An unknown alcohol (M = 74) contains 48 g carbon, 10 g hydrogen, and 16 g oxygen per mole. Determine its molecular formula.

Introduction / Context:Finding a molecular formula from elemental composition and molecular weight is a staple of chemical stoichiometry and fuels characterization.

Given Data / Assumptions:

• Molecular mass = 74 g/mol.
• Per mole: C = 48 g, H = 10 g, O = 16 g.
• Atomic weights: C = 12, H = 1, O = 16.

Concept / Approach:Convert each element’s mass to moles, then assemble the atomic ratios to get the molecular formula that matches the given molecular mass.

Step-by-Step Solution:Moles of C = 48/12 = 4.Moles of H = 10/1 = 10.Moles of O = 16/16 = 1.Empirical ratio → C4H10O; this already sums to 412 + 101 + 16 = 74 g/mol.Write as an alcohol: C4H9OH (equivalent to C4H10O).

Verification / Alternative check:Known C4 alcohols (1-butanol, isobutanol, sec-butanol, tert-butanol) all have formula C4H10O, consistent with the calculation.

Why Other Options Are Wrong:C3H21OH, (C2H4)2H2·OH, and C2H33OH have impossible or non-integer hydrogen counts and do not yield M = 74.

Common Pitfalls:Forgetting that an alcohol’s “OH” adds O and H to the carbon–hydrogen skeleton; mis-summing atomic masses.

Final Answer:C4H9OH