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Chapter 4: Problem 117

A \(2.20-\) 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 mL of 0.500 M 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

Find moles of NaOH used in the titration

To find the moles of NaOH used in the titration, we can use the formula: Moles of NaOH = volume of NaOH (L) × molarity of NaOH Given: Volume of NaOH = 25.0 mL = 0.025 L (conversion from mL to L) Molarity of NaOH = 0.500 M Moles of NaOH = (0.025 L) × (0.500 mol/L) = 0.0125 mol
02

Find moles of the unknown acid

Since we know that 1 mole of the acid reacts with 1 mole of NaOH, the moles of the unknown acid are equal to the moles of NaOH: Moles of unknown acid = Moles of NaOH = 0.0125 mol
03

Find the molar mass of the acid

Now that we know the moles of the unknown acid, we can use the mass of the sample to find the molar mass. Molar mass of the unknown acid = mass of the unknown acid (g) / moles of the unknown acid Given: Mass of the unknown acid = 2.20 g Moles of the unknown acid = 0.0125 mol Molar mass of the unknown acid = 2.20 g / 0.0125 mol = 176 g/mol
04

Find the ratio of the empirical formula molar mass to the molar mass

We need to calculate the ratio of the empirical formula's molar mass (\(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3}\)) to the molar mass found in step 3: Empirical formula molar mass: 3 × (12 g/mol for C) + 4 × (1 g/mol for H) + 3 × (16 g/mol for O) = 88 g/mol Ratio = (Molar mass of the unknown acid) / (Empirical formula molar mass) = 176 g/mol / 88 g/mol = 2
05

Determine the molecular formula

Now we multiply the empirical formula by the ratio found in step 4 to determine the molecular formula: Molecular formula = \(\mathrm{C}_{3}\mathrm{H}_{4}\mathrm{O}_{3} × 2 = \mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}\) The molecular formula of the unknown acid is \(\mathrm{C}_{6}\mathrm{H}_{8}\mathrm{O}_{6}\).

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Most popular questions from this chapter

Citric acid, which can be obtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) . A 0.250 -g sample of citric acid dissolved in 25.0 mL of water requires 37.2 mL of 0.105 M NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?

Balance the following oxidation–reduction reactions that occur in basic solution. a. $\mathrm{Al}(s)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}-(a q)$ b. \(\mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl}^{-}(a q)+\mathrm{OCl}^{-}(a q)\) c. $\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{-}(a q)$

When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of 5.0 M HCl is required to react completely with 3.00 g of magnesium?

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ $$\mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ A 10.00-g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156 mL of 3.00 M silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If 78.0 mL of HCl was added, what was the concentration of the HCl?

A mixture contains only sodium chloride and potassium chloride. A 0.1586-g sample of the mixture was dissolved in water. It took 22.90 mL of 0.1000 M AgNO3 to completely precipitate all the chloride present. What is the composition (by mass percent) of the mixture?
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