Determine the empirical formula for a compound that contains C, H, and O. It contains 52.14% C and 34.73% O by mass.
The Correct Answer and Explanation is :
To determine the empirical formula of a compound containing carbon (C), hydrogen (H), and oxygen (O) with the given mass percentages, we can follow these steps:
- Convert percentages to grams: Assume we have 100 grams of the compound. This means we have 52.14 grams of carbon and 34.73 grams of oxygen. Since the total percentage of C and O is 86.87% (52.14% + 34.73%), we can find the mass of hydrogen by subtracting this from 100 grams: [
\text{Mass of H} = 100 \, \text{g} – (52.14 \, \text{g} + 34.73 \, \text{g}) = 13.13 \, \text{g}
] - Convert grams to moles: We convert the masses of C, H, and O to moles using their respective molar masses (C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol): [
\text{Moles of C} = \frac{52.14 \, \text{g}}{12.01 \, \text{g/mol}} \approx 4.34 \, \text{mol}
]
[
\text{Moles of H} = \frac{13.13 \, \text{g}}{1.008 \, \text{g/mol}} \approx 13.00 \, \text{mol}
]
[
\text{Moles of O} = \frac{34.73 \, \text{g}}{16.00 \, \text{g/mol}} \approx 2.17 \, \text{mol}
] - Find the simplest mole ratio: Next, we divide each value by the smallest number of moles calculated (2.17): [
\text{C} : \frac{4.34}{2.17} \approx 2.00
]
[
\text{H} : \frac{13.00}{2.17} \approx 6.00
]
[
\text{O} : \frac{2.17}{2.17} \approx 1.00
] - Write the empirical formula: The mole ratio of C, H, and O is approximately 2:6:1. Therefore, the empirical formula is: [
\text{C}_2\text{H}_6\text{O}
]
Explanation
The empirical formula represents the simplest whole-number ratio of atoms in a compound. In this case, we started with the mass percentages of carbon and oxygen, calculated the mass of hydrogen, and converted all masses to moles. By identifying the smallest mole ratio, we could establish a clear and concise formula. The empirical formula C₂H₆O is significant as it provides insight into the molecular structure and possible functional groups in the compound, indicating it might be an alcohol or ether based on its composition. This method is essential in chemistry for analyzing compounds and understanding their properties and reactions.