Answer to Question #307906 in General Chemistry for cols

Question

An organic compound was found to contain 54.5 % C, 9.2 % H, and 36.3 % O.

a. What is the empirical formula of the compound?

b. If the molar mass of the compound is 88.1 g/mol, what is its molecular formula?

Answer 1

Assuming 100 g of the compound, the moles of each element are:

$n(C)=\frac{54.5\ g}{12.01\ g/mol}=4.54\ mol$

$n(H)=\frac{9.2\ g}{1.01\ g/mol}=9.11\ mol$

$n(O)=\frac{36.3\ g}{16.0\ g/mol}=2.27\ mol$

a) The lowest whole number ratio of C : H : O is 2 : 4 : 1. Therefore, the empirical formula is C₂H₄O.

b) The molar mass of 1 empirical formula unit is 44.05 g/mol. The molar mass of the compound is twice the molar mass of the empirical formula ($\frac{88.1}{44.05}=2$ ). Therefore, the subscripts should be multiplied by 2 to get the molecular formula:

C₄H₈O₂

Answer: a) C₂H₄O; b) C₄H₈O₂