Calculate mass of solute, normality, volume, or equivalent weight from the formula G = N × V × EW with unit conversions for g, kg, L, mL, g/mol, and mg/mol.
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Normality to Grams Formula
The following formula is used to convert normality to grams:
G = N * V * EW
Variables:
- G is the mass of the solute in grams
- N is the normality of the solution (eq/L)
- V is the volume of the solution in liters
- EW is the equivalent weight of the solute in grams per equivalent (g/eq)
To convert normality to grams, multiply the normality of the solution by the volume in liters and the equivalent weight in g/eq.
| Normality (N) | Grams of Solute (g) |
|---|---|
| 0.01 | 0.400 |
| 0.02 | 0.800 |
| 0.05 | 2.000 |
| 0.10 | 4.000 |
| 0.20 | 8.000 |
| 0.25 | 10.000 |
| 0.30 | 12.000 |
| 0.50 | 20.000 |
| 0.75 | 30.000 |
| 1 | 40.000 |
| 1.25 | 50.000 |
| 1.50 | 60.000 |
| 2 | 80.000 |
| 2.50 | 100.000 |
| 3 | 120.000 |
| 4 | 160.000 |
| 5 | 200.000 |
| 6 | 240.000 |
| 8 | 320.000 |
| 10 | 400.000 |
| Uses G = N x V x EW. Assumes V = 1.00 L and EW = 40.00 g/eq (NaOH). Therefore, 1 N = 40.000 g NaOH per liter. | |
Equivalent Weight of Common Reagents
Equivalent weight (EW) = Molar Mass / n-factor. The n-factor is the number of H+ or OH– exchanged per formula unit (acid-base reactions) or electrons transferred per formula unit (redox reactions). Because normality is reaction-dependent, the same compound can carry different EW values in different contexts; KMnO4 is the clearest example, with three distinct equivalent weights depending on the medium.
| Compound | Formula | Molar Mass (g/mol) | n-factor | EW (g/eq) | g to prepare 1 N, 1 L |
|---|---|---|---|---|---|
| Hydrochloric acid | HCl | 36.46 | 1 | 36.46 | 36.46 |
| Nitric acid | HNO3 | 63.01 | 1 | 63.01 | 63.01 |
| Hydrobromic acid | HBr | 80.91 | 1 | 80.91 | 80.91 |
| Perchloric acid | HClO4 | 100.46 | 1 | 100.46 | 100.46 |
| Acetic acid | CH3COOH | 60.05 | 1 | 60.05 | 60.05 |
| Sulfuric acid | H2SO4 | 98.08 | 2 | 49.04 | 49.04 |
| Sulfurous acid | H2SO3 | 82.07 | 2 | 41.04 | 41.04 |
| Oxalic acid | H2C2O4 | 90.03 | 2 | 45.02 | 45.02 |
| Carbonic acid | H2CO3 | 62.03 | 2 | 31.01 | 31.01 |
| Phosphoric acid | H3PO4 | 97.99 | 3 | 32.66 | 32.66 |
| Sodium hydroxide | NaOH | 40.00 | 1 | 40.00 | 40.00 |
| Potassium hydroxide | KOH | 56.11 | 1 | 56.11 | 56.11 |
| Lithium hydroxide | LiOH | 23.95 | 1 | 23.95 | 23.95 |
| Ammonia | NH3 | 17.03 | 1 | 17.03 | 17.03 |
| Sodium bicarbonate | NaHCO3 | 84.01 | 1 | 84.01 | 84.01 |
| Sodium carbonate | Na2CO3 | 105.99 | 2 | 53.00 | 53.00 |
| Calcium hydroxide | Ca(OH)2 | 74.09 | 2 | 37.05 | 37.05 |
| Magnesium hydroxide | Mg(OH)2 | 58.32 | 2 | 29.16 | 29.16 |
| Barium hydroxide | Ba(OH)2 | 171.34 | 2 | 85.67 | 85.67 |
| n-factor = number of H+ or OH- exchanged per formula unit. For V = 1 L, grams to prepare 1 N = EW numerically. | |||||
| Compound | Formula | Reaction / Medium | Half-Reaction | Molar Mass (g/mol) | n-factor | EW (g/eq) |
|---|---|---|---|---|---|---|
| Potassium permanganate | KMnO4 | Acidic (H2SO4) | Mn+7 to Mn+2 | 158.03 | 5 | 31.61 |
| Potassium permanganate | KMnO4 | Neutral (H2O) | Mn+7 to Mn+4 | 158.03 | 3 | 52.68 |
| Potassium permanganate | KMnO4 | Alkaline (NaOH) | Mn+7 to Mn+6 | 158.03 | 1 | 158.03 |
| Potassium dichromate | K2Cr2O7 | Acidic | 2Cr+6 to 2Cr+3 | 294.19 | 6 | 49.03 |
| Iodine | I2 | Iodometry | I0 to I– | 253.81 | 2 | 126.90 |
| Sodium thiosulfate | Na2S2O3 | Iodometry | S2O32- to S4O62- | 158.11 | 1 | 158.11 |
| Ferrous sulfate | FeSO4 | Redox titration | Fe+2 to Fe+3 | 151.91 | 1 | 151.91 |
| Hydrogen peroxide | H2O2 | Oxidizer | O-1 to O-2 | 34.01 | 2 | 17.01 |
| Oxalic acid (reducer) | H2C2O4 | vs. KMnO4 | C+3 to C+4 | 90.03 | 2 | 45.02 |
| Sodium oxalate (reducer) | Na2C2O4 | vs. KMnO4 | C+3 to C+4 | 134.00 | 2 | 67.00 |
| Chlorine | Cl2 | Disinfection/oxidation | Cl0 to Cl– | 70.90 | 2 | 35.45 |
| The same compound can have multiple valid n-factors depending on the reaction. Always specify the medium when reporting normality of redox reagents. | ||||||
What is Normality?
