0.6 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24g of carbon, 0.04g of hydrogen with the rest being oxygen. If the molecular weight of the substance is 90g/mol what is the emperical and molecular formula

Molecular formula is, C₃H₆O₃

Empirical formula (with the lowest subscript) → CH₂O

Explanation:

We assume that the organic acid's mass is 0.6 g. We know that, in 0.6 g of compound we have 0.24 g of C and 0.04 g of H, then, we have

(0.6 - 0.24 - 0.04) = 0.32 g of O.

We determine each mol:

0.24 g /12 g/mol = 0.02 mol of C

0.04 g/ 1g/mol = 0.04 mol of H

0.32 g/ 16 g/mol = 0.02 mol of O

1 mol of acid weighs 90 g/mol.

In 0.6 g of acid, we have, (0.6 g / 90g/mol) = 0.0067 moles

Let's find out the formula with rules of three:

0.00667 mol of acid have 0.02 moles of C

1 mol of acid may have (0.02 /0.00667) = 3 mol of C

0.00667 mol of acid have 0.04 moles of H

1 mol of acid may have (0.04 /0.00667) =  6 mol of H

0.00667 mol of acid have 0.02 moles of O

1 mol of acid may have (0.02 /0.00667) = 3 mol of O

Molecular formula is, C₃H₆O₃

We confirm by the molar mass: 12g/mol . 3 + 6 . 1g/mol + 32g/mol . 3 = 90

Empirical formula (with the lowest subscript) → CH₂O

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