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  • A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

     

    (i) empirical formula,

     

    (ii) molar mass of the gas, and

     

    (iii) molecular formula.

A welding fuel gas contains carbon and h | Class 11 Chemistry Chapter Some Basic Concepts of Chemistry, Some Basic Concepts of Chemistry NCERT Solutions

Q34.

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

 

(i) empirical formula,

 

(ii) molar mass of the gas, and

 

(iii) molecular formula.

(i) percentage of C can be calculated as follows:

CO2 = C

i.e 44 parts of CO2= 12 parts of C

OR

44g of CO2 = 12 g of C

 

Therefore according to question

3.38 g of CO2 contains C = 12/44 * 3.38 = 0.921 g

18 g of water contains hydrogen = 2g

 

Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g

0.690 g of water will contain hydrogen

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g = 0.9984 g

Now percentage of carbon = weight of carbon/weight of compound * 100

=0.921 /0.998 * 100= 92.32 %

Also percentage of hydrogen = weight of hydrogen/weight of compound *100

=0.0766/0.998 * 100 = 7.68 %

 

(ii) Given,

Weight of 10.0L of the gas (at S.T.P) = 11.6 g

Weight of 22.4 L of gas at STP

= 25.984 g

≈ 26 g

Hence, the molar mass of the gas is 26 g.

 

(iii) empirical formula

 

Element Percentage Atomic mass Atomic ratio Simplest ratio Simplest whole no ratio
C 92.32 12 92.32/12 = 7.69 7.69/7.65 = 1.00 1
H 7.65 1 7.65/1 = 7.65 7.65/7.65 = 1 1

 

Empirical formula of the compound  = CH

Now molecular formula  calculation

Empirical formula mass = 12 + 1 = 13 amu

Also molecular mass  = 26 g (calculated in previous step)

Therefore n = molecular mass/empirical formula mass = 26/13 = 2

Now molecular formula = n x empirical formula = 2 x CH = C2H2

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What is the correct answer to: A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate   (i) empirical formula,   (ii) molar mass of the gas, and   (iii) molecular formula.?

(i) percentage of C can be calculated as follows:

CO2 = C

i.e 44 parts of CO2= 12 parts of C

OR

44g of CO2 = 12 g of C

 

Therefore according to question

3.38 g of CO2 contains C = 12/44 * 3.38 =...

How do you solve A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate   (i) empirical formula,   (ii) molar mass of the gas, and   (iii) molecular formula. step by step?

Step-by-step explanation:
• (i) percentage of C can be calculated as follows:
•
• CO2 = C
•
• i

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Student Discussion

Ritobrata Pal
Class · · , · Sep 07, 2019
Nice answer
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Edrus basha
Class · · , · Aug 16, 2019
Good
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Shubham
Class · · , · Aug 06, 2019
great question backed with a great answer
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Ashwani kumar Mishra
Class · · , · Jul 28, 2019
From where 22.4 value came
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Arya
Class · · , · Jul 05, 2019
Excellent answer
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Bhumika
Class · · , · Jun 26, 2019
tysm, this answer was really helpful.
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Janvy ram
Class · · , · Jun 25, 2019
Thanks for making me understand this
Question.☺
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Noobmaster
Class · · , · Jun 24, 2019
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CENTIMETRE
Class · · , · Jun 19, 2019
Why so lengthy I can explain better
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Ak
Class · · , · Jun 13, 2019
Yes your answer is very simple and helping and you know the style of answering very well.You explained each and every step.
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