60,000 mg/.233 = 257,655 mg = 257.7 grams of gypsum needs to be added to the 10 hL of strike water.

That was the end of the class but not of the term. Although classes “officially” finish today, we actually  have one more class with Kevin Somerville tomorrow afternoon, ostensibly to make up for Thanksgiving Monday.

In the meantime, on to business math for a final review, and then…

Cry “Havoc!”, and let slip the dogs of exams!

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3 Comments on “Day 96”

  1. Canageek Says:

    A percentage? Intrestinggggg.

    Note 1: Molecular Weight/Molar Mass has the units g/mol (Don’t worry about what a mole is)

    Given that, there is a better way of doing this:
    1. Divide your target concentration (g/L) by the mass of the ion (g/mol) giving you moles/L (The molar concentration, can also be written M as in 5 M nitric acid).

    2. Multiply the number of moles of ion by the molecular weight of the molecule it is contained in (g/mol) giving you g/L.

    3. Multiply by the volume of your solution (L), giving you grams.

    —–
    Based on your above example:

    60 mg/L / 40.1 g/mol = 1.5 mmol/L.
    1.5 mmol/L * 172 g/mol = 257.4 mg/L gypsum.
    257.4 mg/L * 1 kL = 257.4 g.

    The plus side is there are no %, so if you blank on what to do on a test you can just follow the units.

  2. Alan Brown Says:

    A mole is a furry little critter that digs tunnels under the lawn, right?

    • Canageek Says:

      A mole is a set number of atoms (Specifically 1 Avogadro’s number, 6.0221415 × 10^23, but the exact number isn’t important). Basically it puts everything into terms of the number of atoms present, regardless of mass. So if you have 1 mole of table salt (NaCl) you have 1 mole each of Na+ and Cl-, despite the fact they weight different amounts, and thus your ppm of each is different.


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