First of all: If you receive a fertiliser recommendation without having explained exactly which plants you are growing, you can safely ignore such recommendations. There are not hundreds of fertiliser types because there is one answer.
Each plant species has individual nutrient requirements that also differ according to the growth phase it is in. Furthermore, indiscriminate fertilising, over-fertilising, under-fertilising, wrong composition etc. can have devastating consequences for many plants, ranging from undersupply to specific plant diseases. In order to achieve the best nutrient mixture for a specific plant, there is no getting around an analysis of the plant itself. For cost reasons alone, we recommend preparing the nutrient composition yourself.
Mixing hydroponic fertiliser yourself ?
The commercially available fertilisers consist of a complete fertiliser supplemented with macronutrients. They are offered by some hydroponics and/or fertiliser companies and vary depending on the hydroponic plant. An example of a fertiliser programme is the hydroponic tomato programme offered by Hydro-Gardens.
In this programme, growers purchase Hydro-Gardens Chem-Gro tomato formula. It has a composition of 4-18-38 and also contains magnesium and micronutrients. To make a nutrient solution, it is supplemented with calcium nitrate and magnesium sulphate, depending on the variety and/or growth stage of the plant.
Advantages of fertiliser programmes
Programmes like these are easy to use. Minimal ordering of fertilisers is required (only 3 in the Hydro-Gardens example). Very little or no mathematical calculations are required to prepare nutrient solutions.
Disadvantages of fertiliser programmes
Fertiliser programmes do not allow for easy adjustments of individual nutrients. For example, if the leaf analysis shows that more phosphorus is needed. When using a fertiliser programme exclusively, it is not possible to simply add phosphorus. Another disadvantage is that fertiliser programmes do not allow farmers to take into account the nutrients already present in the water source. For example, if a water source has a potassium content of 30 ppm, there is no way to adjust the amount of potassium added in the fertiliser programme. And too much potassium can in turn block the uptake of other nutrients.
Fertilizer programs can be more expensive than using Recipes for the production of nutrient solutions.
Mix recipes for nutrient solutions / hydroponics fertilizer yourself
There are also recipes for the production of nutrient solutions. The recipes contain a certain amount of each nutrient to be added to the nutrient solution. They are specifically available for a specific crop and in a variety of sources, e.g. B. at the university advice centers, on the Internet and in specialist journals. One example is the modified Sonovelds solution for herbs (Mattson and Peters, Insidegrower) shown below.
Modified Sonneveld recipe / herbs
element
concentration
Nitrogen
150 ppm
Phosphorus
31 ppm
Potassium
210 ppm
Calcium
90 ppm
Magnesium
24 ppm
Iron
1 ppm
Manganese
0.25 ppm
Zinc
0.13 ppm
copper
0.023 ppm
Molybdenum
0.024 ppm
Boron
0.16 ppm
It is at the discretion of the breeder which fertilizers he uses to produce a nutrient solution according to a recipe. The fertilizers commonly used include:
fertilizer
Dosage, contained nutrients
Calcium nitrate
15.5 – 0 – 0.19% calcium
Ammonium nitrate
34 – 0 – 0
Potassium nitrate
13 – 0 – 44
Sequestrene 330TM
10% iron
Potassium phosphate monobasic
0 – 52 – 34
Magnesium sulfate
9.1% magnesium
Borax (laundry quality)
11% boron
Sodium molybdate
39% molybdenum
Zinc sulfate
35.5% zinc
Copper sulfate
25% copper
Magnesium sulfate
31% manganese
Farmers calculate the amount of fertilizer in the
nutrient solution based on the amount of a nutrient
in the fertilizer and in amount specified in the recipe.
Advantages of nutrient solution recipes
Nutritional solutions allow fertilizers to be adjusted based on the nutrients contained in water sources. An example: A gardener uses a water source with 30 ppm potassium and produces the modified Sonneveld solution for herbs that requires 210 ppm potassium. It would have to add 180 ppm potassium ( 210 ppm - 30 ppm = 180 ppm ) to the water in order to obtain the amount of potassium required in this recipe. With recipes, nutrients can be easily adjusted. When a leaf analysis report indicates that a plant has iron deficiency. It is easy to add more iron to the nutrient solution. Since recipes make it easy to adapt, fertilizers can be used more efficiently than in fertilizer programs. Using recipes can be less expensive than using fertilizer programs.
