The complete combustion of 0.492g of an organic compound containing ‘C’, ‘H’ and ‘O’ given 0.793g of CO2 and 0.442g H2O. The percentage of oxygen composition in the organic compound is __________. (nearest integer)

Moles of C = moles of CO2= <!- [if gte mso 9]><xml> </xml><! [endif]-> 0 . 7 9 3 4 4 m o l Mass of C = <!- [if gte mso 9]><xml> </xml><! [endif]-> 0 . 7 9 3 4 4 × 1 2 g = 0.261gMoles of H = 2 × <!- [if gte mso 9]><xml> </xml><! [endif]-> 0 . 4 4 2 1 8 m o l Mass of H = <!- [if gte mso 9]><xml> </xml><! [endif]-> 2 × 0 . 4 4 2 1 8 × 1 g Total mass of compound = 0.492g (given)So; mass of O = (0.492 – 0.216 – 0.049) g= 0.227g% of O = 0 . 2 2 7 0 . 4 9 2 × 1 0 0 <!- [if gte mso 9]><xml> </xml><! [endif]-> = 46.14%the nearest integer = 46

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Chemistry NCERT Exemplar Solutions Class 12th Chapter Three Alcohol Phenol And Ethers Ncert Solutions Chemistry Class 12th Chemistry Class 12th

The complete combustion of 0.492g of an organic compound containing ‘C’, ‘H’ and ‘O’ given 0.793g of CO₂ and 0.442g H₂O. The percentage of oxygen composition in the organic compound is __________. (nearest integer)

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alok kumar singh | Contributor-Level 10

2 months ago

Moles of C = moles of CO₂

= $\frac{0.793}{44} m o l$

Mass of C = $\frac{0.793}{44} \times 12 g$

= 0.261g

Moles of H = 2 × $\frac{0.442}{18} m o l$

Mass of H = $2 \times \frac{0.442}{18} \times 1 g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492} \times 100$

= 46.14%

the nearest integer = 46

Moles of C = moles of CO2= <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814701944"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>7</mn> <mn>9</mn> <mn>3</mn> </mrow> <mrow> <mn>4</mn> <mn>4</mn> </mrow> </mfrac> <mi>m</mi> <mi>o</mi> <mi>l</mi> </mrow> </math> Mass of C = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814701945"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>7</mn> <mn>9</mn> <mn>3</mn> </mrow> <mrow> <mn>4</mn> <mn>4</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>2</mn> <mi>g</mi> </mrow> </math> = 0.261gMoles of H = 2 × <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1814701946"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>4</mn> <mn>4</mn> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mi>m</mi> <mi>o</mi> <mi>l</mi> </mrow> </math> Mass of H = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1814701947"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mn>2</mn> <mo>×</mo> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>4</mn> <mn>4</mn> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mi>g</mi> </mrow> </math> Total mass of compound = 0.492g (given)So; mass of O = (0.492 – 0.216 – 0.049) g= 0.227g% of O = <math> <mrow> <mfrac> <mrow> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>2</mn> <mn>7</mn> </mrow> <mrow> <mn>0</mn> <mo>.</mo> <mn>4</mn> <mn>9</mn> <mn>2</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1814701948"> </o:OLEObject></xml><! [endif]-> = 46.14%the nearest integer = 46