Ammonium nitrate Percent Solution Calculator

Grams of ammonium nitrate for any % w/v strength, with the molarity and mg/mL equivalents.

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Percentage strength for ammonium nitrate is read as % w/v: grams of solute in 100 mL of solution. So 1% is 1 g per 100 mL, which for this compound works out at 0.12493 mol/L, and 0.9% is 9 g per litre or 0.11244 mol/L. The calculator takes a percentage and a volume and returns the grams to weigh, with the molarity and mg/mL equivalents beside it.

The figure comes from the formula. Ammonium nitrate is 2 × 14.007 (N) + 4 × 1.008 (H) + 3 × 15.999 (O), which is 80.043 g/mol, taking standard atomic weights, and every page here computes that sum rather than quoting it. Oxygen makes up the largest share of the mass: 3 of the 9 atoms in a formula unit are oxygen, and they account for 47.997 g of the 80.043 g, or 59.96% by mass. The derivation table below breaks the whole molecule down element by element.

Three different percentages are in circulation and they are not interchangeable. Per cent w/v is mass in a volume, the one used here and on medical labels; per cent w/w is mass in mass, the one on drums of concentrated reagent; per cent v/v is volume in volume, used for liquids mixed into liquids. For a dilute aqueous solution w/v and w/w are within a percent of each other, but for anything concentrated the density has to come into it.

A nitrogen fertiliser and the salt in instant cold packs, which work because it absorbs heat as it dissolves. At 80.043 g/mol it is heavier than 5 of the 34 salts covered here, and that ranking matters more than it looks: potassium chloride has a molar mass of 74.5483 g/mol, so a gram of it contains 7.37% more formula units than a gram of ammonium nitrate. Weigh by mass and you are not weighing equal amounts of substance.

Purity is the usual gap between the calculation and the balance. A reagent sold at 98% means 1.6009 g of a nominal 80.043 g is something else, so for exact work you divide the weighed mass by the assay figure on the certificate. For most purposes the difference is smaller than the error in reading the meniscus, but it is systematic rather than random, so it does not average out.

The formula

m = p × V ÷ 100
p
The strength you want, as % w/v
V
The volume of solution, in millilitres
m
The mass of ammonium nitrate to weigh out, in grams
80.043
The molar mass of ammonium nitrate in g/mol, derived from its formula

How it works, step by step

  1. Enter the percentage strength you want, as % w/v.
  2. Enter the volume of solution in millilitres.
  3. Grams needed is percentage × volume ÷ 100, since 1% w/v is 1 g per 100 mL.
  4. The molarity that strength corresponds to for ammonium nitrate is shown beneath, using 80.043 g/mol.

Worked examples

0.9% w/v ammonium nitrate in 500 mL

0.9% w/v is 0.9 g per 100 mL, so 500 mL needs 4.5 g. In molar terms that solution is 0.11244 mol/L, and it is 9 mg/mL if you are dosing by volume.

5% w/v ammonium nitrate in 100 mL

5% w/v is 5 g per 100 mL, so 100 mL needs 5 g. In molar terms that solution is 0.62466 mol/L, and it is 50 mg/mL if you are dosing by volume.

10% w/v ammonium nitrate in 1000 mL

10% w/v is 10 g per 100 mL, so 1000 mL needs 100 g. In molar terms that solution is 1.2493 mol/L, and it is 100 mg/mL if you are dosing by volume.

How to read your score

0–0NothingNothing to weigh.
0–0.05Below a sensible weighingUnder 50 mg. Weigh ten or a hundred times this and dilute; the balance error does not shrink with the sample.
0.05–10Weighable directlyFrom 50 mg to 10 g, which is where a two- or three-decimal balance gives better than 1% on the mass.
10–—Large batchMore than 10 g of ammonium nitrate. Worth checking solubility and, for a solid that generates heat on dissolving, adding it in portions.

Frequently asked questions

What is the molar mass of ammonium nitrate (NH4NO3)?

It is 80.043 g/mol. That is the sum of the standard atomic weights of every atom in the formula: 2 × 14.007 (N) + 4 × 1.008 (H) + 3 × 15.999 (O). One mole of ammonium nitrate therefore weighs 80.043 g, and a gram of it is 12.493 mmol.

What percentage of ammonium nitrate is oxygen?

59.96% by mass. Each formula unit contains 3 oxygen atoms contributing 47.997 g of the 80.043 g total, so 100 g of ammonium nitrate contains 59.96 g of oxygen and a kilogram contains 599.64 g of it.

How many grams of ammonium nitrate make a 1% w/v solution?

1 g in 100 mL of solution, 5 g in 500 mL, 10 g in a litre. The percentage is fixed to the volume, not to the compound, so those masses are the same whatever is being dissolved — what changes with the compound is the molarity they come to.

What molarity is a 1% w/v solution of ammonium nitrate?

0.124933 mol/L, because 1% w/v is 10 g/L and 80.043 g/L is one molar. A 0.9% solution is 0.11244 mol/L and a 5% solution is 0.62466 mol/L.

Is % w/v the same as % w/w?

No. Per cent w/v is grams per 100 mL of solution; per cent w/w is grams per 100 g of solution. They coincide only when the solution has a density of 1.00 g/mL, which dilute aqueous solutions approximately do and concentrated ones do not. Labels on concentrated reagents are usually w/w.

What is 1% w/v in mg/mL?

10 mg/mL, since 1 g in 100 mL is 1000 mg in 100 mL. For ammonium nitrate that is also 124.93 mmol/L, which is the form a dose calculation usually wants.

Percentage strengths of ammonium nitrate

How the molar mass of ammonium nitrate is arrived at
ElementAtomsAtomic weightContribution (g/mol)By mass
Oxygen (O)315.99947.99759.96%
Nitrogen (N)214.00728.01435%
Hydrogen (H)41.0084.0325.037%
Total — one mole of ammonium nitrate80.043100%

Standard atomic weights, IUPAC 2021. The contribution column is atoms × atomic weight, and the total is the molar mass this page uses: 80.043 g/mol.

Percentage strengths of ammonium nitrate in grams and mol/L
Strengthper 100 mLper 500 mLper 1 LMolarity (mol/L)
0.1% w/v100 mg500 mg1 g0.012493
0.25% w/v250 mg1.25 g2.5 g0.031233
0.5% w/v500 mg2.5 g5 g0.062466
0.9% w/v900 mg4.5 g9 g0.11244
1% w/v1 g5 g10 g0.12493
2% w/v2 g10 g20 g0.24987
3% w/v3 g15 g30 g0.3748
5% w/v5 g25 g50 g0.62466
10% w/v10 g50 g100 g1.2493
20% w/v20 g100 g200 g2.4987

A 1% w/v solution is 1 g in 100 mL, which for this compound is 0.12493 mol/L because 10 g/L divided by 80.043 g/mol gives that figure.

Ammonium nitrate among other salts, by molar mass
CompoundFormulaMolar mass (g/mol)Millimoles in 1 g
Sodium nitriteNaNO268.9947714.494
Sodium hypochloriteNaClO74.4387713.434
Potassium chlorideKCl74.548313.414
Ammonium nitrate (this page)NH4NO380.04312.493
Sodium bicarbonateNaHCO384.0057711.904
Sodium nitrateNaNO384.9937711.766
Magnesium chlorideMgCl295.20510.504

Ordered by molar mass. The last column is 1000/M, which is the number a weighed gram actually gives you.

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