Disadvantages of nutrient solution recipes
It has to be calculated how much fertilizer has to be added to the nutrient solution. (Link to performing calculations). Some people may feel intimidated by the calculations involved. However, the calculations only require uncomplicated mathematical skills based on multiplication and division. A high-precision scale is also required for the measurement of micronutrients, since the required quantities are very small. Such a scale can be found on Amazon from 30.- €: e.g .: KUBEI 100g / 0.001g.
Calculation of the concentrations of nutrient solutions using the following two equations
The calculation of the amount of fertilizer that has to be added to the nutrient solutions is part of a successful hydroponic production. Only multiplication, division and subtraction are used for the calculations; no advanced mathematical knowledge is required.
If you want to know more about the compositions and concentration information, the article series can be too Stoichiometry and a look at the conversion of Mol and grams When specifying the concentration of the individual elements and connections, it is helpful to better understand the complexity of the topic.
If you master the general process, producing nutrient solutions and adjusting the amount of nutrients is child's play.
Fertilizer recipes for hydroponics are almost always given in ppm (in the long form: parts per million). This may differ from the fertilizer recommendations for growing vegetables and fruits outdoors, which are generally given in lb / acre (pounds per acre).
First you have to convert ppm to mg / l (milligrams per liter) using this conversion factor: 1 ppm = 1 mg / l (1 part per million corresponds to 1 milligram per liter). For example, if 150 ppm nitrogen is required in a recipe, 150 mg / l or 150 milligrams of nitrogen in 1 liter of irrigation water are actually required.
Ppm P (phosphorus) and ppm K (potassium) are also used in recipes for nutrient solutions. This also differs from the fertilizer recommendations for growing vegetables and fruits in the field, which use P2O5 (phosphate) and K2O (potash). The fertilizers are also given as phosphate and potash. Phosphate and potash contain oxygen, which must be taken into account in hydroponic calculations. P2O5 contains 43% P and K2O contains 83% K.
Let us check the previous circumstances:
1 ppm = 1 mg / l P2O5 = 43% P K2O = 83% K
Nutrient solution tanks are usually measured in gal ( gallons ) in the United States. When we convert ppm to mg / l, we work with liters. To convert liters into gallons, use the conversion factor of 3.78 l = 1 gal ( 3.78 liters corresponds to 1 gallon ). The invoice is also given below for continental interested parties.
Depending on the scale you use to weigh fertilizers, it may be useful to convert milligrams into grams: 1,000 mg = 1 g ( 1,000 milligrams correspond to 1 gram ). If your scale measures in pounds, you should use this conversion: 1 lb = 454 g ( 1 pound = 454 grams ).
Let us summarize these circumstances:
3.78 l = 1 gallon 1000 mg = 1 g 454 g = 1 lb
Now we have all the necessary circumstances. Let's look at an example.
How do you determine how much 20-10-20 fertilizer is needed to deliver 150 ppm N with a 5 gallon tank and a fertilizer injector that is at a concentration of 100:1 is set?
First, write down the concentration you know you want to reach. In this case it is 150 ppm N or 150 mg N / l.
Note that we multiply by 1. This allows you to cancel the units that are the same in the numerator and denominator. Now we can paint "mg N" and get the unit g N / l water.
Continue this process by converting liters into gallons. Most containers are still traded in gallons ( 3.78 liters ). Entertaining here: the metric system was invented by the Britten. If you want a metric result, omit this calculation step.
Now there are only grams of nitrogen left per gallon of water. We'll get closer to it. Now we want to convert grams of nitrogen into grams of fertilizer. Remember that our fertilizer is a 20-10-20, which means that it contains 20% nitrogen. It can be imagined that 100 grams of fertilizer contain 20 grams of nitrogen.
So where do we stand now? We calculated how many grams of fertilizer are needed in each gallon of irrigation water. At the moment we have a normally strong solution. Our example prompts us to calculate a concentrated solution of 100: 1. This means that for every 100 gallons of water that are applied, 1 gallon of stock solution is also applied via a fertilizer injector. We also know that our storage tank holds 5 gallons. Below see calculation for metric system (liters).
