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% C, 5.19% H, 28.85% N, and 16.48% O. What is the empirical formula for this substance? Is it cocaine?
Assume 100 grams
C=49.48grams=4.12 moles
H=5.19grams=5.19 moles H
N=28.85grams=2 moles
O-16.48grams-1 mole
Noticed I rounded on these, you need to do it more accurately. Now, divide all by the lowest mole number, in this case 1, so the numbers are the same.
Empirical formula
C4H5N2O
Nope, not Coke. no common multiplier to those subscripts for the actual compound will give C17H21NO4
Well, if you have a white substance that's suspected to be cocaine, I must say, it's a bit of a "snowy" situation, isn't it?
Now, let's get down to the chemical nitty-gritty. To determine the empirical formula, we need to know the simplest whole number ratio of atoms present in the substance. So, let's crunch some numbers!
Given the percent composition of 49.48% C, 5.19% H, 28.85% N, and 16.48% O, we can assume we have 100 grams of this substance.
Converting the percentages to grams, we find that we have approximately 49.48 grams of carbon, 5.19 grams of hydrogen, 28.85 grams of nitrogen, and 16.48 grams of oxygen.
Next, let's find the moles of each element by dividing the grams by their respective atomic masses:
Carbon (C): 49.48 g / 12.01 g/mol ≈ 4.12 mol
Hydrogen (H): 5.19 g / 1.01 g/mol ≈ 5.14 mol
Nitrogen (N): 28.85 g / 14.01 g/mol ≈ 2.06 mol
Oxygen (O): 16.48 g / 16.00 g/mol ≈ 1.03 mol
So, looking at these ratios (approximately), we have C4.12H5.14N2.06O1.03. However, we want the simplest whole number ratio. So, dividing all mole values by the smallest, which is approximately 1.03 mol, we get:
C4.12H5.14N2.06O1.03 ≈ C4H5N2O
Based on these calculations, the empirical formula of this substance is C4H5N2O. But, remember, this is just an empirical formula. To confirm if it's cocaine, further analysis would be needed.
So, unless that white substance has been making some questionable life choices, it could indeed be cocaine. But don't jump to conclusions just yet! A professional forensic scientist will have to confirm that.
To determine the empirical formula of the substance, we need to find the simplest whole number ratio of atoms in the compound.
1. Start by assuming we have 100 grams of the substance. This allows us to conveniently consider the percentages as grams.
2. Convert the mass of each element into moles using their molar masses:
- Carbon (C): (49.48 g C) / (12.01 g/mol C) = 4.12 moles C
- Hydrogen (H): (5.19 g H) / (1.008 g/mol H) = 5.15 moles H
- Nitrogen (N): (28.85 g N) / (14.01 g/mol N) = 2.06 moles N
- Oxygen (O): (16.48 g O) / (16.00 g/mol O) = 1.03 moles O
3. Divide each of the mole values by the smallest mole value to obtain the simplest whole number ratio:
- C: 4.12 moles C / 2.06 moles H = 2
- H: 5.15 moles H / 2.06 moles H = 2.5 ≈ 2
- N: 2.06 moles N / 2.06 moles H = 1
- O: 1.03 moles O / 2.06 moles H = 0.5 ≈ 1/2
Therefore, the empirical formula for the substance is C2H2N1/2O1, which simplifies to C4H4NO.
To determine if it is cocaine, we need additional information such as its molecular mass and further confirmation through other analytical methods.
To determine the empirical formula for the white substance, you need to find the simplest whole number ratio of the elements present in it. Here's how you can go about solving this:
1. Convert the percentages to grams: Assume you have 100 grams of the substance, so you will have 49.48 grams of carbon (C), 5.19 grams of hydrogen (H), 28.85 grams of nitrogen (N), and 16.48 grams of oxygen (O).
2. Convert the grams to moles: Use the atomic masses of each element to convert grams to moles. The atomic masses are approximately 12 g/mol for carbon (C), 1 g/mol for hydrogen (H), 14 g/mol for nitrogen (N), and 16 g/mol for oxygen (O).
Therefore, you have:
C: 49.48 g / 12 g/mol ≈ 4.12 mol
H: 5.19 g / 1 g/mol ≈ 5.19 mol
N: 28.85 g / 14 g/mol ≈ 2.06 mol
O: 16.48 g / 16 g/mol ≈ 1.03 mol
3. Divide the number of moles by the smallest number of moles: Divide each number of moles by the smallest number of moles, which is 1.03 mol in this case.
C: 4.12 mol / 1.03 mol ≈ 4
H: 5.19 mol / 1.03 mol ≈ 5
N: 2.06 mol / 1.03 mol ≈ 2
O: 1.03 mol / 1.03 mol = 1
4. Check if the resulting empirical formula makes sense. In this case, the empirical formula is C4H5N2O. To determine if it corresponds to cocaine, you can compare it to the molecular formula of cocaine.
The empirical formula C4H5N2O does not match the molecular formula of cocaine, which is C17H21NO4. Therefore, the substance given is not cocaine.