Alaboratory analysis of an unknown sample yields 40.5% carbon, 4.5% hydrogen, and 55.0% oxygen. if the sample has a molar mass of 176.0 g/mol, what is the molecular formula of the sample

Answer is: molecular formula is C₆H₈O₆.
n(C) = m(C) ÷ M(C).
n(C) = 40,5 g ÷ 12 g/mol.
n(C) = 3,375 mol.
n(H) = m(H) ÷ M(H).
n(H) = 4,5 g ÷ 1 g/mol.
n(H) = 4,5 mol.
n(O) = m(O) ÷ M(O).
n(O) = 55 g ÷ 16 g/mol.
n(O) = 3,4 mol.
n(C) : n(H) : n(O) = 3,375 mol : 4,5 mol : 3,4 mol / :3,375.
n(C) : n(H) : n(O) = 1 : 1,33 : 1.
n(C) : n(H) : n(O) = 3 : 4 : 3.
M(C₃H₄O₃) = 88 g/mol · 2 = 176 g/mol.

It would be (at) + cu (s)