Question

QUESTION 16What is the empirical formula that corresponds to a compound that contains 57.69% C, 11.54% H, and 30.77% O by mass?O C5H1202O C2.5H6OC3H1202O C58H12031С6HO3

Answer 1

`C_5H_(12)O_2`

Step-by-step explanation

Given:

57.69% C, 11.54% H, and 30.77% O by mass,

Meaning:

Mass of carbon = 0.5769 g

Mass of Hydrogen = 0.1154 g

Mass of Oxygen = 0.3077 g

We know:

Molar masses of:-

C = 12.0107 g/mol

H = 1.00794 g/mol

O = 15.999 g/mol

Required: Empirical formula

Solution:

Step 1: Calculate the number of moles of the 3 atoms

C : 0.5769 g/12.0107 g/mol = 0.04826 mol

H : 0.1154 g/1.00794 g/mol = 0.1145 mol

O : 0.3077 g/15.999 g/mol = 0.0192

Step 2: Divide the moles by the lowest number of moles

In this case divide by 0.0192

C: 2.5

H: 6

O:1

Step 3: multiply by 2 to remove a decimal

C: 5

H: 12

O:2

Therefore the answer is: C5H12O2