what is the empirical formula of a compound composed of 3.25% hydrogen (h), 19.36% carbon (c), and 77.39%…

Explanation:

Step1: Assume 100g of the compound

So we have 3.25g of H, 19.36g of C, and 77.39g of O.

Step2: Calculate the moles of each element

For H: $n_H=\frac{3.25g}{1.01g/mol}\approx3.22mol$ For C: $n_C=\frac{19.36g}{12.01g/mol}\approx1.61mol$ For O: $n_O=\frac{77.39g}{16.00g/mol}\approx4.84mol$

Step3: Divide by the smallest number of moles

The smallest is 1.61mol. For H: $\frac{3.22mol}{1.61mol}=2$ For C: $\frac{1.61mol}{1.61mol}=1$ For O: $\frac{4.84mol}{1.61mol}=3$

Answer:

$CH_2O_3$