What is the empirical formula of a compound that is 62.0% c, 10.4% h, and 27.5% o by mass?

The empirical formula for the given compound is

Explanation:

We are given:

Percentage of C = 62.0 %

Percentage of H = 10.4 %

Percentage of O = 27.5 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 62.0 g

Mass of H = 10.4 g

Mass of O = 27.5 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.72 moles.

For Carbon =

For Hydrogen  =

For Oxygen  =

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is

2b2t

explanation:

2b2t

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.