Normality (N) is a measure of concentration expressed as gram equivalents of solute per liter of solution (eq/L). Its defining characteristic is reaction dependence: the equivalent weight of a substance changes with the specific reaction taking place. A 1 M solution of H2SO4 is 2 N in acid-base neutralization (n-factor = 2, two H+ donated) but 1 N in reactions where only one proton participates. This property makes normality indispensable for titration: at the equivalence point, N1V1 = N2V2 holds for any acid-base or redox pair regardless of their molar masses or formula weights, because 1 equivalent always reacts with exactly 1 equivalent. The relationship to molarity is N = M x n, where n is the n-factor for the specific reaction. A solution does not have a single normality in isolation; it has one molarity and potentially multiple normalities depending on context.
Normality vs Molarity
Molarity counts formula units per liter; normality counts reactive equivalents per liter. For monoprotic acids and monobasic bases, they are numerically identical. The gap widens for polyprotic acids, polybasic bases, and most redox reagents.
| Reagent (1 M solution) | n-factor | Normality | Reaction context |
|---|---|---|---|
| HCl | 1 | 1 N | Acid-base (1 H+) |
| HNO3 | 1 | 1 N | Acid-base (1 H+) |
| CH3COOH | 1 | 1 N | Acid-base (1 H+) |
| H2SO4 | 2 | 2 N | Acid-base (2 H+) |
| H3PO4 | 3 | 3 N | Acid-base (3 H+) |
| NaOH | 1 | 1 N | Acid-base (1 OH–) |
| Ca(OH)2 | 2 | 2 N | Acid-base (2 OH–) |
| Na2CO3 | 2 | 2 N | Acid-base (2 OH– equivalent) |
| KMnO4 | 5 | 5 N | Redox, acidic medium |
| KMnO4 | 3 | 3 N | Redox, neutral medium |
| K2Cr2O7 | 6 | 6 N | Redox, acidic medium |
| FeSO4 | 1 | 1 N | Redox (Fe2+ to Fe3+) |
| Na2S2O3 | 1 | 1 N | Redox, iodometry |
| Normality = Molarity x n-factor. A solution has one fixed molarity; its normality depends on the reaction. | |||
How to Calculate Grams from Normality?
- First, determine the normality (N) of the solution.
- Next, determine the volume (V) of the solution in liters.
- Next, determine the equivalent weight (EW) of the solute in g/eq. Use EW = Molar Mass / n-factor. Refer to the tables above for common reagents.
- Next, apply the formula: G = N x V x EW.
- Finally, calculate the mass of the solute in grams (G).
- After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem 1 (Sulfuric Acid):
How many grams of H2SO4 are needed to prepare 500 mL of a 2.5 N solution for acid-base titration?
Normality (N) = 2.5 N | Volume (V) = 0.500 L | EW = 49.04 g/eq (MW = 98.08, n = 2)
G = 2.5 x 0.500 x 49.04 = 61.3 g H2SO4
Example Problem 2 (KMnO4 in Acidic Medium):
How many grams of KMnO4 are needed to prepare 250 mL of a 0.1 N solution for an acidic redox titration?
Normality (N) = 0.1 N | Volume (V) = 0.250 L | EW = 31.61 g/eq (MW = 158.03, n = 5 in acidic medium)
G = 0.1 x 0.250 x 31.61 = 0.79 g KMnO4
