120g of an organic compound that contains only carbon and hydrogen gives 330g of CO2 and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

Number of moles of C = Number of moles of CO2 = 3 3 0 4 4 molesNumber of moles of H = 2 × no. of moles of H2O = ( 2 7 0 1 8 × 2 ) moles Mass of C = 3 3 0 4 4 × 1 2 gm = 90 gm Mass of H = 2 7 0 1 8 × 2 × 1 g m = 3 0 g m <!- [endif]-><!- [if gte mso 9]><xml> </xml><! [endif]-> % o f C = 9 0 1 2 0 × 1 0 0 % = 7 5 % % o f H = 3 0 1 2 0 × 1 0 0 % = 2 5 %

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Quantitative Analysis Chemistry Organic Chemistry Chemistry Class 11th

120g of an organic compound that contains only carbon and hydrogen gives 330g of CO₂ and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

Option 1 -

25 and 75

Option 2 -

40 and 60

Option 3 -

60 and 40

Option 4 -

75 and 25

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Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 4

Detailed Solution:

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 × no. of moles of H₂O = $(\frac{270}{18} \times 2)$ moles

Mass of C = $\frac{330}{44} \times 12$ gm = 90 gm

Mass of H = $\frac{270}{18} \times 2 \times 1 g m = 30 g m$

$\begin{array}{l} % o f C = \frac{90}{120} \times 100 % = 75 % \\ % o f H = \frac{30}{120} \times 100 % = 25 % \end{array}$

Number of moles of C = Number of moles of CO2 = <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mn>3</mn> <mn>0</mn> </mrow> <mrow> <mn>4</mn> <mn>4</mn> </mrow> </mfrac> </mrow> </math> molesNumber of moles of H = 2 × no. of moles of H2O = <math> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>2</mn> <mn>7</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> moles Mass of C = <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mn>3</mn> <mn>0</mn> </mrow> <mrow> <mn>4</mn> <mn>4</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>2</mn> </mrow> </math> gm = 90 gm Mass of H = <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mn>7</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </mfrac> <mo>×</mo> <mn>2</mn> <mo>×</mo> <mn>1</mn> <mi>g</mi> <mi>m</mi> <mo>=</mo> <mn>3</mn> <mn>0</mn> <mi>g</mi> <mi>m</mi> </mrow> </math> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1819198591"> </o:OLEObject></xml><! [endif]-> <math> <mtable columnalign="left"> <mtr> <mtd> <mrow> <mn>%</mn> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mi>C</mi> <mo>=</mo> <mfrac> <mrow> <mn>9</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>2</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>%</mn> <mo>=</mo> <mn>7</mn> <mn>5</mn> <mn>%</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mn>%</mn> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mi>H</mi> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>2</mn> <mn>0</mn> </mrow> </mfrac> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mn>%</mn> <mo>=</mo> <mn>2</mn> <mn>5</mn> <mn>%</mn> </mrow> </mtd> </mtr> </mtable> </math>

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120g of an organic compound that contains only carbon and hydrogen gives 330g of CO₂ and 270g of water on complete combustion. The percentage of carbon and hydrogen, respectively are

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