Arithmetic and Minerals
Livestock Update, January 1997
Mark L. Wahlberg, Animal and Poultry Sciences
Recently I received a request from an Extension Agent for some assistance with a mineral formulation problem. The problem and method of solution are detailed below.
Situation: A producer is using a commercial mineral product that contains Selenium at 0.0052%. The producer wanted to increase the Selenium level in the mineral using a Selenium premix that contains .06% Selenium. How much of the Selenium premix needs to be added to the commercial mineral to have the final mixture contain 80 Parts per Million?
Background Information: There are a few basic facts needed to solve this problem. They are:
The Solution: First of all we need to calculate how many milligrams of Se are needed in 50 pounds of commercial mineral plus Se-600 in order to have the total mixture contain 80 PPM.
80 PPM = 80 mg per Kg of Mineral
50 Pounds / 2.205 Pounds per Kg = 22.7 Kg
Se in the 50-pound bag = 22.7 Kg X 52 Mg/Kg = 1180 Mg of Se
Mg of Se needed in 50-pounds to provide 80 PPM = 22.7 Kg X 80 Mg/Kg = 1816 Mg
Extra Se needed is 1816 - 1180 = 636 mg additional Se
There are 600 PPM or 600 Mg in 1 Kg of the high Se premix.
636 mg needed/600 mg in 1 kg = 1.06 Kg of high Se premix.
1.06 Kg X 2.205 Pounds in 1 Kg = 2.34 Pounds of high Se premix.
Final Se content of the mixture is 1180 mg in 22.7 Kg + 636 Mg in 1.06 Kg = 1816 mg in 23.76 Kg
1816 mg / 23.76 Kg = 76.4 Mg/Kg or 76.4 PPM.
You will note that the concentration is 76.4 PPM and not 80 because we added 2.34 pounds more material to the 50 pounds of mineral. However, it is certainly close enough to the desired outcome to be considered the proper amount, especially considering that we cannot estimate cow intake of mineral very accurately.