On complete combustion 1.0 g of an organic compound (X ) gave 1.46 g of CO_2 and 0.567 g of H_2O. The empirical formula mass of compound (X ) is g. (Given molar mass in gmol^-1C :12 ,H :1 ,O :16 )

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Correct answer: C. (c) Calculate the Moles of Carbon: Carbon is fully converted to CO_2. Moles of CO_2= 1.46 g44 g/ mol=0.033 mol. Moles of Carbon (C )= Moles of CO_2=0.033 mol. Mass of Carbon =0.033 12 g/ mol=0.396 g. Calculate the Moles of Hydrogen: Hydrogen is fully converted to H_2O. Moles of H_2O= 0.567 g18 g/ mol=0.0315 mol. Moles of Hydrogen (H )= 2 0.0315=0.063 mol. Mass of Hydrogen = 0.063 1 g/ mol=0.063 g. Calculate the Mass of Oxygen: Total mass of compound X=1.0 g. Mass of Oxygen = Total mass - (Mass of Carbon + Mass of Hydrogen) Mass of Oxygen = 1.0 g- (0.396 g+0.063 g )=0.541 g. Moles of Oxygen (O) = 0.54116=0.0338 ≈0.033 mol. Determine the Empirical Formula: The ratio of moles: C=0.033 ,H=0.063, 0=0.033. Simplified, this ratio is approximately 1 :2 :1. Empirical formula: CH_2O. Calculate the Empirical Formula Mass: Empirical formula mass = 12 1+1 2+16 1=30 g/ mol. Therefore, the empirical formula mass of compound X is 30 g/ mol.

$On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO_{2} and 0.567 g of H_{2} O . The empirical formula mass of compound (X) is g. (Given molar mass in gmol^{- 1} C : 12, H : 1, O : 16)$

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