Going from grams to moles for sodium nitrate means dividing by its molar mass, 84.993769 g/mol. One gram is 0.0117656 mol, or 11.7656 mmol; 100 mg is 1.1766 mmol; and a 84.99377 g portion is exactly one mole. The division is the whole calculation, but the molar mass has to be the one for the exact formula you have, NaNO3, which is where these figures come from.
The figure comes from the formula. Sodium nitrate is 22.9898 (Na) + 14.007 (N) + 3 × 15.999 (O), which is 84.993769 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 5 atoms in a formula unit are oxygen, and they account for 47.997 g of the 84.99377 g, or 56.47% by mass. The derivation table below breaks the whole molecule down element by element.
Balance resolution is the practical limit here. On a two-decimal balance the smallest step is 10 mg, which is 0.1177 mmol of sodium nitrate; on a three-decimal balance it is 1 mg, or 11.77 µmol. To hit a target within 1% you therefore need to weigh at least a gram on the first and at least 100 mg on the second, and below that a stock solution measured by pipette beats weighing.
Chile saltpetre, used as a fertiliser and a heat-transfer salt. It is hygroscopic, unlike the potassium salt, which is why curing mixes prefer KNO3. At 84.99377 g/mol it is heavier than 7 of the 34 salts covered here, and that ranking matters more than it looks: sodium bicarbonate has a molar mass of 84.00577 g/mol, so a gram of it contains 1.18% more formula units than a gram of sodium 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 1.6999 g of a nominal 84.99377 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- The mass you weighed, in grams
84.993769- The molar mass of sodium nitrate in g/mol, derived from its formula
n- The amount of substance in moles
How it works, step by step
- Enter the mass you weighed, in grams.
- It is divided by 84.993769 g/mol, the molar mass of sodium nitrate.
- The result is the amount of substance in moles, with millimoles and micromoles beneath it.
- Check it against the table if you want the figure for a standard weighing rather than a typed mass.
Worked examples
0.25 g of sodium nitrate in moles
0.25 ÷ 84.99377 = 0.00294139 mol, or 2.9414 mmol. If the balance was reading to 1 mg, the last digit of that mass is worth 11.77 µmol, so quoting the answer to more than four significant figures claims a precision the weighing did not have.
2 g of sodium nitrate in moles
2 ÷ 84.99377 = 0.0235311 mol, or 23.531 mmol. If the balance was reading to 1 mg, the last digit of that mass is worth 11.77 µmol, so quoting the answer to more than four significant figures claims a precision the weighing did not have.
25 g of sodium nitrate in moles
25 ÷ 84.99377 = 0.294139 mol, or 294.14 mmol. If the balance was reading to 1 mg, the last digit of that mass is worth 11.77 µmol, so quoting the answer to more than four significant figures claims a precision the weighing did not have.
How to read your score
Frequently asked questions
What is the molar mass of sodium nitrate (NaNO3)?
It is 84.993769 g/mol. That is the sum of the standard atomic weights of every atom in the formula: 22.9898 (Na) + 14.007 (N) + 3 × 15.999 (O). One mole of sodium nitrate therefore weighs 84.99377 g, and a gram of it is 11.766 mmol.
What percentage of sodium nitrate is oxygen?
56.47% by mass. Each formula unit contains 3 oxygen atoms contributing 47.997 g of the 84.99377 g total, so 100 g of sodium nitrate contains 56.47 g of oxygen and a kilogram contains 564.71 g of it.
How do I convert grams of sodium nitrate to moles?
Divide the mass in grams by 84.993769 g/mol. So 1 g is 0.0117656 mol, 10 g is 0.117656 mol and 100 g is 1.17656 mol. Nothing else enters the calculation — no volume, no temperature, no concentration.
How many moles is 100 g of sodium nitrate?
1.17656 mol, which is 1176.56 mmol. Read the other way, one mole of sodium nitrate is 84.99377 g, so 100 g is 1.177 of a mole.
What about milligrams and micromoles?
The same division, scaled. A milligram of sodium nitrate is 11.766 µmol, 10 mg is 117.66 µmol and 100 mg is 1.1766 mmol. Working in µmol per mg avoids the string of leading zeros that mol per g produces at this scale.
Does purity change the number of moles?
Yes, in proportion. A 98% pure sample weighing 1 g contains 0.98 g of sodium nitrate, which is 0.0115303 mol rather than 0.0117656 mol. The calculator assumes the mass you enter is the compound itself, so divide by the assay figure first if you need the corrected amount.
Grams to moles reference for sodium nitrate
| Element | Atoms | Atomic weight | Contribution (g/mol) | By mass |
|---|---|---|---|---|
| Oxygen (O) | 3 | 15.999 | 47.997 | 56.47% |
| Sodium (Na) | 1 | 22.98977 | 22.98977 | 27.05% |
| Nitrogen (N) | 1 | 14.007 | 14.007 | 16.48% |
| Total — one mole of sodium nitrate | 84.993769 | 100% |
Standard atomic weights, IUPAC 2021. The contribution column is atoms × atomic weight, and the total is the molar mass this page uses: 84.993769 g/mol.
| On the balance | Moles | Millimoles | Micromoles |
|---|---|---|---|
| 1 mg | 0.000012 | 0.0117656 | 11.7656 |
| 5 mg | 0.000059 | 0.0588278 | 58.8278 |
| 10 mg | 0.000117656 | 0.117656 | 117.656 |
| 25 mg | 0.000294139 | 0.294139 | 294.139 |
| 50 mg | 0.000588278 | 0.588278 | 588.278 |
| 100 mg | 0.00117656 | 1.17656 | 1176.56 |
| 250 mg | 0.00294139 | 2.94139 | 2941.39 |
| 500 mg | 0.00588278 | 5.88278 | 5882.78 |
| 1 g | 0.0117656 | 11.7656 | 11,765.6 |
| 2 g | 0.0235311 | 23.5311 | 23,531.1 |
| 5 g | 0.0588278 | 58.8278 | 58,827.8 |
| 10 g | 0.117656 | 117.656 | 117,656 |
| 25 g | 0.294139 | 294.139 | 294,139 |
| 50 g | 0.588278 | 588.278 | 588,278 |
| 100 g | 1.17656 | 1176.56 | 1176556.832896 |
| 250 g | 2.94139 | 2941.39 | 2941392.082241 |
| 500 g | 5.88278 | 5882.78 | 5882784.164482 |
Every row is the mass divided by 84.993769 g/mol. A balance reading to 1 mg resolves 11.77 µmol of this compound, which is the smallest step the middle columns can really move in.
| Compound | Formula | Molar mass (g/mol) | Millimoles in 1 g |
|---|---|---|---|
| Potassium chloride | KCl | 74.5483 | 13.414 |
| Ammonium nitrate | NH4NO3 | 80.043 | 12.493 |
| Sodium bicarbonate | NaHCO3 | 84.00577 | 11.904 |
| Sodium nitrate (this page) | NaNO3 | 84.99377 | 11.766 |
| Magnesium chloride | MgCl2 | 95.205 | 10.504 |
| Calcium carbonate | CaCO3 | 100.086 | 9.9914 |
| Potassium nitrate | KNO3 | 101.1023 | 9.891 |
Ordered by molar mass. The last column is 1000/M, which is the number a weighed gram actually gives you.