Concentration is the dissolved amount divided by the weight of the whole solution
The concentration of salt water (mass percent) is the weight of salt divided by the total weight of the salt water, times 100. The divisor is not the weight of the water alone but the water plus the salt. So 10 g of salt in 100 g of water gives 10 ÷ 110 × 100 = 9.09 %, and a true 10 % solution needs 10 g of salt in 90 g of water. One liter of water with 50 g of salt comes out at 4.77 %. In US units, 1 gallon of water with 8 oz (by weight) of salt is 5.66 %.
The calculator above solves this as a three-box equation: ① what you add + ② water = ③ the finished solution. Enter what you know and it shows the concentration of ③ in mass %, ppm, g/L, mol/L and more at once. Choose "Make a target concentration" and it works backward, for example 500 g of 10 % salt water needs 50 g of salt and 450 g of water. For salt, sugar and hydrochloric acid it uses 20 °C density tables to find the real volume too.
When you need to calculate concentration
Eyeballing is fine for plenty of jobs, but in many others the concentration changes the result. Typical cases:
- Cooking: brines, pickling liquids and simple syrups are usually given as a percentage. Whether the recipe means "per weight of water" or "per weight of the whole" changes the salt you add. To reach 10 % in 1 L of water you need 110.9 g of salt, not 100 g.
- Diluting disinfectant: the CDC gives bleach amounts per gallon or per quart, but the strength you end up with depends on how strong your bottle is. The calculator turns any mix into ppm.
- Lab solutions: making a solution of a set molarity, or working out how many mL of a concentrated stock such as 36 % hydrochloric acid to measure out.
- Homework and chemistry classes: mixing two salt solutions, boiling water off, or adding more salt. The calculation steps show the dissolved amount and the formula used.
How to use it
- Pick a mode: choose "Find the concentration" when you know what you put in, or "Make a target concentration" when the final strength is fixed.
- Pick a substance: salt, sugar and hydrochloric acid come with density tables and molar masses. For anything else, such as bleach or fertilizer, choose "Custom" and enter a molar mass only if you need mole units. If you know the density of the stock, enter it too.
- Choose ① and ②: ① is either a pure substance (such as table salt) or a solution you already have (the stock). ② is water, another solution or evaporation. To mix two salt solutions, set both ① and ② to salt water; to boil water off, set ② to evaporation.
- Enter amounts and units: pick g, kg, oz or lb for weight, and mL, L, tsp, tbsp, fl oz, cup, qt or gal for volume. Volumes of water and solutions are converted to weight with their 20 °C density. Note that oz here means ounces of weight, not fluid ounces.
- Make a target concentration: enter the target in ③ and fill in one of the three amounts. The box you filled in last is marked "Basis" and the other two are calculated. Fill ③ if you know how much you want to end up with, ② if you know how much water you have, or ① if you know how much stock you have.
- Switch units: click a chip under "In other units" to read ③ in that unit. In target mode, the same concentration is rewritten in that unit.
How it is calculated
The calculator converts every amount to weight and relies on one rule: the weight of dissolved material does not change when you mix.
- Finding the concentration: mass % = dissolved ÷ (dissolved + water) × 100. When two solutions are mixed, the dissolved amounts are found separately, added, and divided by the total weight. Evaporation removes only water and leaves the dissolved amount unchanged.
- Making a target concentration: with a final weight W and target c, the material is W × c and the water is W − W × c. If the water weight w is given instead, the material is w × c ÷ (1 − c).
- Mixing two strengths: to make strength c from ① at c₁ and ② at c₂, ① : ② = (c − c₂) : (c₁ − c) by weight. Diluting with plain water means c₂ = 0.
- Volume-based units: g/L = mass % × 10 × density (g/mL), and mol/L = g/L ÷ molar mass. The same 4.77 % salt water is 47.7 g/L if you assume a density of 1, but 49.24 g/L with the real density of 1.0323 g/mL.
