Dear student,

Given, A hydrocarbon contains 4.8g of carbon per gram of hydrogen.

(a) Since, 12g of carbon = 1g atom of carbon
So, 4.8g of carbon= 4.8 / 12 = 0.4
Therefore, gram atom of carbon is 0.4.

​​​​​​1g of hydrogen = 1g atom of hydrogen
Therefore, gram atom of hydrogen is 1.

(b) Since, vapour density is 29

The molecular mass of the hydrocarbon = vapour density x 2 = 29 x 2 = 58g
Let x be the mass of hydrogen in the hydrocarbon
So, the mass of carbon in the hydrocarbon = 4.8x

Therefore,
x + 4.8x = 58g
5.8x = 58g
x = 10g

Therefore mass of hydrogen in the hydrocarbon = 10g
Mass of carbon in the hydrocarbon = 4.8 (10) = 48g

no. of moles of carbon = mass of carbon / molar mass of carbon = 48g / (12g/mol) = 4moles
No. of moles of hydrogen = mass of hydrogen / molar mass of hydrogen = 10g / (1g/mol) = 10moles

The ratio of C:H is 4:10 = 2:5
Therefore, the empirical formula of hydrocarbon is C2H5

(c) Empirical formula mass ( C2H5) = 12(2) + 1(5) = 29g

Now, molecular formula = n(empirical formula)

n = molecular mass / empirical formula mass = 58g / 29g = 2

molecular formula = 2( C2H5) = C4H10

Regards

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