Question

calculate the empirical formula and molecular formula of a compound containing 76.6% carbon ,6.38% hydrogen and rest oxygen its vapour density is 47

Answer 1

HEY!!!!!!




we work of 100 g of compound and calculate the number of moles of each element

moles of C = 40 / 12 = 3.33 moles
moles of H = 6.67 /1 = 6.67
moles of O = 53.33 / 16 = 3.33 moles

molar ratio of C : H : O = 3.33 : 6.67 : 3.33

= 2 : 2 : 1

Answer 2

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 76.6% = 76,6 g 
H: 6.38% = 6.38 g
O: 47% = 47 g

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{76.6\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.383\:mol$

$H: \dfrac{6.38\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 6.38\:mol$

$O: \dfrac{47\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 2.9375\:mol$

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.383}{2.9375}\to\:\:\boxed{C\approx 2}$

$H: \dfrac{6.38}{2.9375}\to\:\:\boxed{H\approx 2}$

$O: \dfrac{2.9375}{2.9375}\to\:\:\boxed{O = 1}$

Thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{C_2H_2O_1}}\end{array}}\qquad\checkmark$

Thus, the molecular formula found for the compound will be:

$\boxed{\boxed{C_2H_2O}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR!