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Magazine Chapter 3: Problem 103
Thiophene is a liquid compound of the elements C, H, and \(S .\) A sample of thiophene weighing \(7.96 \mathrm{mg}\) was burned in oxygen, giving \(16.65 \mathrm{mg} \mathrm{CO}_{2} .\) Another sample was subjected to a series of reactions that transformed all of the sulfur in the compound to barium sulfate. If \(4.31 \mathrm{mg}\) of thiophene gave \(11.96 \mathrm{mg}\) of barium sulfate, what is the empirical formula of thiophene? Its molecular mass is 84 amu. What is its molecular formula?
Short Answer
Expert verified
The empirical and molecular formula of thiophene is C鈧凥鈧凷.
Step by step solution
01
Determine the Moles of Carbon Produced
First, we need to calculate the moles of carbon dioxide produced. The molar mass of CO鈧� is 44 g/mol.\[\text{Moles of } \mathrm{CO}_2 = \frac{16.65 \text{ mg}}{44 \text{ g/mol}} = \frac{16.65 \times 10^{-3} \text{ g}}{44 \text{ g/mol}} \approx 0.378 \times 10^{-3} \text{ mol}\] Each mole of CO鈧� contains one mole of carbon, so the moles of carbon in the sample is the same as the moles of CO鈧�:
02
Calculate Mass of Carbon in Thiophene
Multiply the moles of carbon by the atomic mass of carbon (12 g/mol) to find the mass of carbon from the sample.\[\text{Mass of carbon} = 0.378 \times 10^{-3} \text{ mol} \times 12 \text{ g/mol} = 4.536 \times 10^{-3} \text{ g} = 4.536 \text{ mg}\]
03
Determine Moles of Sulfur from Barium Sulfate
Calculate the moles of barium sulfate (BaSO鈧�). The molar mass of BaSO鈧� is approximately 233 g/mol.\[\text{Moles of } \mathrm{BaSO}_4 = \frac{11.96 \text{ mg}}{233 \text{ g/mol}} = \frac{11.96 \times 10^{-3} \text{ g}}{233 \text{ g/mol}} \approx 0.0513 \times 10^{-3} \text{ mol}\] Each mole of BaSO鈧� corresponds to one mole of sulfur:
04
Calculate Mass of Sulfur in Thiophene
Multiply the moles of sulfur by the atomic mass of sulfur (32 g/mol) to find the mass of sulfur in the sample.\[\text{Mass of sulfur} = 0.0513 \times 10^{-3} \text{ mol} \times 32 \text{ g/mol} = 1.6416 \times 10^{-3} \text{ g} = 1.6416 \text{ mg}\]
05
Find Mass of Hydrogen in Thiophene
Subtract the masses of carbon and sulfur from the total mass of the thiophene sample to find the mass of hydrogen.\[\text{Mass of hydrogen} = 4.31 \text{ mg} - 4.536 \text{ mg} - 1.6416 \text{ mg} = 0.7824 \text{ mg}\]
06
Calculate Moles of Hydrogen
Calculate the moles of hydrogen as the atomic mass of hydrogen is 1 g/mol.\[\text{Moles of hydrogen} = \frac{0.7824 \times 10^{-3} \text{ g}}{1 \text{ g/mol}} = 0.7824 \times 10^{-3} \text{ mol}\]
07
Determine Empirical Formula
Find the simplest whole number ratio of moles of C, H, and S in the thiophene sample.\[\begin{align*}\text{Moles C} & = 0.378 \times 10^{-3} \text{ mol} \\text{Moles H} & = 0.7824 \times 10^{-3} \text{ mol} \\text{Moles S} & = 0.0513 \times 10^{-3} \text{ mol}\end{align*}\]Divide each by the smallest number of moles (S):\[\begin{align*}\text{Ratio for C} & = \frac{0.378}{0.0513} \approx 7.4 \\text{Ratio for H} & = \frac{0.7824}{0.0513} \approx 15.2 \\text{Ratio for S} & = \frac{0.0513}{0.0513} = 1\end{align*}\] These simplify approximately to a whole number 4:4:1, leading to an empirical formula of C鈧凥鈧凷.
