1. Nicotine,a poisonous compound found in tobacco leaves,is 74% carbon,8,65% hydrogen and 17,35% nitrogen (a)calculate the empirical formula of nicotine (b)what is the molecular formula of nicotine if it has a molar mass of 165 g.mol^-1
Take a 100 g sample which will give you
74 g C
8.65 g H
17.35 g N
Now, mols = g/atomic mass.
Convert these to mols and you redo these because I will estimate.
74/12 = about 6.2
8.65/1 = 8.65
17.35/14 = 1.24
Now find the ratio of these numbers to one another with the smallest being 1.0. The easy way to do that is to divide all of them by the smallest number in the lot. That is 1.24; therefore,
6.2/1.24 = about 5
8.65/1.24 = about 7
1.24/1,24 = 1.00 (because we made it that way).
So the empirical formula is C5H7N.
What is the empirical formula mass? That's (5*12) + (7*1) + (1*14) = about 81
The molar mass is 165; we want to know how may of the "empirical formula pieces" fill fit within that 165. That's simply 165/81 = 2.04. We round to a whole number of 2.00 so the molecular formula is (C5H7N)2 since two of the empirical formulas make up 1 molecular formula. The problem didn't ask but you can find the "real" molar mass (the 165 is an estimate) by adding the atomic masses together. You can do it but I found 81*2 = 162. (The molar mass is determined by boiling point elevation or freezing point depression. Those methods give good estimates of the molar mass but there is some error involved in the "chemistry" of the process due to interactions that the formulas do not take into account. So they are useful for problems like this and that's why we can round the answer to a whole number as above.
Thank u
(a) To calculate the empirical formula of nicotine, we need to find the simplest whole number ratio of the elements present in the compound.
First, we can assume we have 100 grams of nicotine.
Given that nicotine is 74% carbon, we have 74 grams of carbon.
Given that nicotine is 8.65% hydrogen, we have 8.65 grams of hydrogen.
Given that nicotine is 17.35% nitrogen, we have 17.35 grams of nitrogen.
Now, let's convert these masses into moles using their respective molar masses:
Molar mass of carbon (C): 12.01 g/mol
Molar mass of hydrogen (H): 1.008 g/mol
Molar mass of nitrogen (N): 14.01 g/mol
Number of moles of carbon = 74 g / 12.01 g/mol ≈ 6.16 mol
Number of moles of hydrogen = 8.65 g / 1.008 g/mol ≈ 8.58 mol
Number of moles of nitrogen = 17.35 g / 14.01 g/mol ≈ 1.24 mol
Next, divide each of the mole values by the smallest number of moles (in this case, 1.24 mol):
Carbon: 6.16 mol / 1.24 mol ≈ 4.97 (approximately 5)
Hydrogen: 8.58 mol / 1.24 mol ≈ 6.93 (approximately 7)
Nitrogen: 1.24 mol / 1.24 mol = 1
Thus, the empirical formula of nicotine is C5H7N.
(b) To find the molecular formula of nicotine, we need to know its molar mass.
Given that the molar mass of nicotine is 165 g/mol, we can compare it to the empirical formula's molar mass.
The molar mass of one carbon atom (C) is 12.01 g/mol.
The molar mass of one hydrogen atom (H) is 1.008 g/mol.
The molar mass of one nitrogen atom (N) is 14.01 g/mol.
Now, let's calculate the molar mass of the empirical formula:
(5 * 12.01 g/mol) + (7 * 1.008 g/mol) + (1 * 14.01 g/mol) = 72.09 g/mol + 7.056 g/mol + 14.01 g/mol ≈ 93.16 g/mol
To find the number of empirical formula units in the molecular formula, divide the molar mass of the molecular formula by the molar mass of the empirical formula:
165 g/mol / 93.16 g/mol ≈ 1.77
Since we cannot have a fractional number of molecules, we need to round it to the nearest whole number. In this case, it would be 2.
Therefore, the molecular formula of nicotine is approximately (C5H7N)2, which means 2 times the empirical formula.
To calculate the empirical formula of nicotine, we need to find the ratio between the number of moles of each element present.
(a) To determine the empirical formula, we assume we have 100 grams of nicotine. Then we can calculate the number of moles of each element using their respective percentages.
Given:
Carbon (C): 74%
Hydrogen (H): 8.65%
Nitrogen (N): 17.35%
First, convert the percentages to grams of each element:
Carbon (C): 74g
Hydrogen (H): 8.65g
Nitrogen (N): 17.35g
Next, find the number of moles of each element:
Moles of Carbon (C) = 74g / Atomic mass of Carbon (12 g/mol) = 6.17 mol
Moles of Hydrogen (H) = 8.65g / Atomic mass of Hydrogen (1 g/mol) = 8.65 mol
Moles of Nitrogen (N) = 17.35g / Atomic mass of Nitrogen (14 g/mol) = 1.24 mol
Divide each mole value by the smallest mole value (in this case, nitrogen) to get the simplest whole number ratio:
Moles of Carbon (C) = 6.17 mol / 1.24 mol = 4.97 ≈ 5
Moles of Hydrogen (H) = 8.65 mol / 1.24 mol = 6.97 ≈ 7
Moles of Nitrogen (N) = 1.24 mol / 1.24 mol = 1
Therefore, the empirical formula of nicotine is C5H7N.
(b) To find the molecular formula of nicotine, we need to know its molar mass. In the question, we are given that the molar mass of nicotine is 165 g/mol.
The empirical formula mass of C5H7N can be calculated as follows:
C5H7N = (5 * atomic mass of Carbon) + (7 * atomic mass of Hydrogen) + (1 * atomic mass of Nitrogen)
C5H7N = (5 * 12 g/mol) + (7 * 1 g/mol) + (1 * 14 g/mol)
C5H7N = 60 g/mol + 7 g/mol + 14 g/mol
C5H7N = 81 g/mol
To find the molecular formula from the empirical formula, divide the molar mass of the compound by the empirical formula mass:
Molar mass of nicotine / Empirical formula mass of nicotine = 165 g/mol / 81 g/mol
This calculation yields approximately 2.04. Since we need a whole number, we round this to the nearest whole number, which is 2.
Finally, multiply the subscripts in the empirical formula by this whole number to get the molecular formula of nicotine:
Molecular formula = C5H7N * 2
Molecular formula = C10H14N
Therefore, the molecular formula of nicotine is C10H14N.