Question

A 6.40 g sample of a compound is burned to produce 8.37 g CO_2, 2.75 g H_2O, 1.06 g N_2, and 1.23 g SO_2. What is the empirical formula of the compound? Give your answer in the form C#H#N#O#S# where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly).

Answer 1

The empirical formula :

C₁₀H₁₆N₄SO₇

Further explanation

Given

6.4 g sample

Required

The empirical formula

Solution

mass C :

= 12/44 x 8.37 g

= 2.28

mass H :

= 2/18 x 2.75 g

= 0.305

mass N = 1.06

mass S :

= 32/64 x 1.23

= 0.615

mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g

Mol ratio :

= C : H : N : S : O

= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16

= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019

= 10 : 16 : 4 : 1 : 7

The empirical formula :

C₁₀H₁₆N₄SO₇