Question

give the empirical formula of a hydrocarbon that is 52.14% c, 13.13% h, and 34.73% o by mass. write the formula with the atoms in the order cxhyoz . you can ignore the charge and phase.

Answer 1

The empirical formula for a hydrocarbon that is 52.14% C, 13.13% H, and 34.73% O by mass is C2H6O, assuming a 100 g sample and converting each percentage to moles based on their atomic masses.

Step-by-step explanation:

To determine the empirical formula of a hydrocarbon with a composition of 52.14% carbon (C), 13.13% hydrogen (H), and 34.73% oxygen (O), assume a 100 g sample for simplicity.

This translates to 52.14 g of C, 13.13 g of H, and 34.73 g of O. Next, find the number of moles of each element by dividing by the atomic mass (12.01 g/mol for C, 1.01 g/mol for H, and 16.00 g/mol for O). The result would be 4.343 moles of C, 13.001 moles of H, and 2.170 moles of O.

Find the smallest ratio by dividing each by the smallest number of moles obtained, which in this case is 2.170 moles. This results in a ratio of 2 for C, approximately 6 for H, and 1 for O. Therefore, the empirical formula is C2H6O.