In gallons
In the calculator: 150 x 1: 1000 x 3.78 x 100: 20 x 100 x 5 is 1417.5 grams on 5 gallons of water (in the storage tank)
After we have deducted everything, we have a gram of fertilizer left. This is the amount of fertilizer we need to put in our storage tank to apply 150 ppm N at a concentration of 100: 1. Multiply and divide and you get the answer 1417.5 grams of fertilizer.
In liters
In the calculator: 150 x 1: 1000 x 100: 20 x 100 x 10 is 1500 grams per 10 liters of water ( in the storage tank )
After we have deducted everything, we have a gram of fertilizer left. This is the amount of fertilizer we need to put in our storage tank to apply 150 ppm N at a concentration of 100: 1. Multiply and divide and you get the answer 750.0 grams of fertilizer.
This means that for every 100 liters of water that is applied, 1 liter of stock solution is also applied via a fertilizer injector. We also know that our storage tank holds 10 liters.
If we measure in pounds, we have to put 0.75 kg / 1.15 lb fertilizer in our storage tank to apply 150 ppm N with a concentration of 100: 1.
You have just completed one of the two equations. Now let's look at the other one.
We just found that we need to add 750 grams of fertilizer to deliver 150 ppm nitrogen at a concentration of 100: 1. The fertilizer we used was a 20:10:20. In addition to nitrogen, we also add phosphorus and potassium. With the next equation we determine how much phosphorus we supply. This is basically the reversal of the first calculation.
We start with the amount of fertilizer that we put in our tank. The final units are ppm or mg / l. As with the previous calculation, we use our specifications until we receive these units.
Multiply with the concentration of the nutrient solution.
Multiply to convert to liters.
Next, convert milligrams of fertilizer into milligrams of phosphate.
Next we will convert grams of phosphate into grams of phosphorus, assuming that phosphate contains 43% phosphorus.
Finally, we convert grams of phosphorus into milligrams of phosphorus.
When we calculate this, we find that we have added 32.25 mg / l P or 32.25 ppm P. This is the second equation. We can also use them to determine how much potassium we have added.
We added 124.5 mg / l K or 124.5 ppm K.
With these two basic calculations, you can use any nutrient solution recipe program. How they are used to calculate a recipe can be seen in this article:
Now that you have the two basic equations for the production of nutrient solutions, we want to use them to calculate the amounts of fertilizer required for a nutrient solution recipe.
If you are not familiar with the two equations, read this first: Hydroponic systems: Calculating the concentrations of nutrient solutions using the two equations.
Here is our problem: We want to use a modified Sonneveld solution (Matson and Peters, Insidegrower) for herbs in an NFT system. We use two 5-gallon containers and injectors set to a concentration of 100: 1 and call them storage tank A and storage tank B. How much of each fertilizer do we have to put in each storage tank ?
You may be asking: why two storage tanks? This is due to the fact that certain chemicals in our fertilizer solution react with each other as soon as they come into contact with each other. In all nutrient solutions ( fertilizer mixtures ) you have calcium, phosphates and sulfates - among other things, these three chemicals for all plants vital are. The last two react with calcium and are no longer present in the form we need in our nutrient solution. They connect to each other and fall to the bottom of the container as white flakes ( precipitates ). Therefore, phosphates and sulfates must be kept separate from calcium and, when introduced into the nutrient solution of the ( system, saved from direct mixing by means of a dosing pump or measuring cup ).
Modified Sonneveld recipe for herbs
element
concentration
nitrogen
150 ppm
phosphorus
31 ppm
potassium
210 ppm
calcium
90 ppm
magnesium
24 ppm
iron
1 ppm
manganese
0.25 ppm
zinc
0.13 ppm
copper
0.023 ppm
Molybdenum
0.024 ppm
boron
0.16 ppm
These are the fertilizers that we will use. Some fertilizers contain more than one nutrient in the recipe, while others contain only one. Here is a small overview Commercial fertilizer from which you can put together your recipe
Fertilizer
Contained nutrients
(Nitrogen phosphate potassium and other nutrients)
Calcium nitrate
15.5-0-0, 19% Ca (calcium)
Ammonium nitrate
34-0-0
Potassium nitrate
13-0-44
Potassium phosphate monobasic
0-52-34
Magnesium sulfate
9.1% mg (magnesium)
Sequestrene 330 TM
10% Fe (iron)
Manganese sulfate
31% Mn (Mangan)
Zinc sulfate
35.5% Zn (zinc)
Copper sulfate
25% Cu (copper)
Boron
11% B (Boron)
Sodium molybdenum
39% Mo (molybdenum)
The first thing you notice is that we have three sources of nitrogen (calcium nitrate, ammonium nitrate and potassium nitrate), have two sources of potassium (potassium nitrate and potassium phosphate monobasic) and one source of calcium (calcium nitrate) and phosphorus (single-base potassium phosphate). We can start calculating the calcium or phosphorus in the recipe because only one fertilizer provides each nutrient. Let's start with calcium.
