Problem: Tartaric acid is the white, powdery substance that coats sour candies such as Sour Patch Kids. Combustion analysis of a 12.01-g sample of tartaric acid-which contains only carbon, hydrogen, and oxygen-produced 14.08 g CO2 and 4.32 g H2 O.Find the empirical formula for tartaric acid.

Recall that in combustion analysis, a compound reacts with excess O₂ to form products. 

For a compound composed of C, H, and O, the reaction looks like this:

C_xH_yO_z + O₂ (excess) → x CO₂ + y H₂O

Calculate the mass of C and H.

Mass C:

molar mass of CO₂ = 44.01 g/mol

There is 1 mole of C in 1 mole of CO₂. Finding the mass of C:

$\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{08}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{C}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{C}}{\mathbf{1}\cancel{\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{C}}}$

mass of C = 3.8423 g C

Mass H: 

molar mass of H₂O = 18.02 g/mol

There are 2 moles of H in 1 mole of H₂O. Finding the moles of H:

$\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{32}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{H}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{H}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{H}}}$

mass of H = 0.4842 g H

Calculate the mass of O:

 We have 0.8009 g of the sample compound. Subtracting the masses of C and H from this will give us the mass of O:

mass of O = 12.01 g – (3.8423 g C + 0.4842 g H) = 7.6834 g O

Calculate the moles of each element and determine the lowest whole number ratio. 

$\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{8423}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{C}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{C}}{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{C}}}$

moles of C = 0.3199 mol C