The molar mass of ethylene glycol, C2H6O2, is 62.068 g/mol. That single number is what connects a mass you can weigh to an amount of substance you can react: one mole is 62.068 g, and a gram of it is 9.702e+21 formula units. The calculator turns any weighed mass into that count, and reports the amount in moles, millimoles and the mass of a single formula unit alongside it.
The figure comes from the formula. Ethylene glycol is 2 × 12.011 (C) + 6 × 1.008 (H) + 2 × 15.999 (O), which is 62.068 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: 2 of the 10 atoms in a formula unit are oxygen, and they account for 31.998 g of the 62.068 g, or 51.55% by mass. The derivation table below breaks the whole molecule down element by element.
The numbers a bench actually uses sit at the millimole scale. One millimole of ethylene glycol is 62.068 mg and one micromole is 62.068 µg, so a balance reading to a milligram places you within 16.11 µmol. A single formula unit weighs 1.0307e-22 g, which is why the count in the gauge runs into the sextillions for any mass you can see.
Antifreeze and the feedstock for PET. Sweet-tasting and lethal, which is why coolant is dyed and why propylene glycol replaces it where pets are about. At 62.068 g/mol it is heavier than 5 of the 21 organic compounds covered here, and that ranking matters more than it looks: isopropyl alcohol has a molar mass of 60.096 g/mol, so a gram of it contains 3.28% more formula units than a gram of ethylene glycol. 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 1.2414 g of a nominal 62.068 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 of ethylene glycol in grams
62.068- The molar mass of ethylene glycol in g/mol, derived from its formula
N- The number of formula units that mass contains
How it works, step by step
- Enter a mass of ethylene glycol in grams.
- It is divided by the molar mass, 62.068 g/mol, giving the amount in moles.
- That amount is multiplied by the Avogadro constant, 6.02214076 × 10²³ per mole, to give the number of formula units.
- The panel also reports the amount in moles and millimoles and the mass of one formula unit.
Worked examples
0.5 g of ethylene glycol as a molecule count
0.5 g divided by 62.068 g/mol is 0.00805568 mol, and multiplying by the Avogadro constant gives 4.851e+21 formula units of ethylene glycol. The same mass is 8.0557 mmol, which is the figure a reaction is actually planned in.
5 g of ethylene glycol as a molecule count
5 g divided by 62.068 g/mol is 0.0805568 mol, and multiplying by the Avogadro constant gives 4.851e+22 formula units of ethylene glycol. The same mass is 80.557 mmol, which is the figure a reaction is actually planned in.
50 g of ethylene glycol as a molecule count
50 g divided by 62.068 g/mol is 0.805568 mol, and multiplying by the Avogadro constant gives 4.851e+23 formula units of ethylene glycol. The same mass is 805.57 mmol, which is the figure a reaction is actually planned in.
How to read your score
Frequently asked questions
What is the molar mass of ethylene glycol (C2H6O2)?
It is 62.068 g/mol. That is the sum of the standard atomic weights of every atom in the formula: 2 × 12.011 (C) + 6 × 1.008 (H) + 2 × 15.999 (O). One mole of ethylene glycol therefore weighs 62.068 g, and a gram of it is 16.111 mmol.
What percentage of ethylene glycol is oxygen?
51.55% by mass. Each formula unit contains 2 oxygen atoms contributing 31.998 g of the 62.068 g total, so 100 g of ethylene glycol contains 51.55 g of oxygen and a kilogram contains 515.53 g of it.
How many molecules are in one gram of ethylene glycol?
About 9.702e+21. One gram is 0.0161114 mol, and each mole contains 6.02214076 × 10²³ formula units by definition of the mole, so the count is that constant divided by 62.068.
How much does a single molecule of ethylene glycol weigh?
1.0307e-22 g, which is 62.068 daltons. It is the molar mass divided by the Avogadro constant, and the number in daltons is numerically the same as the molar mass in g/mol — that equivalence is what makes the mole convenient.
Is molar mass the same as molecular weight?
They are the same number in ordinary use but not the same quantity. Molecular weight, properly relative molecular mass, is a ratio and has no unit: for ethylene glycol it is 62.068. Molar mass carries grams per mole: 62.068 g/mol. Because the mole is defined so those agree, you can read one off the other.
Why is the molar mass not a whole number?
Because standard atomic weights are averages over the isotopes found in nature. Oxygen is quoted as 15.999 rather than a whole number for that reason, and the same applies to the other elements here, which is why ethylene glycol comes out at 62.068 g/mol rather than 62.
Molar mass, composition and mole equivalents of ethylene glycol
| Element | Atoms | Atomic weight | Contribution (g/mol) | By mass |
|---|---|---|---|---|
| Oxygen (O) | 2 | 15.999 | 31.998 | 51.55% |
| Carbon (C) | 2 | 12.011 | 24.022 | 38.7% |
| Hydrogen (H) | 6 | 1.008 | 6.048 | 9.744% |
| Total — one mole of ethylene glycol | 62.068 | 100% |
Standard atomic weights, IUPAC 2021. The contribution column is atoms × atomic weight, and the total is the molar mass this page uses: 62.068 g/mol.
| Amount | Mass (g) | Mass (mg) | Formula units |
|---|---|---|---|
| 1 µmol | 0.000062 | 0.062068 | 6.022e+17 |
| 10 µmol | 0.00062068 | 0.62068 | 6.022e+18 |
| 100 µmol | 0.0062068 | 6.2068 | 6.022e+19 |
| 1 mmol | 0.062068 | 62.068 | 6.022e+20 |
| 10 mmol | 0.62068 | 620.68 | 6.022e+21 |
| 50 mmol | 3.1034 | 3103.4 | 3.011e+22 |
| 100 mmol | 6.2068 | 6206.8 | 6.022e+22 |
| 250 mmol | 15.517 | 15,517 | 1.506e+23 |
| 500 mmol | 31.034 | 31,034 | 3.011e+23 |
| 750 mmol | 46.551 | 46,551 | 4.517e+23 |
| 1 mol | 62.068 | 62,068 | 6.022e+23 |
| 2 mol | 124.136 | 124,136 | 1.204e+24 |
| 5 mol | 310.34 | 310,340 | 3.011e+24 |
| 10 mol | 620.68 | 620,680 | 6.022e+24 |
Mass is the amount multiplied by 62.068 g/mol. The last column is that amount multiplied by the Avogadro constant, 6.02214076 × 10²³ per mole.
| Compound | Formula | Molar mass (g/mol) | Millimoles in 1 g |
|---|---|---|---|
| Acetone | C3H6O | 58.08 | 17.218 |
| Urea | CH4N2O | 60.056 | 16.651 |
| Isopropyl alcohol | C3H8O | 60.096 | 16.64 |
| Ethylene glycol (this page) | C2H6O2 | 62.068 | 16.111 |
| Glycine | C2H5NO2 | 75.067 | 13.321 |
| Benzene | C6H6 | 78.114 | 12.802 |
| Glycerol | C3H8O3 | 92.094 | 10.858 |
Ordered by molar mass. The last column is 1000/M, which is the number a weighed gram actually gives you.