The recipe provides 90 ppm calcium. We calculate how much calcium nitrate we need to use to achieve this by using the first of our two equations.
We need to add 895.3 g calcium nitrate to get 90 ppm calcium. However, calcium nitrate also contains nitrogen. We use the second equation to determine how much nitrogen should be added in ppm.
We add 73.4 mg N / l or 73.4 ppm nitrogen. Our recipe provides 150 ppm nitrogen. If we subtract 73.4 ppm nitrogen from it, we have to add 76.6 ppm nitrogen.
Let us now calculate how much single-base potassium phosphate we have to use to deliver 31 ppm phosphorus.
We need to add 262 g of potassium phosphate monobed to get 31 ppm phosphorus. However, potassium phosphate also contains single-base potassium. We use the second equation to determine how much potassium should be added in ppm.
We add 39 mg K / l or 39 ppm potassium. Our recipe provides 210 ppm potassium. If we subtract 39 ppm of potassium from it, we see that we still have to add 171 ppm of potassium.
We have only one other source of potassium, namely potassium nitrate. Let's calculate how much we have to use of it.
We need to add 885 g of potassium nitrate to get 171 ppm of potassium. However, potassium nitrate also contains nitrogen. We use the second equation to determine how much nitrogen should be added in ppm.
We add 61 mg N / l or 61 ppm nitrogen. Our recipe provides 150 ppm nitrogen. We supplied 73.4 ppm nitrogen from calcium nitrate and had to add 76.6 ppm nitrogen. Now we can subtract 61 ppm nitrogen. We still have to add 15.6 ppm nitrogen. The only source of nitrogen that we have is ammonium nitrate.
Let us now calculate how much ammonium nitrate we have to use to deliver 15.6 ppm nitrogen.
We need to add 86.7 g of ammonium nitrate to get 15.6 ppm nitrogen.
At this point we have completed the nitrogen, phosphorus, potassium and calcium part of the recipe. For the other nutrients, we only need to use the first equation, since the fertilizers that we use for their supply contain only one nutrient in the recipe.
We need to add 498.5 grams of magnesium sulfate to get 24 ppm magnesium.
We need to add 18.9 grams of Sequestren 330 to get 1 ppm of iron.
We need to add 1.5 grams of manganese sulfate to get 0.25 ppm manganese.
It is easier to weigh small amounts of fertilizers in milligrams. The conversion from milligrams to grams is therefore carried out as follows
We need to add 692 milligrams of zinc sulfate to get 0.13 ppm zinc.
We need to add 0.17 milligrams of copper sulfate to get 0.023 ppm copper.
We need to add 2.8 milligrams of borax to get 0.16 ppm borax.
We need to add 0.12 milligrams of sodium molybdate to get 0.024 ppm molybdenum.
Summary:
Element
Addition
Nutrient Solution
Calcium
895.3 g calcium nitrate
90 ppm calcium
Phosphorus
262 g of potassium phosphate monobasic
31 ppm phosphorus
Potassium
885 g potassium nitrate
171 ppm potassium
Nitrogen
86.7 g ammonium nitrate
15.6 ppm nitrogen
Magnesium
498.5 grams of magnesium sulfate
24 ppm magnesium
Iron
18.9 grams of sequestrene 330
1 ppm iron
Manganese
1.5 grams of manganese sulfate
0.25 ppm manganese
Zinc
692 milligrams of zinc sulfate
0.13 ppm zinc
Copper
0.17 milligrams of copper sulfate
0.023 ppm copper
Boron
2.8 milligrams of borax
0.16 ppm boron
Molybdenum
0.12 milligrams of sodium molybdate
0.024 ppm molybdenum
Now all calculations have been completed. Now we have to decide in which storage tank, A or B, we give the individual fertilizers. In general, the calcium should be kept in a tank other than the sulfates and phosphates, as they can form precipitates that can clog the drip bodies of the irrigation system. Using this guideline, we can put the calcium nitrate in one tank and the monobasic potassium phosphate, magnesium sulfate, manganese sulfate, zinc sulfate and copper sulfate in the other tank. The rest of the fertilizers can be placed in both tanks.
