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Chapter 4: Problem 109

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

Short Answer

Expert verified
The molecular formula of the unknown acid is \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\).

Step by step solution

01

Calculate the moles of NaOH used in the titration

First, we need to find out how many moles of NaOH were used in the titration. This can be done using the given volume and concentration of NaOH. Moles of NaOH = volume × concentration Moles of NaOH = \(25.0 \times 10^{-3} L \times 0.500 M\) Moles of NaOH = \(1.25 \times 10^{-2} mol\)
02

Calculate the moles of the unknown acid

Since one mole of the unknown acid reacts with one mole of NaOH, then the moles of the unknown acid will be equal to the moles of NaOH. Moles of unknown acid = moles of NaOH = \(1.25 \times 10^{-2} mol\)
03

Calculate the molar mass of the unknown acid

Now, we can find out the molar mass of the unknown acid using the given mass of the sample and the moles calculated before. Molar mass of unknown acid = mass of sample / moles of unknown acid Molar mass of unknown acid = \(2.20 g / (1.25 \times 10^{-2} mol)\) Molar mass of unknown acid = \(176 g/mol\)
04

Calculate the empirical mass of the unknown acid

Next, we will find the empirical mass of the unknown acid using the given empirical formula (\(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\)) by adding up the atomic masses of the constituent elements. Empirical mass = \(3 \times 12.01 g/mol_{C} + 4 \times 1.01 g/mol_{H} + 3 \times 16.00 g/mol_{O}\) Empirical mass = \(36.03 + 4.04 + 48.00\) Empirical mass = \(88.07 g/mol\)
05

Determine the molecular formula of the unknown acid

Now we can use the molar mass and empirical mass to determine the molecular formula of the unknown acid. Divide the molar mass by the empirical mass to find out how many empirical units are present in the molecular formula. Number of empirical units = molar mass / empirical mass Number of empirical units = \(176 g/mol / 88.07 g/mol\) Number of empirical units = 2 Since there are two empirical units in the molecular formula, we can multiply the subscripts in the empirical formula by 2 to find the molecular formula. Molecular formula = \((\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3})_{2}\) Molecular formula = \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\)

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Most popular questions from this chapter

What volume of \(0.0200 M\) calcium hydroxide is required to neutralize \(35.00 \mathrm{~mL}\) of \(0.0500 M\) nitric acid?

A \(25.00\) -mL sample of hydrochloric acid solution requires 24\. 16 mL of \(0.106 M\) sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

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