Propenoic acid, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2},\) is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An unlabeled container is thought to contain this liquid. A \(0.275-g\) sample of the liquid is combusted to produce \(0.102 \mathrm{~g}\) of water and \(0.374 \mathrm{~g}\) carbon dioxide. Is the unknown liquid propenoic acid? Support your reasoning with calculations.

First, we need to determine the moles of each element present in the sample. We are given that combustion produces \(0.374 \mathrm{~g}\) of carbon dioxide and \(0.102 \mathrm{~g}\) of water. Let's start by finding the moles of carbon and hydrogen. Moles of carbon (in 0.374 g CO₂)= \(\frac {0.374 g}{44.01 g/mol} =0.0085 mol\) Moles of hydrogen (in 0.102 g H₂O)= \(\frac{2×0.102 g}{18.02 g/mol} = 0.0113 mol\) Now, let's find the mass of oxygen in the sample. To do this, we subtract the masses of carbon and hydrogen from the total mass of the sample (\(0.275 g\)): Mass of oxygen = \(0.275 g - (0.0085 mol × 12.01 g/mol) - (0.0113 mol × 1.008 g/mol)\) = \(0.191 g\) Now, we can calculate the moles of oxygen: Moles of oxygen = \(\frac{0.191 g}{16.00 g/mol} = 0.0119\) We have the moles of each element in the sample: Carbon: \(0.0085 \mathrm{mol}\) Hydrogen: \(0.0113 \mathrm{~mol}\) Oxygen: \(0.0119 \mathrm{~mol}\)

To find the ratio of the elements in the sample, we must determine the smallest whole number ratio between moles of carbon, hydrogen, and oxygen. We start by finding the smallest mole value among the three elements, which is 0.0085 in this case (carbon). Next, divide all the mole values by this smallest value: Carbon (\(\mathrm{C}\)) ratio = \(\frac{0.0085}{0.0085} = 1\) Hydrogen (\(\mathrm{H}\)) ratio = \(\frac{0.0113}{0.0085} \approx 1.33\) Oxygen (\(\mathrm{O}\)) ratio = \(\frac{0.0119}{0.0085} \approx 1.4\) Since the ratios aren't close to whole numbers, we can try multiplying all ratios by the smallest ratio (in this case, 3) to obtain whole number ratios: Carbon (\(\mathrm{C}\)) ratio = \(1 × 3 = 3\) Hydrogen (\(\mathrm{H}\)) ratio = \(1.33 × 3 \approx 4\) Oxygen (\(\mathrm{O}\)) ratio = \(1.4 × 3 \approx 4\) This gives us the approximate empirical formula: \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4\).

Now that we have determined the empirical formula of the unknown liquid sample, we can compare it to the given formula for propenoic acid, which is \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_2\). Our empirical formula is \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4\), which is different from the given propenoic acid formula \(\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_2\). Based on this comparison, we can conclude that the unknown liquid is not propenoic acid.