You should also consider the amount of nutrients in irrigation water. For example, if we use irrigation water that contains 10 ppm magnesium, we only need to add 14 ppm more with our fertilizer (24 ppm Mg, which are required in the recipe, minus 10 ppm Mg in water). This is a great way to use nutrients more efficiently and fine-tune your fertilizer plan.
With some micronutrients, you have to decide for yourself what you want to add. You could do a small experiment to find out whether you need to add 0.12 milligrams of sodium molybdate to your stock solution, for example, or whether you are satisfied with the performance of your plants without this addition.
One last point to consider. Sometimes the calculations don't work as well as here for fertilizers that contain more than one required nutrient, and you may need to add more of a nutrient, than is provided in the recipe to provide the other nutrient.
For example, if you apply calcium nitrate to meet calcium needs, the solution may not contain enough nitrogen. In such cases, you have to decide which nutrient you want to give priority to. For example, you could apply calcium nitrate to meet the plants' nitrogen needs because the excess amount of calcium does not harm the plants. Or you choose to apply it based on the plant's calcium needs because the lack of nitrogen is just a few ppm.
First, identifying the elements that make up the compound.
Example: the compound NaHCO 3 consists of four elements: sodium (Na), hydrogen (H), carbon (C) and oxygen (O).
Then determine the number of atoms each element contributes to the compound.
Example: H 2 O has two hydrogen and one oxygen atoms. If an index follows a bracket in a compound, each element in the bracket is multiplied by the index. For example, (NH 4 ) 2 S consists of two nitrogen atoms, one hydrogen atom and one sulfur atom.
Record the atomic weight of each element. A periodic table is the easiest way to determine the atomic weight of an element. Once you've found the element on a periodic table, the atomic weight is usually listed below the element symbol. For example, the atomic weight of oxygen is 15.99.
Calculate Molecular Mass: The molecular mass of a substance is calculated by multiplying the number of atoms of each element by its respective atomic weight.
To convert grams to moles, you need to know the molecular mass of the compound.
Multiply the number of atoms of each element by its atomic weight. Add the total weights of all the elements in the compound together. Here you will find a periodic table to read the values.
Example:
Let's say you have 2g of water, or H2O , and you want to know how much that is in moles. The molecular mass of H 2 O is 18g/mol. Divide 2 by 18 and get 0.1111 mol H 2 O. In the periodic table under H you will find the weight 1.0080 for hydrogen (top right in the box) and the weight 15.999 for oxygen (O). That's two times 1.0080 plus one time 15.999. Around 18 u or ame. Here you can find out more about the details of the atomic weight (u, or ame).
First step
Second step
Result
Another example:
(NH 4 ) 2 S has the molecular mass of (2 x 14.01) + (8 x 1.01) + (1 x 32.07) = 68.17 g/mol.
Molecular mass was previously also referred to as molecular weight.
The number of moles in a compound can be calculated by dividing the number of grams of the compound by the molecular mass of the compound.
The formula looks like this: moles = grams of the compound : molecular mass of the compound.
Once you have set up the formula, you can insert the calculations into the appropriate place in the formula. An easy way to check that everything is in the right place is to look at the units. You should be able to reduce all units so that only moles remain.
Divide the number of grams by the molecular mass. The result is the number of moles in your element or compound.
This shortened overview serves as an aid in estimating the magnitude of the analytical technology required when analyzing and controlling the nutrients with which the plants are fertilized.
The quality of analysis in chemistry has already reached a level of precision that is unnecessary for our purposes of controlled fertilization. In order not to shoot at sparrows when selecting the various analysis methods and analysis devices, we have listed here a very shortened overview of the necessary accuracies that are sufficient for checking the individual additives. The technology used in the chosen analysis method has a major influence on the overall operating costs.
