Answer:
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
