Empirical and Molecular Formulas from Stoichiometry
TLDRThe script details a chemistry experiment involving the combustion of an unknown liquid hydrocarbon, CxHy, with oxygen to produce carbon dioxide and water. Given the mass of reactants and products, the empirical formula is deduced to be C3H7 by comparing molar ratios. Utilizing the molar mass of the compound, the molecular formula is determined to be C6H14, illustrating the process of deriving both empirical and molecular formulas from experimental data.
Takeaways
- 🔍 The script discusses a chemistry problem involving the combustion of an unknown liquid hydrocarbon CxHy.
- 🔥 The hydrocarbon is burned in an apparatus with oxygen, producing carbon dioxide and water as products.
- 📝 Given data includes the initial 1.125 grams of the hydrocarbon, 3.447 grams of CO2, and 1.647 grams of H2O produced.
- 🧪 The molar mass of the hydrocarbon is determined to be 86.2 grams per mole in a separate experiment.
- 🔑 The goal is to find the empirical and molecular formulas for the unknown hydrocarbon.
- 📉 The empirical formula is the simplest whole number ratio of atoms in a molecule, while the molecular formula represents the actual number of atoms.
- ⚖️ The script calculates the moles of CO2 and H2O produced using their respective molar masses.
- 📊 From the moles of CO2 and H2O, the script deduces the moles of carbon and hydrogen in the original hydrocarbon.
- 🔍 The ratio of hydrogen to carbon atoms in the products is used to infer the ratio in the reactants, leading to the empirical formula.
- 📐 The empirical formula is determined to be CH2, with a slight adjustment to get whole numbers, resulting in C3H7.
- 🔢 Using the molar mass of 86.2 g/mol, the script concludes that the molecular formula must be twice the empirical formula, resulting in C6H14.
- 🎓 The problem-solving process involves step-by-step calculations and logical deductions based on stoichiometry and molar mass.
Q & A
What is the problem presented in the transcript about?
-The problem involves determining the empirical and molecular formulas for an unknown liquid hydrocarbon, CxHy, given its combustion in an apparatus that produces carbon dioxide and water, and knowing the molar mass of the compound.
What are the products of the combustion of the unknown hydrocarbon?
-The combustion of the unknown hydrocarbon produces 3.447 grams of carbon dioxide and 1.647 grams of water.
How is the molar mass of the unknown hydrocarbon determined in the problem?
-The molar mass of the unknown hydrocarbon is determined to be 86.2 grams per mole from a separate experiment.
What is the significance of the molar mass in determining the molecular formula?
-The molar mass helps to determine the molecular formula by providing the total mass of one mole of the compound, which can be compared to the mass of the empirical formula to find the exact number of atoms in the molecule.
What is the empirical formula and how is it derived?
-The empirical formula is the simplest whole number ratio of atoms in a compound. It is derived by comparing the moles of carbon in CO2 and the moles of hydrogen in H2O produced from the combustion of the hydrocarbon.
How many moles of carbon dioxide are produced from 3.447 grams of the unknown hydrocarbon?
-0.07834 moles of carbon dioxide are produced, calculated by dividing the mass of CO2 (3.447 grams) by its molar mass (44 grams/mole).
How many moles of water are produced from 1.647 grams of the unknown hydrocarbon?
-0.0915 moles of water are produced, calculated by dividing the mass of H2O (1.647 grams) by its molar mass (18 grams/mole).
What is the ratio of moles of hydrogen to moles of carbon in the products?
-The ratio is approximately 2.336 moles of hydrogen for every 1 mole of carbon, which is then approximated to a whole number ratio of 7:3 after adjusting.
How is the molecular formula determined from the empirical formula and the molar mass?
-The molecular formula is determined by comparing the molar mass of the empirical formula to the given molar mass of the compound. If the molar mass is a multiple of the empirical formula's molar mass, the molecular formula is the empirical formula multiplied by that multiple.
What is the molecular formula of the unknown hydrocarbon based on the given information?
-The molecular formula of the unknown hydrocarbon is C6H14, which is twice the atoms in the empirical formula (C3H7), based on the molar mass of 86.2 grams per mole.
