Determination of the molecular formula of alcohols

Exercise 2

    

An organic compound of molar mass $58 \; \frac{g}{mol}$ contains the elements $C, H $ and $ O $ Elemental analysis gave the following percentages: $ C: \: $ 62.1% $ H: \: $ 10.3% 1) Determine the molecular formula! 2) If it would be an alcohol, what can you say about its hydrocarbon chain?

1) General formula: $C_xH_yO_z$ $\%_C$ $=$ $62.1$ $\frac{x\cdot 12}{58}$ $=$ $0.621$ $x$ $\approx$ $3$ $\%_H$ $=$ $10.3$ $\frac{y\cdot 1}{58}$ $=$ $0.103$ $y$ $\approx$ $6$ $\%_O$ $=$ $100-62.1-10.3$ $=$ $27.6$ $\frac{z\cdot 16}{58}$ $=$ $0.276$ $z$ $\approx$ $ 1$ Molecular formula: $C_3H_6O$ 2) The molecular formula has the general type : $C_nH_{2n}O$ Beeing an alcohol (no $C = O$), the chain must be either cyclic or aliphatic with one double bond.