Molar Mass of Carbonic acid

H2CO3 — 62.024 g/mol, worked out element by element.

Molar mass of Carbonic acid, H₂CO₃

62.024 g/mol

Molecular weight 62.024 · 6 atoms per formula unit

Also known as: h2co3.

Formed when CO₂ dissolves in water; the buffer that holds blood pH near 7.4.

How 62.024 g/mol is worked out

Mass contribution of each element
ElementAtomsAtomic massMass% of total
Oxygen315.99947.99777.38%
Carbon112.01112.01119.37%
Hydrogen21.0082.0163.25%
Total662.024100%

M = 3 × 15.999 + 1 × 12.011 + 2 × 1.008 = 62.024 g/mol

Grams to moles for Carbonic acid

Mass to moles conversions
MassMolesFormula units
1 g0.0161 mol9.709 × 10^21
10 g0.1612 mol9.709 × 10^22
25 g0.4031 mol2.427 × 10^23
50 g0.8061 mol4.855 × 10^23
100 g1.6123 mol9.709 × 10^23
250 g4.0307 mol2.427 × 10^24
62.024 g1 mol6.022 × 10^23

For any other amount, use the molar mass calculator, or the molarity calculator to make up a solution of a given concentration.

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Frequently Asked Questions

What is the molar mass of Carbonic acid?

The molar mass of Carbonic acid (H2CO3) is 62.024 g/mol. That is the sum of the standard atomic weights of all 6 atoms in one formula unit: 3 × 15.999 + 1 × 12.011 + 2 × 1.008 = 62.024.

What is the molecular weight of Carbonic acid?

62.024 — the same number as the molar mass, but as a dimensionless ratio rather than in g/mol. The two terms describe the same arithmetic, so a molecular weight of 62.02 and a molar mass of 62.02 g/mol are the same answer.

How many moles are in 100 g of Carbonic acid?

100 ÷ 62.024 = 1.6123 mol. To convert any mass to moles, divide by the molar mass; to go the other way, multiply.

What is the percent composition of Carbonic acid?

By mass, Carbonic acid is 77.38% oxygen, 19.37% carbon, 3.25% hydrogen. Each element's share is its total mass in the formula divided by 62.024.

How many atoms are in one formula unit of Carbonic acid?

6: 3 oxygen, 1 carbon, 2 hydrogen. One mole contains 6.02214076 × 10²³ formula units, so a mole of Carbonic acid contains 6 × 6.022 × 10²³ atoms in total.