- Dilution factor: finished volume ÷ stock volume, which equals stock g/L ÷ finished g/L. Stock volume = finished volume ÷ factor.
mass % = dissolved ÷ whole solution × 100 · ppm = mass % × 10,000 · mol/kg = moles dissolved ÷ kg of water
Salt in 1 L of water: concentration table
This is salt dissolved in 1 L (998.2 g) of water at 20 °C. The volumes come from the CRC density table, so dissolving 100 g of salt brings the volume to 1,032 mL, only about 32 mL more than the water alone. For reference, 1 US gallon is 3,785 mL.
| Salt | Mass % | g/L | Volume (mL) | mol/L |
|---|---|---|---|---|
| 10 g | 0.9919 | 9.971 | 1,003 | 0.1706 |
| 30 g | 2.918 | 29.73 | 1,009 | 0.5087 |
| 50 g | 4.77 | 49.24 | 1,015 | 0.8426 |
| 100 g | 9.106 | 96.89 | 1,032 | 1.658 |
| 150 g | 13.06 | 142.9 | 1,050 | 2.445 |
| 200 g | 16.69 | 187.2 | 1,068 | 3.203 |
| 250 g | 20.03 | 229.9 | 1,087 | 3.934 |
| 300 g | 23.11 | 271.1 | 1,107 | 4.638 |
| 350 g | 25.96 | 310.7 | 1,126 | 5.316 |
Salt and water for a target concentration
The salt you need depends on whether you fixed the weight of the finished solution or the amount of water you have. Starting from 1 L of water, the finished weight is more than 1 kg because the salt adds to it.
| Target | Salt (g) · 1 kg finished | Water (g) · 1 kg finished | Salt (g) · 1 L water | Finished volume (mL) · 1 L water |
|---|---|---|---|---|
| 1 % | 10 | 990 | 10.08 | 1,003 |
| 2 % | 20 | 980 | 20.37 | 1,006 |
| 3 % | 30 | 970 | 30.87 | 1,009 |
| 5 % | 50 | 950 | 52.54 | 1,016 |
| 10 % | 100 | 900 | 110.9 | 1,036 |
| 15 % | 150 | 850 | 176.2 | 1,059 |
| 20 % | 200 | 800 | 249.6 | 1,087 |
| 25 % | 250 | 750 | 332.7 | 1,120 |
The same salt water in different units
Mass % and ppm are ratios of weights, so they do not depend on temperature. g/L and mol/L are per liter of solution, so density comes into it. Molality (mol/kg) is moles per kg of water and uses weights only. As NOAA explains, about 3.5 % of the weight of seawater is dissolved salts, and the 0.9 % on a saline bag is a weight-per-volume (w/v) figure: 0.9 g per 100 mL of solution.
| Mass % | ppm | g/L | mol/L | mol/kg | Density (g/mL) |
|---|---|---|---|---|---|
| 0.1 % | 1,000 | 0.9989 | 0.01709 | 0.01713 | 0.9989 |
| 0.5 % | 5,000 | 5.009 | 0.08571 | 0.08598 | 1.0018 |
| 1 % | 10,000 | 10.05 | 0.172 | 0.1728 | 1.0053 |
| 3.5 % | 35,000 | 35.81 | 0.6128 | 0.6206 | 1.0232 |
| 5 % | 50,000 | 51.7 | 0.8846 | 0.9006 | 1.0340 |
| 10 % | 100,000 | 107.1 | 1.832 | 1.901 | 1.0707 |
| 20 % | 200,000 | 229.6 | 3.928 | 4.278 | 1.1478 |
| 26 % | 260,000 | 311.3 | 5.326 | 6.012 | 1.1972 |
Volumes do not simply add up
Mixing 1 kg of sugar with 1 kg of water gives 2 kg, but only about 1,627 mL, not 2 L. The dissolved molecules fit in between the water molecules. This is why a 1:1 simple syrup measured by weight changes strength if you measure it by the cup instead. The table shows how much volume actually grows when sugar is dissolved in 1 kg of water.
| Sugar | Mass % | Total weight (g) | Volume (mL) | Volume gained (mL) |
|---|---|---|---|---|
| 250 g | 20 | 1,250 | 1,156 | 154.5 |
| 500 g | 33.33 | 1,500 | 1,312 | 310.4 |
| 1,000 g | 50 | 2,000 | 1,627 | 624.9 |
| 1,500 g | 60 | 2,500 | 1,943 | 941.6 |
| 2,000 g | 66.67 | 3,000 | 2,261 | 1,259 |
A 2:1 syrup (2 kg of sugar to 1 kg of water) is 66.67 %. PubChem says 1 g of sugar dissolves in 0.5 mL of water, which is that same near-saturation point, so at room temperature some sugar tends to stay undissolved.
