Anyone Here Good at Math? - Math and Medications

[casportpony]

Quote: Yes, good idea!

-Kathy

 

[casportpony]

Exercise Number Four


What are the per liter doses for the following:

• 0.024%
• 0.012%
• 0.006%

-Kathy

 

[casportpony]

Exercise Number Five

How many milligrams (mg) of amprolium are there in:

• One liter at the 0.024% level = ??? mg
• One liter at the 0.012% level = ??? mg
• One liter at the 0.006% level = ??? mg

-Kathy

 

[casportpony]

I start by converting to metric.

9.6% Corid = 16 ounces = 473.176 ml
@ 0.024%, 473.176 ml makes 50 gallons, so 473.176 ÷ 50 = 9.46352 ml and I round up to 9.5 ml for one gallon.

The 0.012% amount is 1/2 that (4.75ml) and the 0.006% amount is 1/4 the 0.024% amount (2.375 ml)

-Kathy

 

[casportpony]

Quote:
So how many ml's for each?

-Kathy

 

[Free Spirit]

So how many ml's for each?

-Kathy

I was just starting to work on that before I saw you post this very question. As it dawned on me that someone reading that would confuse mg and ml.

 

[Free Spirit]

So how many ml's for each?

-Kathy

Measuring 9.6% amprolium in mg = 0.096 mg x 1000 = 96g which is equal to 96ml (equal density 1 to 1 water mass/volume weight conversion). then the amount of amprolium in each dose in milliliters is...

2 ounce chick = dose of 1.12 mg of amprolium = 1.12mg divided by 96g = 0.011667g or 0.012ml
7 ounce chick = 3.969 mg dose of amprolium = 113.4 mg divided 96g = 0.041344g or 0.04ml
1 pound bird = 9.072 mg of amprolium = 9.072 mg divided by 96g = 0.0945g or 0.095ml
10 pound bird = 90.72 mg of amprolium 90.72 mg divided by 96g = 0.945g or 0.95ml


Ok figured out where I went wrong the first time and corrected it

.

 

[casportpony]

Exercise Number Two

Second up is how to calculate an oral drench of 9.6% amprolium based on this:

Source: http://www.manorvets.co.uk/ckfinder/userfiles/files/Manor chickens.pdf

What the above picture says is that the dose should be 20 mg of amprolium per 1kg (kilogram) of bird weight.

Calculate doses for the following:

• 2 ounce chick
• 7 ounce chick
• 1 pound bird
• 10 pound bird

Show your work!

-Kathy

Edited to change 200 ounce to 7 ounce

Edited by casportpony - Today at 12:51 pm

The formula I use is weight of bird in kg, times number of mg per kg, divide by number of mg per ml = number of ml to give.

2 ounces is 56.6990463 grams, so I round up to 57 grams. 57 grams is 0.057 kg, the liquid has 96 mg amprolium per ml, so it's
0.057 x 20 ÷ 96 = 0.011875 ml and I round up to 0.12 ml.

7 ounces is 198.447 grams, so I round up to 200 grams. 200 grams is 0.2kg.
0.2 x 20 ÷ 96 = 0.04166666666 ml - round up to 0.042 ml

For pounds I divide the pound by 2.2 to convert to kg.
1 pound bird - 1 ÷ 2.2 x 20 ÷ 96 = 0.09469696969 - round up to 0.095 ml
10 pound bird - 10 ÷ 2.2 x 20 ÷ 96 = 0.94696969697 - round up to 0.95 ml

-Kathy

 

[casportpony]

Note that I made a big typo... I meant to have a 7 ounce chick, not a 200 ounce chick, lol.

-Kathy

 

[Free Spirit]

Exercise Number Three


How many milligrams (mg) of amprolium are there in:

• One gallon at the 0.024% level = ??? mg
• One gallon at the 0.012% level = ??? mg
• One gallon at the 0.006% level = ??? mg

-Kathy

Using these formulas to convert from ounces to milliliters to grams for each of the 3 solutions:
1oz = 29.5735ml = 29.5735g
16oz = 473.176ml = 473.176g
8oz = 236.588ml = 236.588g
4oz = 118.294ml = 118.294g

0.024%: 16oz amp/50 gal = 473.176g divided by 50gal = 9.46352g per gallon
0.012%: 8oz amp/50 gal = 236.588g divided by 50gal = 4.73176g per gallon
0.006%: 4oz amp/50 gal = 118.294g divided by 50gal = 2.36588g per gallon

Next convert g/gal back to milligrams by multiplying by 1000 (shown in to formulas below).

Then find out how much amprolium there is in each gallon of solution. Convert the 9.6% amprolium into a decimal by moving two decimal places to the left (0.096). Then multiply by the amount in milligrams of each solution.

0.024%: 9463.52mg x 0.096 = 908.49792 mg amprolium in one gallon
0.012%: 4731.76mg x 0.096 = 454.24896 mg amprolium in one gallon
0.006%: 2365.88mg x 0.096 = 227.12448 mg amprolium in one gallon

Should be correct this time

.