About pesticide calculation

1. JODEN M. ADIOVA (jeyemey@gmail.com)
Department of Crop Protection

2. Pesticide calculations is the “easiest” part
of the exam.
1. You don’t need to memorize many
things
2. The problems are simple.
3. There are different ways you can come
up with the answer
4. I was assigned to handle this review
session =)
Why?

3. Analyze the problem.
1. What is asked?
2. What are the given?

4. Example:
You wish to apply 320 liters of spray
solution/ha to a 0.5 hectare field. The
recommended spray concentration of
the 45% EC pesticide is 0.04%. How
many liters of the commercial
formulation are required for the
treatment?

5. Example:
You wish to apply 320 liters of spray
solution/ha to a 0.5 hectare field. The
recommended spray concentration of
the 45% EC pesticide is 0.04%. How
many liters of the commercial
formulation are required for the
treatment?
1.What is being asked?
2.What are the given?

6. Example:
You wish to apply 320 liters of spray
solution/ha to a 0.5 hectare field. The
recommended spray concentration of the 45%
EC pesticide is 0.04%. How many liters of the
commercial formulation are required for the
treatment?
Asked: volume of pesticide needed
Given: vol of spray solution = 320 liters
area to be treated = 0.5 hectare
percent active ingredient = 45%
concentration of spray = 0.04%

7. REMEMBER!!!!!
Area: 1 hectare (ha) = 10,000 m²
Weight: 1 kilogram (kg) = 1,000 g
Volume: 1 liter (l) = 1,000 ml
1 gallon (gal) = 3.8 liters
1 tablespoon = 10 ml
1 g water = 1 ml water
1 kg water = 1 L water

8. REMEMBER!!!!!
Important terms
• % active ingredient
• Percent concentration
• Parts per million
• Dilution

9. Pesticide = active ingredient + other compounds
Active ingredient - the actual poison in the
pesticide expressed in percent as:
ent weight/weight as in wettable powder, granules,
water soluble powders
Percent weight/volume in liquid concentrates (e.g. g/l ,
kg/l)
% active ingredient ( % a.i.) tells us how much of the
actual poison is present in a formulated pesticide
Active Ingredient:
Diazinon ------------- 5.0 %
Inert Ingredient ----- 95.0 %
TOTAL ---------------- 100.0 %

10. Pesticide = active ingredient + other compounds
Active ingredient - the actual poison in the
pesticide expressed in percent as:
ent weight/weight as in wettable powder, granules,
water soluble powders
Percent weight/volume in liquid concentrates (e.g. g/l ,
kg/l)
% active ingredient ( % a.i.) tells us how much of the
actual poison is present in a formulated pesticide
PESTICIDE IN LIQUID FORM
1. Organic solvent – dissolves
pesticide
2. Emulsifier – help pesticide mix
with water
3. Spreaders and stickers – help
pesticide cover and stick to target
area better

11. Pesticide formulations:
1. Dusts
2. Granules
3. Liquid formulations
4. Wettable powders
5. Soluble powders

12. Pesticide = active ingredient + other compounds
Active ingredient - the actual poison in the
pesticide expressed in percent as:
a. Percent weight/weight as in wettable powder,
granules, water soluble powders
b. Percent weight/volume in liquid concentrates (e.g.
g/l , kg/l)
% active ingredient ( % a.i.) tells us how much of the
actual poison is present in a formulated pesticide

13. Pesticide = active ingredient + other compounds
Active ingredient - the actual poison in the
pesticide expressed in percent as:
ent weight/weight as in wettable powder, granules,
water soluble powders
Percent weight/volume in liquid concentrates (e.g. g/l ,
kg/l)
% active ingredient ( % a.i.) tells us how much of the
actual poison is present in a formulated pesticide
Active Ingredient:
Diazinon ------------- 5.0 %
Inert Ingredient ----- 95.0 %
TOTAL ---------------- 100.0 %

14. A 1 kg of Herbicide X contains 50% glyphosate.
How many grams of the active ingredient
does Herbicide X has?
a. If it has 25% a.i.
b. If it has 30% a.i.
Amt of a.i. = % a.i. x wt or vol of pesticide formulation
= 0.25 kg or 250 g a.i.
= 0.3 kg or 300 g a.i.

