Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is :
Solution
C<sub>x</sub>H<sub>y</sub>O<sub>z</sub> + O<sub>2</sub> $\to$ xCO<sub>2</sub> + ${y \over 2}{H_2}O$
<br><br>2.64 g of CO<sub>2</sub> contains 0.72 g C.
<br><br>1.08 g of H<sub>2</sub>O contains 0.12 g H.
<br><br>$\therefore$ Mass of oxygen present = 1.80 – (0.72
+0.12) = 0.96 g
<br><br>% of O = ${{0.96} \over {1.80}} \times 100$ = 53.33 %
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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