A sample of a pure compound contains 68.4% chromium and 31.6% oxygen by mass. What is the empirical formula of this compound? (numbers entered will be considered subscripts)
The correct answer and explanation is :
To determine the empirical formula of a compound based on its mass percent composition, follow these steps:
1. Convert the mass percentages to moles.
We are given that the compound consists of 68.4% chromium (Cr) and 31.6% oxygen (O). We will assume we have 100 grams of the compound, making the mass of chromium 68.4 grams and the mass of oxygen 31.6 grams.
For Chromium (Cr):
- The molar mass of chromium is approximately 52.00 g/mol.
- Moles of Cr = ( \frac{68.4 \, \text{g}}{52.00 \, \text{g/mol}} = 1.315 \, \text{mol} ).
For Oxygen (O):
- The molar mass of oxygen is approximately 16.00 g/mol.
- Moles of O = ( \frac{31.6 \, \text{g}}{16.00 \, \text{g/mol}} = 1.975 \, \text{mol} ).
2. Find the simplest ratio of moles.
Next, divide each of the moles of elements by the smaller number of moles (in this case, the moles of chromium):
- Ratio of Cr = ( \frac{1.315}{1.315} = 1 )
- Ratio of O = ( \frac{1.975}{1.315} = 1.5 )
3. Adjust to whole numbers.
The ratio of chromium to oxygen is 1:1.5. To eliminate the decimal, multiply both ratios by 2:
- Cr = 1 × 2 = 2
- O = 1.5 × 2 = 3
4. Write the empirical formula.
The simplest whole number ratio of chromium to oxygen is 2:3. Thus, the empirical formula of the compound is:
Cr₂O₃
Explanation:
The empirical formula represents the simplest whole number ratio of the elements in the compound. In this case, we started by converting the mass percentages of chromium and oxygen to moles, then determined their ratio. After adjusting to whole numbers, we found that the simplest ratio was 2:3, corresponding to the empirical formula Cr₂O₃. This formula indicates that the compound contains two chromium atoms for every three oxygen atoms.