Proper fertilization is critical to the healthy and vigorous growth
of a lawn. While general fertility guidelines can aid in maintaining
high quality turfgrass, even the best recommendations are of little
value if one cannot accurately apply the fertilizer recommended.
In order to intelligently purchase and apply turfgrass fertilizers,
one must be able to read and understand a fertilizer label.
Many different types of fertilizer are available for use on turfgrass.
Complete fertilizers contain nitrogen (N), phosphorous (P)
and potassium (K) and are widely used for turfgrass fertilization.
For certain situations, incomplete fertilizers containing
some combination of N, P and K, but not all three elements, may
be the best choice. Every fertilizer material, whether complete
or incomplete, must carry a label stating the guaranteed analysis
of the material. The exact label information may vary from state
to state, as no uniform countrywide regulations exist; the manufacturer
is, however, usually required to include the following label information:
1. Name, brand, or trade mark.
2. Guaranteed chemical analysis.
3. Potential acidity (CaCO3 equivalent)
4. Manufacturer's name and address.
5. Net weight of fertilizer in the container.
GUARANTEED ANALYSIS (or fertilizer GRADE) is a listing of nutrients
contained in the bag, by weight. The first number of the analysis
lists % N, the second number represents % P205
(phosphate), and the third number % K2O (potash).
a fourth number may appear prominently on the label. In turf fertilizers,
this fourth number usually represents either iron (Fe) or sulfur
(S), two supplemental nutrients that may be of value under special
The RATIO of a fertilizer is an important characteristic
to understand. The ratio of a fertilizer is the relationship between
the N - P2O5 -and K2O
content of a fertilizer. A fertilizer with a 3-1-2 ratio contains
twice as much N as K2O and three times more
N than P2O5.
The ratio does not usually reflect N, P and K content. Only N is
expressed on an actual elemental basis. P and K are expressed on
an oxide basis; that is, they are contained in the phosphate and
potash compounds. Phosphate contains 44% P and potash contains 83%
K. Thus, a 18-6-12 grade fertilizer contains 18% N, 6% P2O5
and 12% K2O. A 50 lb. bag of 18-6-12 contains
9 lbs N, 3 lbs P2O5
and 6 lbs K2O. It also contains only 1.3 lbs
of actual P and 5 lbs of actual K. Fertilizer grades of 45-15-30,
36-12-24 and 9-3-6
also have a 3-1-2 ratio. Pound for pound, a 36-12-24 fertilizer
contains twice the nutrients of a 18-6-12 fertilizer. Thus, to supply
equal amounts of nutrients, one could use 1/2 as much 36-12-24 as
18-6-12 to obtain a similar response.
possible, a soil test should be performed to determine the ratio
and amount of phosphate and potash which should be applied. If P
and K levels are adequate, (more than 10 ppm P, more than 100 ppm
K), an incomplete fertilizer containing only N would be a good choice.
If a soil test is not immediately available, a fertilizer ratio
of about 3-1-2 to 5-1-2 is normally recommended for turfgrass. It
is not necessary that the ratio be exactly 3-1-2 or 5-1-2 if it
is close to that ratio. For late summer and early fall fertilization,
a fertilizer with a balanced ratio of N and K (that is, one part
nitrogen for every one part potassium) is recommended to minimize
potential winter injury.
addition to listing % N, P2O5
and K2O by weight, the label of a turf fertilizer
further describes how much of the total N is water insoluble
nitrogen (WIN). WIN is ntrogen which is slowly released for
use by the turf over a long period of time (several weeks, months
or years) as opposed to quickly available water soluble nitrogen
(WSN). For lawn maintenance, a fertilizer containing both WSN and
WIN is desirable. WSN is quickly available to the turf and thus
provides improved color and growth very soon after application.
However, it is more likely to cause foliar burn at rates exceeding
2 lbs. of N/1000 sq. ft. or during periods of hot, dry weather.
At normal rates of 1 lb. of N/1000 sq. ft., the response from WSN
will last approximately 4-6 weeks depending upon climatic conditions.
If a longer period of response (and thus fewer applications) is
desired, a fertilizer containing some of the N as WIN should be
is slowly released by one of several mechanisms and, therefore,
is less likely to cause foliar burn; it provides a longer lasting
response than quickly available WSN. It is also more expensive per
pound of N applied, especially if the WIN is derived from natural
organic sources such as sewage sludge, plant extracts, proteins,
order to determine how much fertilizer to apply to an area, one
must know: square footage of turf to be treated, recommended
application rate and the analysis of the fertilizer.
Green-Way fertilizer (18-6-12) has been chosen to provide 1 lb.
of total N per 1000 sq. ft. of area. The turf area is 7,000 sq.
ft. How much Green-Way fertilizer is needed to fertilize the area?
to be fertilized) x (recommended rate of N) = Total lbs. of N
sq. ft.) x (1 lb. N per 1000 sq ft) = 7 lbs. N needed for the
nutrient needed) / (percent nutrient in analysis)= lbs. of actual
lbs. N needed)/(.18N/lb. of fertilizer) = 38.9 lbs. of fertilizer
needed for the job
about 39 lbs. of 18-6-12 are needed to supply 1 lb. of N per 1000
sq. ft. to a 7000 sq. ft. lawn.This same calculation can be used
to determine how much of ANY fertilizer to purchase to apply ANY
nutrient if you know: sq. ft. to be treated, fertilizer analysis
and recommended rate of the nutrient to be applied.
fertilizers, especially complete fertilizers, tend to cause an acidic
reaction in soils. The acidification is often a result of the oxidation
of the ammonium (NH4+ ) which provides the
fertilizer's N. Phosphorous and potash fertilizers commonly have
little effect on pH unless they also contain nitrogen. The CaCO3
equivalent is a measure of the acidifying potential of a fertilizer.
It expresses how much CaCO3 (calcium carbonate;
limestone) would have to be applied to the turf area to counteract
the acidifying effects after one ton (2000 lbs.) of the fertilizer
had been applied to the area.
Green-Way fertilizer is applied at a rate of 5 lbs./1000 sq. ft.
to a lawn. How much CaCO3 (limestone) would
have to be applied to counteract the effect of5 lbs. fertilize r/1
000 sq. ft. applied to a 40,000 sq. ft. area?
Lbs. of fertilizer) / (1000 sq. ft) x (40,000 sq.ft.) = 200 lbs
of fertilizer applied
lbs. of fertilizer applied) x (250 lbs CaCO3)
/ (2000 lbs.fertilizer indicated on label) = 25 lbs. CaCO3
Thus, 25 lbs. of limestone would be needed per 40,000 sq. ft.
to neutralize the acidifying nature of 200 lbs. of applied fertilizer.
The acidfying nature of a fertilizer is rarely of critical concern,
but in the absence of a soil test, it is a way of partially estimating
lime requirements over a long period of time.
from the University of Massachusetts Cooperative Extension, 2001