Math Problem Statement
Solution
The problem you're dealing with is a combustion analysis, where a hydrocarbon is burned in excess oxygen to produce carbon dioxide () and water.
To find the mass of hydrogen in the original sample, we'll need to focus on the amount of water produced during the reaction. However, based on the problem, we only have information on the amount of , which relates directly to the carbon content. Let's go step by step:
Step-by-Step Solution:
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Determine the moles of carbon: Since 9.728 g of is produced, we can calculate how much carbon this represents.
- Molar mass of = 44.01 g/mol. Each mole of contains 1 mole of carbon, so the moles of carbon are also 0.221 mol.
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Mass of carbon in the hydrocarbon: The mass of carbon can be found using its atomic mass (12.01 g/mol):
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Mass of hydrogen: The original sample has a total mass of 3.546 g. The rest of the mass (after accounting for the carbon) must be hydrogen, since the combustion products are and water, and oxygen comes from the atmosphere.
So, the mass of hydrogen in the original sample is 0.892 g.
Would you like more details on any of these steps?
Here are five related questions to expand your understanding:
- How do you determine the amount of oxygen consumed in this combustion reaction?
- How would the calculation change if the hydrocarbon contained oxygen atoms?
- What if you were given the mass of water produced instead of ?
- How does stoichiometry apply when determining the empirical formula of a hydrocarbon?
- What assumptions are made in combustion analysis when determining element mass?
Tip: When performing combustion analysis, always remember to account for the total mass of the products to ensure mass conservation.
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Math Problem Analysis
Mathematical Concepts
Stoichiometry
Combustion Analysis
Formulas
Moles of CO2 = mass of CO2 / molar mass of CO2
Mass of Carbon = moles of CO2 × molar mass of Carbon
Mass of Hydrogen = total mass of sample - mass of Carbon
Theorems
Law of Conservation of Mass
Suitable Grade Level
Grades 10-12 (High School Chemistry)
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