Question

In a compound, mass of carbon present is four
times of hydrogen while mass of oxygen present
is 1.5 times of carbon. Empirical formula of the
compound will be
(1) C4H20
mc)= 4(H)
(2) C8H160
8
mode lor(c)
(3) C8H2409
m(oo=UCH)
(4) C8H3002​

Answer 1

Explanation:

1 is correct answer c4h20

Answer 2

Answer:

correct answer is option 3 (C8H24O9)

Explanation:

here,

no of carbon =x

atomic mass of carbon =12

no of hydrogen =y

atomic mass of hydrogen = 1

no of oxygen =z

atomic mass of oxygen =16

condition 1

mass of carbon is 4 times mass of hydrogen

therefore based of above statement we can say x(12)= 4(y(1))

we will check our answer (option 3 )

in which we have 8(x) carbon atoms and 24(Y) hydrogen

x(12)= 4(y(1))

8(12)=4(24(1))

96=96

lhs=rhs

it satisfy our first condition

condition 2

mass of oxygen present is 1.5 times of carbon

therefore based of above statement  we can say z(16)=1.5(x(12))

we will check our answer (option 3 )

in which we have 8(x) carbon atoms and 9(z) oxygen

z(16)=1.5(x(12))

9(16)= 1.5(8(12))

144=144

lhs=rhs

satisfy our condition two

hence correct answer is option 3 (C8H24O9)