Consider the following step of a reaction mechanism.
A+B → D + E (slow)

Which statement identifies a reaction intermediate?

1.A, if it is used in a later step
2.B, if it is produced in an earlier step
3. D. if it is used in an earlier step
4. E, if it is produced in a later step

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