Problem 41
Question
An organic dibasic acid is used in volumetric analysis. The percentage of elements are \(\mathrm{C}=26.7 \%\), \(\mathrm{H}=2.2 \%\) and \(\mathrm{O}=71.1 \%\). Find the empirical formula of the acid. (a) \(\mathrm{CH}_{2} \mathrm{O}\) (b) \(\mathrm{CHO}_{2}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{4} \mathrm{O}\)
Step-by-Step Solution
Verified Answer
The empirical formula of the acid is \( \mathrm{CHO}_{2} \).
1Step 1: Converting percentages to masses
Assume you have 100 grams of the dibasic acid. This makes it straightforward to convert the percentages to grams:
- Carbon: 26.7 grams
- Hydrogen: 2.2 grams
- Oxygen: 71.1 grams.
2Step 2: Converting masses to moles
Convert the masses of each element to moles using the molar mass of each element: - Moles of Carbon: \( \frac{26.7}{12} = 2.225 \)- Moles of Hydrogen: \( \frac{2.2}{1} = 2.2 \)- Moles of Oxygen: \( \frac{71.1}{16} = 4.44375 \)
3Step 3: Finding the smallest mole ratio
Divide each mole quantity by the smallest number of moles calculated: - Ratio for Carbon: \( \frac{2.225}{2.2} = 1.011 \)- Ratio for Hydrogen: \( \frac{2.2}{2.2} = 1 \)- Ratio for Oxygen: \( \frac{4.44375}{2.2} = 2.019 \)
4Step 4: Simplifying to whole numbers
Round each ratio to the nearest whole number to find the subscripts for each element in the empirical formula: - Carbon: Rounded from 1.011 to 1 - Hydrogen: Already 1- Oxygen: Rounded from 2.019 to 2Thus, the empirical formula is \( \mathrm{CHO}_{2} \).
Key Concepts
Dibasic AcidVolumetric AnalysisElemental Composition Analysis
Dibasic Acid
A dibasic acid is a type of acid that can donate two protons or hydrogen ions per molecule in a chemical reaction. This characteristic is what defines its term "dibasic."
The ability of an acid to donate protons is a fundamental concept in acid-base chemistry, and it plays a crucial role in various chemical processes.
When analyzing a dibasic acid, it's essential to understand that its reactions and stoichiometry are often more complex compared to monobasic acids.
For instance, in a complete neutralization with a base like sodium hydroxide, a dibasic acid would require two moles of the base to neutralize one mole of the acid.
Here are some key properties:
The ability of an acid to donate protons is a fundamental concept in acid-base chemistry, and it plays a crucial role in various chemical processes.
When analyzing a dibasic acid, it's essential to understand that its reactions and stoichiometry are often more complex compared to monobasic acids.
For instance, in a complete neutralization with a base like sodium hydroxide, a dibasic acid would require two moles of the base to neutralize one mole of the acid.
Here are some key properties:
- They often show two distinct endpoints in a titration curve.
- Common examples include sulfuric acid (\(H_2SO_4\)) and oxalic acid (\(H_2C_2O_4\)).
- The presence of two acidic hydrogens impacts their behavior in many chemical reactions.
Volumetric Analysis
Volumetric analysis is a technique in chemical analysis that involves measuring the volume of a solution as a way to determine the concentration of a solute.
This method is commonly used in titrations, where a solution of known concentration, called a titrant, is added to a solution with an unknown concentration to determine its content.
Titrations play a significant role in various fields, from industrial to academic research, due to their accuracy and ease of execution.
Here are key steps in a typical titration:
This method is commonly used in titrations, where a solution of known concentration, called a titrant, is added to a solution with an unknown concentration to determine its content.
Titrations play a significant role in various fields, from industrial to academic research, due to their accuracy and ease of execution.
Here are key steps in a typical titration:
- Preparation of solutions: Ensure that the solutions of the titrant and the sample are correctly prepared.
- Execution: Slowly add the titrant to the analyte until the reaction reaches completion at the endpoint, usually indicated by a color change.
- Calculation: Finally, calculate the concentration of the unknown using the volume of titrant used and its concentration.
Elemental Composition Analysis
Elemental composition analysis involves determining the percentage or proportion of elements within a compound. This is a crucial step when working with empirical formulas, as it allows chemists to understand the basic makeup of a compound.
The empirical formula represents the simplest whole-number ratio of the elements within a compound. Calculating it involves:
This is how you derive the empirical formula \(CHO_2\), enabling chemists to predict and understand the chemical reactions and properties of the compound.
This step forms the foundation for analyzing and synthesizing new chemical compounds in the lab.
The empirical formula represents the simplest whole-number ratio of the elements within a compound. Calculating it involves:
- Converting percentage composition to grams, based on a 100-gram sample for simplicity.
- Converting grams to moles using the atomic masses of the elements.
- Finding the simplest whole-number ratio among the moles of elements.
This is how you derive the empirical formula \(CHO_2\), enabling chemists to predict and understand the chemical reactions and properties of the compound.
This step forms the foundation for analyzing and synthesizing new chemical compounds in the lab.
Other exercises in this chapter
Problem 39
The purity of organic compounds is determined by (a) mixed melting point (b) molecular weight (c) melting point (d) density
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In Kjeldahl's method of estimation of nitrogen, copper sulphate acts as (a) catalytic agent (b) silver spiral (c) hydrolyzing agent (d) oxidizing agent
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The molecular formula weight by empirical formula weight of a compound is 3. The empirical formula is \(\mathrm{CH}_{2}\). The vapour density of the compound is
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Chromatographic technique is used for these paration of (a) plant pigments (b) dye stuffs (c) small samples of mixtures (d) all of these
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