In addition to checking the necessary substances, monitoring is also necessary to prevent over-fertilization. The nutrients produced by fish farming must not exceed a certain concentration, otherwise this will impair the optimal growth of the plants.
There are now a very large number of analysis methods on the market, which differ greatly in both the technology used and the on-site application. This overview will help you, even without our advice , to obtain offers from different manufacturers that exactly meet your needs. Here is a random selection of manufacturers.
Here you will find the essential compounds required for plant growth. Depending on the plant and/or growth phase, the form of administration, the chemical compound in which the desired “substance” is bound, can or must vary. In the previous cultivation method (in the soil), the microorganisms and fungi caused the necessary compounds to be broken down. Since no microorganisms take on this task in hydroponics, this is still a current area of basic research.
Compounds and trace elements / orders of magnitude in nutrient solutions
K
potassium
0.5 - 10 mmol/L
Approx
calcium
0.2 - 5 mmol/L
S
sulfur
0.2 - 5 mmol/L
P
phosphorus
0.1 - 2 mmol/L
Mg
magnesium
0.1 - 2 mmol/L
Fe
iron
2 - 50 µmol/L
Cu
copper
0.5 - 10 µmol/L
Zn
zinc
0.1 - 10 µmol/L
Mn
manganese
0 - 10 µmol/L
b
boron
0 - 0.01 ppm
Mo
molybdenum
0 - 100 ppm
NO2
nitrite
0 – 100 mg/L
NO3
nitrate
0 – 100 mg/L
NH4
ammonia
0.1 - 8 mg/L
KNO3
Potassium nitrate
0 - 10 mmol/L
Ca(NO3)2
Calcium nitrate
0 - 10 mmol/L
NH4H2PO4
Ammonium dihydrogen phosphate
0 - 10 mmol/L
(NH4)2HPO4
Diammonium hydrogen phosphate
0 - 10 mmol/L
MgSO4
Magnesium sulfate
0 - 10 mmol/L
Fe-EDTA
Ethylenediaminetetraacetic acid
0 – 0.1 mmol/L
H3BO3
Boric acid
0 – 0.01 mmol/L
KCl
Potassium chloride
0 – 0.01 mmol/L
MnSO4
Manganese (II) sulfate
0 – 0.001 mmol/L
ZnSO4
Zinc sulfate
0 – 0.001 mmol/L
FeSO4
Iron(II) sulfate
0 – 0.0001 mmol/L
CuSO4
Copper sulfate
0 - 0.0002 mmol/L
MoO3
Molybdenum oxide
0 – 0.0002 mmol/L
When it comes to nutrient solutions, you will always find concentration information that is given either in mg/l, ppm or moles. Here is a little help on how these values are converted into one another. You will often find measuring ranges given with a second citation form, for example nitrate as nitrate (NO 3 ) and as nitrate-nitrogen (NO 3 -N).
Conversion: Mol and PPM
A technical definition of ppm
What is ppm? And how can something called "parts per million" be represented by mg/L? Parts per million indicates the number of "parts" of something in a million "parts" of something else. The "part" can be any unit, but when mixing solutions, ppm usually represents units of weight. In this context, ppm indicates how many grams of a solute there are per million grams of solvent (e.g. water).
1 g dissolved / 1,000,000 g solvent
When dealing with water at room temperature, it is common to assume that the density of the water is equal to 1 g/ml. Therefore we can describe the relationship as follows:
1 g dissolved in 1,000,000 ml of water
Then we divide ml by 1000 ml:
1 g dissolved in 1,000 L water
By dividing both units by 1000, the ratio becomes:
1 mg dissolved in 1 L water
Therefore, one can say 1 mg in 1 L of water is the same as 1 mg in 1,000,000 mg of water, or 1 part per million (assuming both room temperature and an atmospheric pressure of 1 atmosphere).
Take the molarity mol/L and multiply by its molar mass g/mol to get g/L. Multiply by 1000 again to convert grams to milligrams and you have mg/L for aqueous solutions.
Example: Prepare a NaOH solution
You have a stock solution of 1 molar NaOH. How do you go about creating a 1L solution of 200 ppm NaOH? NaOH has a molar mass of 39,997 g/mol.