08
Determine Molecular Formula
Find the molecular formula by comparing molecular mass and empirical formula mass.\[\begin{align*}\text{Empirical mass of } C_4H_4S & = 4 \times 12 + 4 \times 1 + 1 \times 32 = 84 \\text{Molecular mass} & = 84 \\end{align*}\]Since these are identical, the empirical formula is the molecular formula: C鈧凥鈧凷.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molecular Formula
The molecular formula, a fundamental concept in chemistry, represents the actual number of each type of atom present in a molecule of a compound. In our exercise, thiophene was examined to determine both its empirical and molecular formulas. The empirical formula gives the simplest whole-number ratio of atoms in a compound.
To find the molecular formula, one must compare the molecular mass with the empirical formula mass. For thiophene, both these values are 84 amu, confirming that the empirical formula (C鈧凥鈧凷) is indeed the molecular formula.
This relationship can sometimes reveal how multiple empirical units exist within the molecular formula, indicating the complexity of larger organic molecules. Understanding molecular formulas helps in predicting properties, structures, and reactions of chemical compounds.
To find the molecular formula, one must compare the molecular mass with the empirical formula mass. For thiophene, both these values are 84 amu, confirming that the empirical formula (C鈧凥鈧凷) is indeed the molecular formula.
This relationship can sometimes reveal how multiple empirical units exist within the molecular formula, indicating the complexity of larger organic molecules. Understanding molecular formulas helps in predicting properties, structures, and reactions of chemical compounds.
Elemental Analysis
Elemental analysis, a crucial technique in chemistry, involves determining the elemental composition of a compound, specifically how much of each element is present. In thiophene's case, a sample underwent combustion analysis to determine carbon content from produced carbon dioxide.
Furthermore, a separate analysis involved converting sulfur into barium sulfate for quantification. These experiments reveal the distribution of elements, leading to a better understanding of the compound's composition.
Furthermore, a separate analysis involved converting sulfur into barium sulfate for quantification. These experiments reveal the distribution of elements, leading to a better understanding of the compound's composition.
- Carbon levels were correlated to the amount of CO鈧� formed.
- Sulfur amounts were related to the formation of BaSO鈧�.
- Hydrogen was deduced by the remaining mass after considering carbon and sulfur.
Stoichiometry
Stoichiometry is the math behind chemical reactions. It involves calculating reactants and products in chemical equations. For thiophene, stoichiometry was used to analyze combustion and reactions involving barium sulfate for sulfur quantification.
Through stoichiometry, we determined the moles of (CO鈧�) produced and (BaSO鈧�) formed, directly correlating to the moles of carbon and sulfur in the sample, respectively.
Through this, stoichiometry helped to:
Through stoichiometry, we determined the moles of (CO鈧�) produced and (BaSO鈧�) formed, directly correlating to the moles of carbon and sulfur in the sample, respectively.
Through this, stoichiometry helped to:
- Calculate moles from masses based on molar masses.
- Relate these moles to individual elements in thiophene.
- Adjust ratios to solve for empirical and thus molecular formulas.
Chemical Formulas
Chemical formulas are symbols representing the composition of chemical compounds. They provide insight into the types and numbers of atoms in compounds, and they are essential for understanding chemical properties and reactions.
In our exercise, the chemical formula C鈧凥鈧凷 indicates:
Through the process of stoichiometry and elemental analysis, understanding chemical formulas involves deducing accurate ratios leading to correct empirical and molecular formulas. This enables predicting compound behavior under different conditions and reactions.
In our exercise, the chemical formula C鈧凥鈧凷 indicates:
- 4 Carbon (C) atoms
- 4 Hydrogen (H) atoms
- 1 Sulfur (S) atom
Through the process of stoichiometry and elemental analysis, understanding chemical formulas involves deducing accurate ratios leading to correct empirical and molecular formulas. This enables predicting compound behavior under different conditions and reactions.
