Combustion analysis of a 12.01 g sample of tartaric acts - which contains only carbon, hydrogen, and oxygen - produced 14.08 g CO2 and 4.32 g H2O. What is the empirical formula for tartaric acid?

Jun 9, 2018

$\text{Empirical formula of tartaric acid} \equiv {C}_{2} {H}_{3} {O}_{3}$

Explanation:

We address the molar quantities of carbon, and hydrogen .... convert these into masses, and then address the empirical formula.

$\text{Moles of carbon dioxide} = \frac{14.01 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} = 0.3183 \cdot m o l$

And thus mass of carbon...$\equiv 0.3183 \cdot m o l \times 12.011 \cdot g \cdot m o {l}^{-} 1 = 3.824 \cdot g$

$\text{Moles of water} = \frac{4.32 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 0.2399 \cdot m o l$

And thus mass of hydrogen...$\equiv 2 \times 0.2399 \cdot m o l \times 1.00794 \cdot g \cdot m o {l}^{-} 1 = 0.4835 \cdot g$

The balance of the mass....$12.01 \cdot g - 0.4835 \cdot g - 3.824 \cdot g = 7.702 \cdot g$...represents the mass of oxygen....a molar quantity of $\frac{7.702 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 0.4814 \cdot m o l$

And so we divide thru by the LEAST molar quantity, that of OXYGEN, to get a trial empirical formula of...

${C}_{\frac{0.3183 \cdot m o l}{0.2399 \cdot m o l}} {H}_{\frac{0.4814 \cdot m o l}{0.2399 \cdot m o l}} {O}_{\frac{0.4814 \cdot m o l}{0.2399 \cdot m o l}}$ $\equiv$ ${C}_{1.33} {H}_{2.00} {O}_{2}$..

But by definition, the $\text{empirical formula}$ $\text{IS THE SIMPLEST WHOLE NUMBER ratio...}$ and so we multiply the calculated empirical formula by a factor of THREE...

$3 \times \left\{{C}_{1.33} {H}_{2.00} {O}_{2}\right\} \equiv {C}_{4} {H}_{6} {O}_{6} \equiv {C}_{2} {H}_{3} {O}_{3}$