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Combustion analysis of a 125 g sample of an unknown compound produces 392 g of CO₂ and 161 g of H₂O. The compound's molecular mass, as determined by mass spectrometry, is 56.1072 amu. What is the empirical formula of the compound?

User Hatef
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Final answer:

By analyzing the mass of CO₂ and H₂O produced from combustion and using their respective molar masses, we calculate the molar amounts of carbon and hydrogen in the compound, leading us to an empirical formula of CH. The molecular mass provided suggests a multiple of this empirical formula to determine the actual molecular formula.

Step-by-step explanation:

To find the empirical formula of the compound, we start by calculating the molar amounts of carbon and hydrogen. For carbon, we can determine the number of moles from the mass of CO₂ produced by using the molar mass of CO₂ (44.01 g/mol). Since each molecule of CO₂ contains one atom of carbon, the moles of carbon are equal to the moles of CO₂:

392 g CO₂ × (1 mol CO₂ / 44.01 g CO₂) = 8.91 mol C

For hydrogen, we calculate the number of moles from the mass of H₂O produced by using the molar mass of H₂O (18.02 g/mol). Each molecule of H₂O contains two atoms of hydrogen, so the moles of hydrogen are twice the moles of H₂O:

161 g H₂O × (1 mol H₂O / 18.02 g H₂O) = 8.94 mol H

We then get the empirical ratio by dividing these amounts by the smallest value. The ratio for C: H is 8.91:8.94, which simplifies to approximately 1:1. Since both elements have nearly the same molar amount and we have a 1:1 ratio, the empirical formula is CH.

The ratio of moles of carbon to moles of hydrogen gives us that the empirical formula of the compound is CH, with each element having an approximately equal number of moles.

The molecular mass given is 56.1072 amu. This means the compound's molecular formula may require multiple units of the empirical formula. The ratio of the molecular mass to the empirical formula mass (approximately 13 g/mol for CH) gives the number of empirical units in the molecular formula. A molecular mass of 56.1072 amu suggests there may be four units of CH (4 x 13 g/mol), leading us to a possible molecular formula of C4H4. However, an exact ratio and molecular formula determination would require further calculations based on the molecular mass provided.

User Soloturn
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