A forensic scientist is given a white substance that is thought to be cocaine. The substance is found to have a percent composition: 49.48% H, 5.19% H, 28.85% N, and 16.48% O. What is the empirical formula for this substance? Is it cocaine?
To determine the empirical formula of the substance, we need to find the ratio of the different elements present in it. The percent composition given tells us the mass percent of each element.
To start, we can assume we have 100 grams of the substance to make calculations easier. This means we have:
- Hydrogen (H): 49.48 grams
- Carbon (C): 5.19 grams
- Nitrogen (N): 28.85 grams
- Oxygen (O): 16.48 grams
Next, we need to convert these masses to moles. To do this, we divide the mass of each element by its molar mass (g/mol).
The molar masses of the elements are approximately:
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol
Converting the masses to moles, we get:
- Hydrogen (H): 49.48 g / 1 g/mol = 49.48 mol
- Carbon (C): 5.19 g / 12 g/mol ≈ 0.433 mol
- Nitrogen (N): 28.85 g / 14 g/mol ≈ 2.061 mol
- Oxygen (O): 16.48 g / 16 g/mol ≈ 1.030 mol
Now, we need to find the simplest whole-number ratio of these moles to determine the empirical formula. To do this, we divide each mole value by the smallest mole value (which is approximately 0.433 mol in this case).
Dividing the moles by 0.433 mol, we get:
- Hydrogen (H): 49.48 mol / 0.433 mol ≈ 114
- Carbon (C): 0.433 mol / 0.433 mol = 1
- Nitrogen (N): 2.061 mol / 0.433 mol ≈ 5
- Oxygen (O): 1.030 mol / 0.433 mol ≈ 2.38
Since we are dealing with whole numbers for the empirical formula, we need to round the results to the nearest whole number.
Rounding off, we have:
- Hydrogen (H): 114
- Carbon (C): 1
- Nitrogen (N): 5
- Oxygen (O): 2
Therefore, the empirical formula for the substance is C1H114N5O2.
To determine if this substance is cocaine, we need to compare its empirical formula to the actual empirical formula of cocaine.
The empirical formula of cocaine is C17H21NO4, which is quite different from C1H114N5O2. Hence, based on the given information, we can conclude that the substance is not cocaine.