Question

A welding fuel gas contains carbon and hydrogen only burning a sample of it in oxygen gives 3.38 g carbon dioxide, 0.69 g of water and no other products. A volume of 10 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 (iii) Molecular formula. Are yrr koi to batao yrr Answer​

Answer 1

Answer:

(i) Mass of carbon in 3.38 g of CO2

= 3.38 g/44 x 12 = 0.922 g

Mass of hydrogen in 0.690 g of H2O

= 0.690 g/18 x 2 = 0.077 g

Total mass of the sample burnt = 0.922 g + 0.077 g = 0.999 g

Percentage of carbon in the fuel = {0.922}/{0.999 g} x 100 = 92.29%

Percentage of hydrogen in the fuel = {0.077 g}/{0.999 g} x 100 = 7.71%

1) Element = Carbon(C)

Mass Percent = 92.29

Atomic mass = 12.0

Relative number of atoms = 92.29/12.0 = 7.69

Simple atomic ratio = 7.69/7.69 = 1

2) Element = Hydrogen(H)

Mass Percent = 7.71

Atomic mass = 1.0

Relative number of atoms = 7.71/1.0

Simple atomic ratio = 7.71 7.71/7.69 = 1

Therefore, empirical formula of the compound = CH

(ii) Volume of the gaseous fuel = 11.6 g

Molar mass of the fuel = {11.6 g}/{10.0 L} x 22.4 L/ml = 26.0 g mol-1

(iii) Empirical formula mass of the fuel = (12 + 1) g mol-1 = 13 g mol-1

Molar mass of the fuel = 26.0 g mol-1

n = {26.0 g mol-1}/{13 g mol-1} = 2

Molecular formula of the fuel = 2 x Empirical formula

= 2 x CH = C2H

Answer 2

Answer:

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