15. To control weeds, you were assigned to spray
herbicide to 1 ha cornfield with a recommended
rate of 1 kg a.i./ ha. You were given Herbicide X
which contains 50% a.i. How much Herbicide x
(in kg) are you going to use for the treatment?
A. If Herbicide X contains 25% a.i.?
B. If Herbicide X contains 43% a.i.???
amt needed
= 1 kg a.i./ha x 1 ha= 2.33 kg
0.43 a.i.
=
r.r. F x area to be treated
% a.i.
Recommended rate
( kg a.i. /ha)

16. 1. You were assigned to spray herbicide to 1.5 ha
cornfield with a recommended rate of 1 kg a.i./
ha. You were given Herbicide X which contains
20% a.i. How much Herbicide x (in kg) are you
going to use for the treatment?
2. If the recommended rate is 0.5 kg a.i./ha, how
much are you going to use in a 12,000 m² field
if the herbicide contains 25% a.i.? How much
a.i. are you going to use per sq m?

17. Percent concentration – number of parts by
weight or volume of active ingredient in 100
parts of water
% conc =
𝑣𝑜𝑙 𝑜𝑟 𝑤𝑡 𝑜𝑓 𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑔𝑟𝑒𝑑𝑖𝑒𝑛𝑡
𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
x 100

18. A 16 liter sprayer contains 2tbsp of glyphosate.
What is the percent concentration of the
solution?
% conc = 2 tbsp X (10 ml / 1 tbsp) x 100
16 liter x (1,000 ml / liter)
= 0.125 %

19. An 8 liter sprayer has 0.04% concentration of
butachlor. How many ml of active ingredient
are there in the solution?
% conc = active ingredient x 100
vol or wt of solution
= % concentration x total vol of solution
= 0.0004 x (8 liter x 1,000 ml / liter)
= 3.2 ml

20. A farmer has 20ml of glyphosate, what volume of
water will he need if he wishes to make a
solution that has 0.04% concentration?
% conc = active ingredient x 100
vol of water
vol of water = active ingredient
% concentration
= 20 ml / 0.0004
= 50,000 ml

21. 1. A 16 liter sprayer has 0.05% concentration of
2,4-D. How many ml of active ingredient are
there in the solution?
2. A farmer has 40ml of glyphosate, what volume
of water will he need if he wishes to make a
solution that has 0.04% concentration?
3. A 8 liter sprayer contains 2.5 tbsp of
glyphosate.What is the percent concentration
of the solution?

22. Parts per million (ppm) – number of parts by
weight or volume of active ingredient in
1,000,000 parts of water
Example:
1ppm = 1 ml /1,000,000 ml
20ppm = 20 g /1,000,000 g
ppm = percent x 10,000
Percent = ppm m
10,000

23. InsecticideY has 3% active ingredient, express
the concentration of insecticideY in parts
per million
ppm = %concentration x 10,000
= 3 x 10,000
= 30,000 ppm

24. An 8 liter sprayer solution has 400ppm concentration of active
ingredient. How many ml of active ingredient are there in the
solution?
Express ppm into % conc : % conc = ppm / 10,000
400ppm / 10,000
= 0.04%
= % conc x total vol of solution
= 0.0004 x 8 𝑙𝑖𝑡𝑒𝑟 𝑥
1,000 𝑚𝑙
𝑙𝑖𝑡𝑒𝑟
= 3.2 ml
% conc =
𝑣𝑜𝑙 𝑜𝑟 𝑤𝑡 𝑜𝑓 𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑔𝑟𝑒𝑑𝑖𝑒𝑛𝑡
𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
x 100

25. Suggestion: Consider using ratios to
solve calculation problems. No formulas
to remember! Logical!
400 f
1,000,000
= x f
8,000 ml
400 x 8,000 ml
1,000,000
= x
3.2 ml = x

26. A 16 liter sprayer contains 2tbsp of glyphosate.
What is the concentration of the solution in
ppm?
You can first look for % conc  ppm
= 2 tbsp (10ml/tbsp) x 100%
16 l (1000 ml / l)
= 0.125%
= 0.125% x 10,000
= 1,250 ppm
% conc =
𝑣𝑜𝑙 𝑜𝑟 𝑤𝑡 𝑜𝑓 𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑔𝑟𝑒𝑑𝑖𝑒𝑛𝑡
𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
x 100
Ppm = percent x 10,000

