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A welding fuel gas contains carbon and h...

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 litre (Measured at STP) of this welding gas is found weigh `11.6 g`. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

Text Solution

AI Generated Solution

To solve the problem step by step, we will determine the empirical formula, molar mass, and molecular formula of the welding fuel gas based on the given data. ### Step 1: Calculate the mass of carbon in the carbon dioxide produced. Given: - Mass of CO₂ = 3.38 g To find the mass of carbon (C) in CO₂: ...
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A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gas 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. Calcuate

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 S.T.P.) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical fomula (ii) molar mass of the gas, and (iii) molecular formula.

Knowledge Check

  • A gaseous hydrocarbon on complete combustion gives 3.38 g of CO_(2) and 0.690 g of H_(2)O and no other product. The empirical formula of the hydrocarbon is

    A
    `CH`
    B
    `CH_(2)`
    C
    `CH_(3)`
    D
    The data is not complete

  • A 10 g sample of oxygen gas is taken in a container of volume 1 litre and is found to exert a pressure of 3 bar. Which of the following options is correct regarding speed of the molecuels?

    A
    All the molecules are moving at a same speed which is equal to 310 m/sec.
    B
    `U_("avg")=300m//"sec"`
    C
    `U_("mps")=300xxsqrt((2)/(5)` m/sec.
    D
    `U_("mps")=310` m/sec.

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