A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis. Combustion of a 50.86-mg sample yielded 150.0 mg CO2 and 46.05 mg H2O. Analysis for nitrogen showed that the compound contained 9.39% N by mass. The formula of cocaine is C17H21NO4. Can the forensic chemist conclude that the suspected compound is cocaine?
It says 21 moles of Hydrogen , not 2 !
In the cocaine formula for every N (14 g/mole) atom here are 17 C (12g) atoms and 2 H (1g) atoms
Therefore thee ratios of masses of those three elements must be:
N -- 14
C --17*12 = 204
H --2*1 =2
Now how many moles of C in 150 g CO2?
(I am going to use grams instead of milligrams for everything because we are only interested in ratios in the end)
mole of CO2 = 12 +32 = 44 g/mole
so we have 150/44 = 3.41 moles CO2
and therefore 3.41 moles of C
which is 3.41 * 12 = 40.9 grams of C
Now how many moles of H in 46.05 g of H2O?
mole of H2O = 2+16 = 18 g/mole
so we have 46.05/8 = 2.56 moles H2O
and thus 2*2.56 = 5.12 moles of H
Now check ratios of either moles or grams
moles C/moles H = 3.41/5.12 = .666
but we know that we need 17 moles C to every 2 moles H if it is cocaine!
check my arithmetic and do more similar analysis for Nitrogen as well
46/18 not 46/8
The calculation is right above but I typed 8 instead of 18
In order to determine whether the suspected compound is cocaine, we need to compare the results obtained from the elemental analysis with the expected values based on the formula of cocaine (C17H21NO4).
Let's analyze the given information step by step to see if we can draw a conclusion:
1. Combustion of a 50.86-mg sample yielded 150.0 mg CO2 and 46.05 mg H2O.
This information tells us that the carbon (C) and hydrogen (H) content of the compound can be calculated by comparing the masses of CO2 and H2O produced during combustion.
To calculate the moles of carbon and hydrogen, we need to convert the masses of CO2 and H2O to moles using their respective molar masses.
- Moles of CO2 = mass of CO2 / molar mass of CO2 = 150.0 mg / 44.01 g/mol = 0.00340 mol CO2
- Moles of H2O = mass of H2O / molar mass of H2O = 46.05 mg / 18.02 g/mol = 0.00256 mol H2O
From the balanced equation of combustion, we know that 1 mole of CO2 is produced for each mole of carbon burned and 1 mole of H2O is produced for each 2 moles of hydrogen burned. Therefore:
- Moles of C = 0.00340 mol CO2
- Moles of H = 2 × 0.00256 mol H2O = 0.00512 mol H
2. Analysis for nitrogen showed that the compound contained 9.39% N by mass.
This information tells us the nitrogen (N) content of the compound, which is given as a percentage by mass. To calculate the moles of nitrogen, we need to convert the mass percentage to mass and then to moles.
The mass of nitrogen can be calculated as follows:
Mass of N = total mass × percentage of N = 50.86 mg × 0.0939 ≈ 4.77 mg
To convert the mass of N to moles, we use the molar mass of nitrogen:
- Moles of N = mass of N / molar mass of N = 4.77 mg / 14.01 g/mol ≈ 0.341 mol N
Now, let's compare the obtained mole ratios with those expected from the formula of cocaine:
- C17H21NO4: 17 moles of C, 21 moles of H, 1 mole of N
The mole ratios obtained from the elemental analysis are as follows:
- Moles of C = 0.00340 mol
- Moles of H = 0.00512 mol
- Moles of N = 0.341 mol
By comparing the obtained mole ratios with those expected from the formula of cocaine, we can see that the mole ratio of nitrogen is significantly higher than expected. Therefore, the forensic chemist cannot conclude that the suspected compound is cocaine based on the elemental analysis alone.
It is important to note that this analysis assumes that all the nitrogen in the compound originates from the cocaine molecule. However, other compounds containing nitrogen may also contribute to the measured value. To further confirm the identity of the compound, additional tests or analyses may be necessary.