27. We can use ratio to solve this
concentration problem.
20ml f
16,000 ml
= x f
1,000,000
20 x 1,000,000
16,000
= x
1,250 = x

28. A farmer has 20ml of glyphosate, what volume of
water will he need if he wishes to make a solution
that has 400ppm concentration?
1,000,000
f 400
= x f
20 ml
20 x 1,000,000
400
= x
50,000 ml = x

29. You prepared a spray solution by mixing 20ml of
fungicide X in a 20 liters of water. Fungicide X
contains 50 EC (50% a.i. emulsifiable concentrate).
What is the percent concentration of the spray
solution?
20ml x 0.5 x 100
20 l (1,000 ml/l)
= 0.05 %

30. 15 grams of Herbicide X with 50% a.i. was added to 5
gallons of water. What is the percent concentration
of the solution?
15 g x 0.5 x 100
5 gal (3,800 ml/ gal)
= 0.04 %

31. You wish to apply 320 liters of spray solution/ha to a 0.5 ha
area.The recommended spray concentration of 45% EC
pesticide is 0.04%. How many liters of the commercial
formulation are required for the treatment?
Step 1 : Knowing the desired concentration, how much a.i.
should a 320 liters of spray solution have?
Step 2: Knowing the % a.i. of the pesticide and the amount of
a.i. in the solution, compute for the required amount of
pesticide.

32. You wish to apply 320 liters of spray solution/ha to a 0.5 ha
area.The recommended spray concentration of 45% EC
pesticide is 0.04%. How many liters of the commercial
formulation are required for the treatment?
Amt of a. i. = % concentration x total volume
= 0.0004 x 320 liters
Amt of a. i. = 0.128 liter
Vol of pesticide = r.r. / % a. i.
= 0.128 l / 0.45
= 0.284 liter
Vol needed for 0.5 ha = 0.142 liter

33. 1. How many square meters can a 62.5 ml of a 40 %
formulation cover if the recommended rate is 0.5
kg a.i. /ha?
Sol’n:
62.5 ml x 0.40 = 25 a.i.
10,000 m² f
500 g
= x f
25 g
10,000 m² x 25 g
500 g
= x
500 m² = x

34. 2. You put 0.588 kg of Pesticide X to 1,000 L to make
a 0.05% spray solution. What is the percent a.i. of
Pesticide X?
0.0005 x 1,000 L = 0.5
0.5 kg x 100 = 85%
0.588 kg

35. 3. A 2.875 kg WP insecticide is required for the
control of fruit and shoot borer in a 2 ha eggplant
field.The recommended rate is 0.5 kg a.i. /ha.
What is the % a.i. formulation ?
2.875 kg / 2 = 1.44 kg WP / ha
% a.i. = 0.5 kg x 100%
1.44 kg
= 34.8% a.i.

36. 5. Insecticide K has a dilution rate to water of 1:320.
What is its recommended dosage per 16 L of
water?
1 f
320
=
x f
16 L
0.05 L = x
50 ml = x

37. 6. If the recommended rate of insecticide R is 0.3 kg
a.i. / ha, how much of a 30% formulation, in liters,
is needed to cover an area of 650 sq m?
0.0195 kg = a.i. to cover 650 m²
total vol = 0.0195
0.3
= 0.065 L
0.3 kg f
10,000 m²
=
x f
650 m²

38. 7. Juan bought an insecticide for his 1.17 ha tomato
field. How much insecticide will he need if the
recommended dosage is 3 tbsp / 16 L of water and
spray volume of 160 L per hectare?
300 ml = needed in 1 ha
300 ml x 1.17 ha = 351 ml
30 ml f
16 L
=
x f
160 L

39. 8. A 1,000 L of 0.07% spray solution is needed to
control corn borer.The wettable powder to be
used contains 80% active ingredient.What is the
required weight ofWP formulation
0.0007 x 1,000 L = 0.7 kg  a.i. in 1,000 L
weight of WP = 0.7 kg  a.i.
0.80  % a.i.
= 0.875 kg

40. 9. A 0.6 kg a.i. / ha of insecticide J is needed to
control cabbage webworm. How much 40% WP
formulation is required for 1.75 ha cabbage field?
0.6 kg a.i. / ha x 1.75 ha = 2.625 kg
0.4

