Give the empirical formula of a hydrocarbon that is 52.14% c, 13.13% h, and 34.73% o by mass. Write the formula with the atoms in the order cxhyoz . You can ignore the charge and phase.
The Correct Answer and Explanation is:
To determine the empirical formula of a hydrocarbon that is 52.14% carbon (C), 13.13% hydrogen (H), and 34.73% oxygen (O) by mass, we start by converting these percentages into moles of each element.
Step 1: Convert mass percentages to moles
Assuming we have 100 grams of the compound:
- Carbon (C): ( 52.14 \, \text{g} \div 12.01 \, \text{g/mol} = 4.34 \, \text{moles} )
- Hydrogen (H): ( 13.13 \, \text{g} \div 1.008 \, \text{g/mol} = 13.02 \, \text{moles} )
- Oxygen (O): ( 34.73 \, \text{g} \div 16.00 \, \text{g/mol} = 2.17 \, \text{moles} )
Step 2: Determine the mole ratio
Next, we find the simplest whole-number ratio by dividing each mole value by the smallest value among them:
- Carbon: ( 4.34 \, \text{moles} \div 2.17 = 2 )
- Hydrogen: ( 13.02 \, \text{moles} \div 2.17 = 6 )
- Oxygen: ( 2.17 \, \text{moles} \div 2.17 = 1 )
Step 3: Write the empirical formula
The simplest whole-number ratio of C, H, and O is 2:6:1. Therefore, the empirical formula is:
[
\text{C}_2\text{H}_6\text{O}
]
Explanation
An empirical formula represents the simplest whole-number ratio of atoms in a compound, providing insight into its elemental composition without specifying the exact molecular structure. This compound’s composition—52.14% C, 13.13% H, and 34.73% O—corresponds to a formula of ( \text{C}_2\text{H}_6\text{O} ). This is consistent with molecules like ethanol (( \text{C}_2\text{H}_6\text{O} )), a common hydrocarbon derivative. Ethanol’s structure has a two-to-six-to-one atomic ratio of carbon, hydrogen, and oxygen, fitting the calculated values precisely. This empirical formula is an important first step in characterizing unknown compounds because it reveals essential details about the molecule’s possible identity and further investigation.