What is the role of oxygen in the combustion process described in the script?
-Oxygen is the oxidizing agent required for the combustion of the hydrocarbon. It reacts with the hydrocarbon to produce carbon dioxide and water.
Outlines
🔍 Mystery Hydrocarbon Combustion Analysis
The script begins by introducing a problem involving the combustion of an unknown liquid hydrocarbon, CxHy, with 1.125 grams as the initial mass. The combustion process is described, yielding 3.447 grams of carbon dioxide and 1.647 grams of water. The script emphasizes the need for oxygen and assumes an excess is present, making the hydrocarbon the limiting reactant. The molar mass of the compound is revealed to be 86.2 grams per mole, and the goal is to determine the empirical and molecular formulas of the hydrocarbon. The process involves calculating the moles of produced carbon dioxide and water to deduce the ratio of carbon to hydrogen atoms in the original compound.
📚 Calculating Moles from Grams
This paragraph delves into the calculation of moles from grams for both carbon dioxide and water produced in the combustion reaction. The molar mass of carbon dioxide (44 grams per mole) is used to determine 0.07834 moles of CO2 from the 3.447 grams obtained. Similarly, the molar mass of water (18 grams per mole) is applied to find 0.0915 moles of H2O from the 1.647 grams produced. The explanation includes the concept of molar mass and how to convert between grams and moles, highlighting the importance of understanding Avogadro's number in these calculations.
🔬 Deduction of Empirical Formula
The script continues by focusing on deducing the empirical formula of the hydrocarbon. It explains that the carbon atoms in the carbon dioxide and the hydrogen atoms in the water must have originated from the hydrocarbon, establishing a direct relationship between the moles of these elements in the products and the reactants. By calculating the moles of carbon from CO2 and doubling the moles of hydrogen from H2O (due to two hydrogens per water molecule), the script approximates a ratio of 7 moles of hydrogen to 3 moles of carbon, suggesting an empirical formula of CH2 with a 7:3 ratio.
🧬 Determining the Molecular Formula
The final paragraph addresses the determination of the molecular formula using the empirical formula and the given molar mass of the compound. It points out the discrepancy between the calculated molar mass of the empirical formula (43 grams per mole) and the provided molar mass of 86.2 grams per mole. The conclusion is that the molecular formula must have twice the number of atoms as the empirical formula while maintaining the same ratio, leading to the molecular formula C6H14. This section illustrates the process of scaling the empirical formula to match the actual molar mass and emphasizes the importance of this step in fully characterizing the compound.
Mindmap
Keywords
💡Hydrocarbon
💡Combustion
💡Carbon Dioxide (CO2)
💡Water (H2O)
💡Empirical Formula
💡Molecular Formula
💡Molar Mass
💡Limiting Reactant
💡Moles
💡Atomic Weight
💡Ratio
Highlights
The mystery liquid hydrocarbon CxHy is burned in an apparatus, producing 3.447 grams of carbon dioxide and 1.647 grams of water.
The hydrocarbon is assumed to be the limiting reactant, with an excess of oxygen for complete combustion.
The empirical formula is determined by the simplest whole number ratio of atoms in the molecule.
The molar mass of the compound is found to be 86.2 grams per mole in a separate experiment.
Calculating moles of carbon dioxide and water produced from their respective masses.
The molar mass of carbon dioxide is determined to be 44 grams per mole.
The molar mass of water is 18 grams per mole, used to calculate moles of water produced.
The ratio of hydrogens to carbons in the products is used to deduce the empirical formula.
The empirical formula is approximated to a whole number ratio of 7 hydrogens for every 3 carbons.
The molecular formula is deduced by comparing the calculated empirical formula molar mass to the given molar mass of 86.2 grams per mole.
The molecular formula is determined to be C6H14, based on the molar mass information.
The process involves an approximation technique to find the simplest whole number ratio of atoms.
The importance of understanding that a mole is a count of entities, not a measure of weight.
The practical application of stoichiometry in determining the composition of an unknown hydrocarbon.
The problem-solving approach emphasizes step-by-step calculations and logical reasoning.
The final determination of the hydrocarbon's molecular formula as C6H14 concludes the problem-solving process.
Transcripts
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