What is the empirical formula and empirical formula mass for the following compound: C3H6O3?

Jan 4, 2016

The empirical formula of ${C}_{3} {H}_{6} {O}_{3}$ is $C {H}_{2} O$. Why?

Explanation:

By definition, the empirical formula is the SIMPLEST whole number ratio that defines constituent atoms in a species. The ratio of atoms in that species is 1:2:1 with respect to carbon, hydrogen, and oxygen.
The molecular formula is ALWAYS a mulitple of the empirical formula. Here ${\left(E N\right)}_{n}$ $=$ $M F$ $=$ $\left(C {H}_{2} O\right) \times 3$ $=$ ${C}_{3} {H}_{6} {O}_{3}$ as required. And of course its empirical formula mass is 30, and its molecular mass is ???

Microanalysis can tell you percentage compositions of a compound by mass, and we can use this percentage composition to give an empirical formula. We need an idea of molecular mass before we can calculate the molecular formula of an unknown.

Jan 4, 2016

The empirical formula is $\text{CH"_2"O}$.
The empirical formula mass is "30.026.

Explanation:

An empirical formula represents the simplest whole number ratio of elements in a compound.

The given formula, $\text{C"_3"H"_6"O"_3}$, is not an empirical formula because the subscripts can be divided by the whole number $3$ to simplify the formula to $\text{CH"_2"O}$, which is the empirical formula.

Determine the empirical formula mass by multiplying each element's subscript by its atomic weight on the periodic table and adding them together.

$\text{Empirical formula mass}$$=$$\left(1 \times 12.0107\right) + \left(2 \times 1.00794\right) + \left(1 \times 15.999\right) = 30.026$