Question

a compound has the following percentage composition by mass carbon 54.55% hydrogen 9.09% oxygen 36.26% vapour density is 44 find the empirical and molecular formula of the compound{H=1;C=12;O=16}​

Answer 1

Answer : The empirical and molecular of the compound is, $C_2H_4O_1$  and $C_4H_8O_2$

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 54.55 g

Mass of H = 9.09 g

Mass of O = 36.26 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{54.55g}{12g/mole}=4.56moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.09g}{1g/mole}=9.09moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{36.26g}{16g/mole}=2.26moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{4.56}{2.26}=2.01\approx 2$

For H = $\frac{9.09}{2.26}=4.02\approx 4$

For O = $\frac{2.26}{2.26}=1$

The ratio of C : H : O = 2 : 4 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_2H_4O_1$

The empirical formula weight = 2(12) + 4(1) + 1(16) = 44 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}\\\\n=\frac{2\times \text{Vapor density}}{\text{Empirical formula weight}}=\frac{2\times 44}{44}=2$

Molecular formula = $(C_2H_4O_1)_n=(C_2H_4O_1)_2=C_4H_8O_2$

Therefore, the empirical and molecular of the compound is, $C_2H_4O_1$  and $C_4H_8O_2$