Diluting a stock and the dilution factor
When you dilute a stock with water, the amount of dissolved material stays the same and only the volume grows. The CDC says that when the bottle has no directions, you can mix 5 tablespoons (1/3 cup) of bleach per gallon of room temperature water, or 4 teaspoons per quart. It also notes that most household bleach contains 5%–9% sodium hypochlorite, and that a bleach product outside this range or with no percentage stated should not be used. The strength you get depends on your bottle: at 6 %, 5 tablespoons in a gallon is about 1,150 ppm (roughly a 52× dilution), at 5 % it is about 960 ppm, and at 9 % about 1,720 ppm. 4 teaspoons per quart at 6 % is about 1,220 ppm. The CDC adds: "Make a new diluted bleach solution daily."
| Target | Stock (mL) | Water (mL) | Dilution factor | Stock : water |
|---|---|---|---|---|
| 200 ppm | 4 | 996 | 250 | 1 : 249 |
| 500 ppm | 10 | 990 | 100 | 1 : 99 |
| 1,000 ppm | 20 | 980 | 50 | 1 : 49 |
| 5,000 ppm | 100 | 900 | 10 | 1 : 9 |
By the same method, a 1,000× dilution is 20 mL of stock for 20 L finished, but 20.02 mL if it is added to 20 L of water. The bigger the factor, the smaller that gap. Stock strength differs from product to product, so enter the percentage printed on your label.
Lab solutions: molarity and diluting strong acid
For a solution of set molarity (mol/L), the most accurate method is to weigh the solute, dissolve it, and add water up to the volume mark. For 1 L of 1 mol/L sodium chloride, dissolve 58.44 g and make up to 1 L. To dilute a concentrated stock, first find the stock's molarity from its mass % and density. In the CRC table, 36 % hydrochloric acid has a density of 1.1791 g/mL at 20 °C and 11.64 mol/L, so 500 mL of 1 mol/L needs 42.95 mL of stock.
| Target | Finished mass % | Stock (mL) | Stock (g) | Finished density (g/mL) |
|---|---|---|---|---|
| 0.1 mol/L | 0.3646 | 8.59 | 10.13 | 1.0000 |
| 0.5 mol/L | 1.81 | 42.95 | 50.64 | 1.0072 |
| 1 mol/L | 3.589 | 85.9 | 101.3 | 1.0159 |
| 2 mol/L | 7.059 | 171.8 | 202.6 | 1.0330 |
| 6 mol/L | 19.93 | 515.4 | 607.7 | 1.0976 |
Reagents that carry water of crystallization have a different molar mass. Calculating copper(II) sulfate pentahydrate (CuSO₄·5H₂O, about 249.7 g/mol) with the molar mass of the anhydrous form (CuSO₄, about 159.6 g/mol) gives a concentration about 36 % too low. For a custom substance, enter the molar mass of the formula printed on the bottle. Handle strong acids with protective equipment and follow your lab's safety rules.
Common mistakes
- Dividing by the weight of the water: 10 g of salt in 100 g of water is 9.09 %, not 10 %. The divisor includes the salt.
- Treating mL and g as the same: water at 20 °C is 0.9982 g per mL, close enough, but 1 mL of 10 % salt water is 1.0707 g and 1 mL of 50 % sugar water is 1.2295 g. Measuring a strong solution by the cup changes its weight.
- Mixing up oz and fl oz: oz measures weight and fl oz measures volume. 8 oz of salt is 226.8 g, while 8 fl oz of water is about 236.6 mL.
- Using ppm and mg/L interchangeably: they are nearly equal in dilute water and drift apart as the solution gets stronger. A 4.77 % salt water is 47,700 ppm but 49,240 mg/L.
- Reading "1:50" as a 50× dilution: 1 part stock plus 50 parts water is a 51× dilution. If the instructions give the amounts of stock and water, calculate from those.
- Going past the solubility limit: at 25 °C salt dissolves only up to about 26.47 % and sugar up to about 66.67 %. A higher calculated value just means the rest stays undissolved.
- Simply adding volumes: it is tempting to count 50 g of salt as 50 mL and expect 1,050 mL, but 1 L of water with 50 g of salt actually comes to about 1,015 mL. When a volume-based strength matters, dissolve first and then top up to the volume mark.
Limits
- 20 °C basis: water and solution densities are 20 °C values. At other temperatures volumes and volume-based strengths (g/L, mol/L) shift slightly, while mass %, ppm and mol/kg do not.
- Density table range: the tables reach 26 % for salt, 80 % for sugar and 40 % for hydrochloric acid. Stronger values extend the last segment of the table as an estimate.
- Custom substances: if the density is unknown, 1 mL of solution is taken to weigh the same as water; if you enter the stock density, a straight line between water and the stock is used. The difference is small for the dilute solutions used in cleaning, but volume-based values for strong solutions may differ from reality.
- Saturation: the saturation figures are 25 °C values. Amounts that would not dissolve only trigger a warning, and the concentration is still calculated from what you entered, as in textbook problems.
- Not a substitute for measurement: for disinfectants and chemicals, follow the product label and official guidance. When an exact concentration matters, check with a scale, a volumetric flask or a hydrometer.
Sources
- 01CRC Handbook of Chemistry and Physics: Concentrative Properties of Aqueous SolutionsDensity, molality and molarity by mass % at 20 °C. Sodium chloride 10 %: 1.0707 g/mL (1.901 mol/kg, 1.832 mol/L); sucrose 50 %: 1.2295 g/mL; hydrogen chloride 36 %: 1.1791 g/mL (11.642 mol/L)
- 02NIST Chemistry WebBook: WaterThermophysical data; density of water 0.998207 g/mL at 20 °C and 0.101325 MPa
- 03PubChem: Sodium Chloride (CID 5234)Solubility "36.0 g/100 g of water at 25 °C"
- 04PubChem: Sucrose (CID 5988)Solubility "1 g dissolves in 0.5 ml water"
- 05CDC: Cleaning and Disinfecting with Bleach5 tablespoons (1/3 cup) of bleach per gallon of room temperature water or 4 teaspoons per quart; most household bleach contains 5%–9% sodium hypochlorite; make a new diluted bleach solution daily
- 06NOAA National Ocean Service: Why is the ocean salty?Average salinity about 35 parts per thousand; about 3.5 percent of the weight of seawater comes from dissolved salts
- 07DailyMed: 0.9% Sodium Chloride Injection (Baxter) label250 mL contains 2.25 g of sodium chloride and 500 mL contains 4.5 g (0.9 g per 100 mL of solution, w/v)
Frequently asked questions
How do I calculate the concentration of salt water?
Divide the weight of the salt by the total weight of the salt water (salt + water) and multiply by 100. 10 g of salt in 100 g of water is 10 ÷ 110 × 100 = 9.09 %, not 10 %. 1 L of water (998.2 g at 20 °C) with 50 g of salt is 50 ÷ 1,048.2 × 100 = 4.77 %.
How much salt and water make a 10 % salt solution?
10 % of the finished weight is salt. For 500 g of salt water you need 50 g of salt and 450 g of water. If you start from 2 L of water (1,996 g), the salt is 1,996 × 10 ÷ 90 = 221.8 g, the finished weight is 2,218 g and the volume is about 2,072 mL.
Are ppm and mg/L the same unit?
Here ppm means mg per kg of solution (by weight) and mg/L means mg per liter of solution. In dilute water 1 L is almost 1 kg, so the two are close, but they drift apart as the solution gets stronger. A 4.77 % salt solution is 47,700 ppm and 49,240 mg/L.
How strong is bleach mixed at 5 tablespoons per gallon?
The CDC says to use 5 tablespoons (1/3 cup) of bleach per gallon of room temperature water, or 4 teaspoons per quart, when the bottle has no directions, and that most household bleach is 5%–9% sodium hypochlorite. At 6 %, 5 tablespoons in 1 gallon is about 1,149 ppm (a 52.2× dilution); at 5 % it is 958 ppm and at 9 % 1,724 ppm. The CDC also says to make a new solution daily.
Does 1 kg of sugar plus 1 kg of water make 2 L?
No. A 50 % sugar solution of 2,000 g has a 20 °C density of 1.2295 g/mL (CRC table), so the volume is about 1,627 mL. Dissolved material fits between the water molecules, so volumes do not simply add. Likewise 1 L of water with 50 g of salt comes to about 1,015 mL, only about 15 mL more than the water.
How do I find molarity (mol/L)?
Molarity is the moles of solute in 1 L of solution. A 4.77 % salt solution (58.44 g/mol) holds 49.24 g per liter, which is 0.8426 mol/L. It depends on solution volume, so density is needed; molality (0.8571 mol/kg here) is per kg of water and needs weights only.
How much 36 % hydrochloric acid makes 500 mL of 1 mol/L?
In the CRC table, 36 % hydrochloric acid has a 20 °C density of 1.1791 g/mL and is 11.64 mol/L. You need 0.5 mol (18.23 g) of hydrogen chloride, so measure 42.95 mL of stock into water and make up to 500 mL. Handle strong acid with protective equipment and your lab's safety rules.
What happens to the concentration when I mix two salt solutions or boil water off?
Add up the salt and divide by the total weight. Mixing 300 g of 8 % salt water (24 g of salt) with 200 g of 12 % (24 g of salt) gives 48 ÷ 500 × 100 = 9.6 %. Evaporation leaves the salt unchanged, so boiling 40 g of water out of 200 g of 8 % salt water gives 16 ÷ 160 × 100 = 10 %.