A compound is 53.31 % C , 11.18 % H , and 35.51 % O by mass. What is its empirical formula? Insert subscripts as needed.

The empirical formula is C2H5O

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100.0 grams

The compound contains:

53.31 % C = 53.31 grams C

11.18 % H = 11.18 grams H

35.51 % O = 35.51 grams O

Step 2: Calculate moles

moles = mass / molar mass

Moles C = 53.31 grams / 12.01 g/mol

Moles C = 4.439 moles

Moles H = 11.18 grams / 1.01 g/mol

Moles H = 11.07 moles

Moles O = 35.51 grams / 16.0 g/mol

Moles O = 2.22 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 4.439 / 2.22 = 2

H: 11.07 / 2.22 = 5

O: 2.22/2.22 = 1

C2H5O

The empirical formula is C2H5O

> 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.