Question

a compound contains 54.2% carbon 9.2% Hydrogen and 36.6% Oxyen. Determine its empirical formula of its molecular weight 88 u (Atomic mass of C = 12 u O = 16 u H = 1 u)

Answer 1

Refer the photo for steps.
Empirical formula is C2H4O

Answer 2

Answer : The empirical formula of a compound is $C_2H_4_O_1$.

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 = 54.2 g

Mass of H = 9.2 g

Mass of O = 36.6 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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{54.2g}{12g/mole}=4.516moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{9.2g}{1g/mole}=9.2moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{36.6g}{16g/mole}=2.287moles$

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

For C = $\frac{4.516}{2.287}=1.97\approx 2$

For H = $\frac{9.2}{2.287}=4.022\approx 4$

For O = $\frac{2.287}{2.287}=1$

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 = $C_2H_4_O_1$

Therefore, the empirical formula of a compound is $C_2H_4_O_1$.