1. Convert 200 ppm to molarity.
First let's assume 200 ppm = 200 mg/L. Then divide the result by 1000 and you get g/L: 200 mg/L divided by 1000 mg/g equals 0.2 g/L.
Next, divide 0.2 g/L by the molar mass of NaOH (Na=22.9 O=16 H=1) to get the molarity: 0.2 g/L divided by 39,997 g/mol which is 0.005 mole /L.
2. Calculate the dilution recipe.
From step 1 we know the target molarity of 0.005 mol/L. To calculate the dilution we use the dilution equation: m1⋅v1=m2⋅v2
where: • m1— the concentration of the stock solution; • m2— the concentration of the diluted solution; • v1—the volume of the stock solution; and • v2 - The volume of the diluted solution
We can enter the numbers for all variables except the volume of the stock solution:
1 M ⋅ v1 = 0.005 M ⋅ 1 L
By rearranging the equation, we find the required volume of the stock solution: v1 = 0.005 M / 1 M ⋅ 1 L = 0.005 L
Therefore we need to dilute 0.005 L (or 5 ml) stock solution to a final volume of 1 L and so we get 200 ppm NaOH solution.
How do I calculate ppm from volume concentration?
How to get volume ppm:
Take the molar concentration of the solutions in mol/L. Multiply it by the molar mass in g/mol. Divide it by the density of the solute in g/cm³. Multiply everything by 1000 mg/g. The resulting ppm volume unit is typically μL/L.
You can find a slightly more detailed example here for both conversion directions:
The molar volume of a substance is a substance-specific property that indicates the volume filled by one mole of a substance. For an ideal gas, one mole occupies a volume of 22.414 liters under normal conditions (273.15 K, 101325 Pa). For real gases, solids and liquids, however, the molar volume depends on the substance.
Molar mass
Molar mass M is the quotient of the mass and the amount of a substance. In the unit g/mol it has the same numerical value as the atomic or molecular mass of the substance in the unit u (atomic mass unit). Its meaning is equivalent to the earlier “atomic weight” in chemistry.
Calculation of substance quantities
Formula: n = m / M
Here n denotes the amount of substance, m the mass and M the molar mass. M can be taken from tables for chemical elements and can be calculated from such values for chemical compounds of known composition.
The atomic mass given in tables for each chemical element refers to the natural isotope mixture. For example, the atomic mass for carbon is given as 12.0107 u. This value cannot be used, for example, for material enriched in 13 C. While for stable elements the deviations from isotope mixtures as they occur in nature are relatively small, particularly for radioactive elements the isotope mixture can depend heavily on the origin and age of the material.
Use of the mole unit for concentration information
Concentrations (salinity of solutions, acidity of solutions, etc.). One of the most common uses is the x-molar solution (the x stands for any rational positive number).
Examples
A 2.5 molar A solution contains 2.5 moles of solute A in 1 liter of the solution.
Helium has a mass of approximately 4 u (u is the atomic mass unit; a helium atom has 2 protons and 2 neutrons). Helium gas is monatomic, so in the following example the mole refers to He atoms without the need for specific mention.
1 mol of helium has a mass of about 4 g and contains about 6,022 e 23 helium atoms.
The mass of a nuclear particle is approximately 1 .6605 e -24 g.
1 water molecule usually has the mass 18 · 1 .6605 e -24 g.
The mass of 1 mol of water is 6 .022 e 23 times the mass of a water molecule.
The mass of 1 mol of water is therefore 6 .022 e 23 · 18 · 1 .6605 e -24 g = 18 g (the numerical value is equal to the molecular mass in u).
If you take the more precise atomic masses instead of the number of nucleons, the result is a slightly higher value of 18.015 g.
Production of lithium hydroxide from lithium and water
When LiOH is formed, two water molecules are split by two lithium atoms into one H and one OH part. Because there are the same number of particles in every mole of every substance (see above), you need, for example, 2 moles of lithium and 2 moles of water (or any other amount of substance in a 2:2 ratio).
For example, 6.94 g of lithium twice and 18 g of water twice react to form 2 g of hydrogen and 47.88 g of lithium hydroxide.
The mass in grams of a mole of a substance (that is, the mass in grams per mole) is called the molar mass of that substance.
The molar mass (in g/mol) of a substance is numerically always equal to the formula weight of the substance (in ame = atom mass unit or also called u = unit). The atomic mass can be found at the top right of every periodic table under atomic weight .