41. 10. How many liters of insecticide X are required to
treat a 0.5 ha area of tomato?The recommended
concentration is 0.04% , spray volume desired is
320 per ha and the insecticide contains 45% a.i.
0.0004 x 320 x 0.5 = 0.142 L
0.45

42. 11. How many sprayer loads are required to spray the
0.5 hectare if the spray volume required is 320
L/ha? Assume that the spray load holds 8 liters.
320 L / ha x 0.5 ha = 20 sprayer loads
8 L

43. 12. An 800 sq m of land is planted with eggplant. How
much insecticide Z 40% EC is needed to cover this
area with the recommended rate of 0.5 kg a.i. / ha?
0.5 a.i. / ha = 1.25 L / ha
0.40
1.25 L = x s
10,000 m² 800 m²
0.1 L = x

44. 1. If insecticide X contains the active ingredient imidacloprid at 300
g/liter of the formulated product, what percentage of the product
is its active ingredient?
2. During severe corn borer infestation, Furadan 3G should be
applied at 33.3 kg/ha. How much of the product is needed for
15,000 sq. m of corn in case of severe infestation?
3. If 33.3 kg Furadan 3G had been used and it has 30 g Carbofuran
active ingredient / kg product, how many kilograms of Carbofuran
was applied?
4. At the application rate of 33.3 kg Furadan 3G / ha, how much
would be applied per plant if there were 75,000 plants in the
15,000 m² farm?

45. 5. How much of Confidor SL 100 is needed to prepare 2,500
liters of spray solution if the dilution rate is 25 ml / 100 L
of water?
6. If the only available packaging in a farmer’s favorite store
is 250 ml bottle of Confidor SL 100, how many bottles
should the farmer buy if he needs 2.25 liters?
7. Confidor SL 100 costs P750 per 250 ml bottle. How much
does a mango farmer has to pay for 2.25 liters of the
product?
8. Sumicidin 3EC has to be mixed at 5 tbsp / 16 L water.
What is the dilution rate of Sumicidin 3EC to water?

46. 9. If Sumicidin 3EC has to be mixed at 5 tbsp / 16 L water
what is the percentage of Sumicidin 3EC in the spray
solution?
10. At seedling stage of the rice plant, insecticide M should
be applied at 19 16-liter sprayer loads per hectare. How
many liters of spray solution have to be prepared for a
1.25 hectare rice farm at seedling stage?
11. A carbamate insecticide, effective aginst thrips, has 500
g Methiocarb per kg formulated product. What is the
concentration in percent of the formulated product?

47. 12. For severe infestation, pyrethroid insecticide with 25% a.i.
should be used at 5 tbsp /16 L water.What is the concentration in
percent of the spray solution?
13. If 150 liters of the spray solution should be prepared , how many
kgs of the formulated prodcut are needed with the
recommendation that it has to be applied at 5 tbsp /16 L water?
14. A slurry, good for 1kg of seeds, is prepared by mixing 12.5 ml of
insecticide in 10 ml water. How much of the slurry is needed for
4.5 tons of seeds?
15. How much water is needed in the preparation of a slurry for 4.5
tons of seeds if the dilution rate is 12.5 ml of insecticide per 10 ml
water per kilogram of seeds?

48. 16. If a farmer would do his own seed treatment how much
of insecticide ST would he need to treat 20 kg of seeds
enough to plant a hectare given a dilution rate of 12.5 ml
of insecticide in 10 ml water?
17. A farmer used 360 ml of insecticideY in his farm. He
followed the recommended dosage of 3 tbsp / 16 L water
and spray volume of 160 liters per hectare.What is the
area of his farm?
18. What is the rate of application in amount of formulated
product per hectare if the recommended dosage is 3
tbsp / 16 L water with spray volume of 160 L/ha?

49. 19. What is the rate of application in terms of g a.i. / ha if
InsecticideY which has 2.5% a.i. has to be diluted at
4 tbsp / 16 L of water and applied at 240 L / ha?
20. A farmer wants to know why Herbicide X, (containing
400g ai/liter product) failed to provide adequate
weed control . The recommended rate of application
of the herbicide is 0.4 kg ai/hectare. According to the
farmer he placed 60ml of Herbicide X into each
tankload and that one tankload of spray was
sufficient to cover 2 plots measuring 25m x 20m
each. Why was his treatment failed to control weeds
as expected?