Question

A hydrocarbon contains 4.8 g of carbon per gram of hydrogen. Calculate: the g atom of each, find the emperical formula, find molecular formula, if its vapour density is 29

Answer 1

The molecular mass of the hydrocarbon = Vapour density x 2

                                                            = 29 x 2 = 58 g

Let x be the mass of hydrogen in the hydrocarbon

Therefore, mass of carbon = 4.8 x

Now, 4.8 x + x = 58g

5.8 x = 58 g

 x = 10 g

Therefore, mass of hydrogen in the hydrocarbon = 10 g

Mass of carbon in the hydrocarbon = 4.8 x 10 = 48 g

Moles/ g atoms of carbon = 48/12 = 4

Moles/ g atoms of hydrogen = 10/1 = 10

Ratio of moles of carbon and hydrogen = 4:10

  = 2:5

So, empirical formula of the hydrocarbon = C2H5

Empirical formula mass = 24 + 5 = 29 g

Now,

Molecular formula = (empirical formula)n

Where,

n = Molecular mass / Empirical formula mass

  = 58/29

  = 2

So, molecular formula = (C2H5)2 = C4H10

Answer 2

Answer:

(a)0.4 g atom of carbon,1 g atom of hydrogen.

(b)c2h5

(c)c4h10

Explanation:

(a)12g of carbon= 1 g atom of carbon

So, 4.8 g of carbon=(1/12)*4.8=

0.4g atom of carbon ans

1 g of hydrogen=

1 g atom of hydrogen ans

(b) for empirical formula, check the image.

(c) empirical formula mass

=2*12+5=29

2*vd=n*emperical formula mass

2*29=n*29

n=2

Therefore,

molecular formula=C4H10

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