The substance NaCl has e.g. B. a formula weight of 58.5 ame and a molar mass of 58.5 g/mol. The table below contains further examples of calculations using the mole unit.
The entries for N and N 2 in the table make it clear that it is important to precisely name the chemical form of a substance when specifying an amount of substance in moles. Suppose it is stated that 1 mol of nitrogen is produced in a certain reaction. You might conclude that this means 1 mol of nitrogen atoms (14.0 g). However, unless otherwise stated, this probably means 1 mol of nitrogen molecules N2 ( 28.0 g), because N2 is the common chemical form of the element. To avoid such misunderstandings, the chemical form of the substance should be explicitly stated. By specifying the chemical formula N 2 , such misunderstandings are avoided.
Substance name
formula
Formula weight
in name
Molar mass
in (g/mol)
Number & type of particles present in a mole
Atomic nitrogen
N
14.0
14.0
6.022 * 10 23 N atoms
Molecular nitrogen
N2 _
28.0
28.0
6.022 * 10 23 N 2 molecules
2 * (6.022 * 10 23) N atoms
Silver
Ag
107.9
107.9
6.022 * 10 23 Ag atoms
Silver ions
Ag +
107.9
107.9 (1
6.022 * 10 23 Ag + ions
Barium chloride
BaCl2 _
208.2
208.2
6.022 * 10 23 BaCl 2 units
6.022 * 10 23 Ba 2+ ions
2 * (6.022*1023) Cl – ions
1) Remember that the mass of the electron can be neglected and therefore ions and atoms have essentially the same mass.
Example:
What is the mass in grams of 1,000 mol of glucose/sugar, C6 H12 O6 ?
Solution:
First: analysis. The chemical formula is given and we are supposed to calculate the molar mass from it.
Procedure: The molar mass of a substance can be calculated by adding the atomic weights of the atomic components together.
Glucose has a formula weight of 180.0 ame. One mole of this substance has a mass of 180.0 g, so the substance C 6 H 12 O 6 has a molar mass of 180.0 g/mol.
Verification: The order of magnitude of our answer seems plausible and g/mol is the correct unit for specifying molar mass.
6 C – Atoms = 6 (12.0 ame) = 72.0 ame 12 H – Atoms = 12 (1.0 ame) = 12.0 ame 6 O – atoms = 6 (16.0 ame) = 96.0 ame -------------------------------------- 180.0 ame or 180.0 u written
tl;dr: forget about the decimal places in atomic weights - as long as it's about fertilizer.
In order to be complete, one more detail needs to be explained. The information shown here assumes an ideal atomic weight, which cannot be found in the periodic table - with a few exceptions. If you look at the atomic weight of hydrogen, it should be exactly 1.0 u (or 1.0 ame). However, 1.0080 is given.
This is where reality gets in the way. There is almost no element in nature that occurs without isotopes. The atoms in the periodic table are "sorted" according to the number of protons. But the number of neutrons can vary. For magnesium, for example, only about 78.6% contains 12 neutrons in any sample (i.e. no matter where you find magnesium on Earth). 10.1% have 13 neutrons and 11.3% of them have 14 protons. This is how you get an atomic weight of magnesium of 24.327 u. It works out like this:
786 24 Mg isotopes with a mass of 24 u provide a mass of 18864 u.
101 25 Mg isotopes with a mass of 25 u provide a mass of 2525 u.
113 26 Mg isotopes with a mass of 25 u provide a mass of 2938 u.
The total gives the weight of 1000 Mg atoms: 24,327 u. So statistically one Mg atom weighs 24.327 u.
If you put together fertilizer according to your own formulas, this inaccuracy should only come into play for very (very) large quantities. This last paragraph only serves to remind you of the chemistry lessons you may have already forgotten at school and to clear up the confusion about the crooked numbers.
Context: You often need the periodic table to calculate fertilizer solutions, as the quantities of an existing fertilizer solution usually have to be calculated in relation to the amount of additional fertilizers added. See the Fertilizer article series .
By Antonsusi, Public Domain, https://commons.wikimedia.org/w/index.php?curid=82871392
We need to add 18.9 grams of Sequestren 330 to get 1 ppm of iron.
