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Empirical formula of 26.5 g carbon, 2.94 g hydrogen, 70.6 g oxygen

User Spodger
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1 Answer

5 votes

Answer:

C3H4O6

Step-by-step explanation:

See the attached table. Since we have the masses of each element, we can calculate the moles of each, by dividing by that element's molar mass (grams/mole). The moles of each are shown on the table. They are not in whole ratio values, but if we multiply each by 1.36 we obtain whole numbers, which would be the ratio of each element in the formula.

Atoms

C 3

H 4

O 6

The empirical formula is C3H4O6

Possibly Dihydroxymalonic acid

Empirical formula of 26.5 g carbon, 2.94 g hydrogen, 70.6 g oxygen-example-1
User Danae
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