0.092 g of a compound with the molecular formula $C_3{H}_8{O}_3$on reaction with excess of $C{H}_{3}MgI$gives 67.00 ml of methane at STP. The number of active hydrogen atoms present in a molecule of the compound is.

ROH+CH₃MgI→CH₄+Mg(OR)I

According to the above reaction every one mole of a compound contaning one active H atom per molecule releases one mole of CH₄ gas at STP. Thus, number of moles of CH₄ gas realesed per mole of compound should be an indicator of the number of active hydrogen atoms per molecule of compound.

no. of moles of CH₄= $\frac{\mathrm{Vol}.\mathrm{of} {\mathrm{CH}}_4 \mathrm{at} \mathrm{STP}}{22,400{\mathrm{mLmol}}^{-1}}=0.003mol$

${\mathrm{n}}_{\mathrm{compound}}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}=\frac{0.092\mathrm{g}}{92{\mathrm{gmol}}^{-1}}=0.001\mathrm{mol}$

This implies that 1 mol of compound releases 3 mols of CH₄ gas i.e. there are 3 active hydrogen atoms present in a molecule of the compound.