Question

percentage of carbon = 52.17, percentage of hydrogen = 13.02 and rest is oxygen. the molecular mass is 92 . determine the empirical and molecular formula.

Answer 1

Here is your answer

First step, divide percentage by the molar mass of the atom.

in 100 g of the compound, there will be: 52.17 g of Carbon

in 100 g of the compound, there will be: 13.02 g of hydrogen

in 100 g of the compound, there will be: 100 - 52.17 - 13.02 = 34.81 % of oxygen

Now, $\frac{52.17}{12}$, $\frac{13.02}{1}$ and $\frac{34.81}{16}$ = 4.2 (approx), 13.02, 2.1 (approx)

Now, take approximate ratios

4.2 : 13.02 : 2.1

2 : 6 : 1

= C₂H₆O

Empirical formula

Step 2:  Now, find the molar mass of the empirical formula:

C₂H₆O = 46 g

Step 3: Find n factor:

Divide the original molar mass by the empirical formula mass and you will get n factor

92/46 = 2

So, n = 2

Step 4: multiply the n factor

ₙ(C₂H₆O)

Applying n, we get: C₄H₁₂O₂

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