What is the empirical formula for nicotine?

Question

Nicotine, a component of tobacco, is composed of C, H, and N. A 5.775 mg sample of nicotine was combusted, producing 15.666 mg of CO2 and 4.491 mg of H2O. What is the empirical formula for nicotine?

how did you get those first numbers irishkar?

Answer 1

ok, you first figure out how much C is present

15.666 mg CO2 x 12/44 (MW C / MW CO2) = 4.275 mg C
4.491 mg H2O x 2/18 (MW H / MW H2O) = 0.5 mg H
5.775 mg sample - 3.454 mg C - 0.5 mg H = 1 mg N

4.275mg / 12 g/mol = 0.356 millimoles C
0.5 mg / 1 g/mol = 0.500 millimoles H
1.0 mg / 14 g/mol = 0.071 millimoles N

since you have the least amount of N, bring that to be equal to 1 by multiplying each by diving by 71 and multiplying by 1000.

this gives you the ratio of

C 5
H 7
N 1

the actual structure of nicotine is C10H14N2, so this checks out

Answer 2

Nicotine Chemical Formula

Answer 3

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I think you have a typo. Mass of nicotine should be 0.00525 g. The law of conservation of mass tells you that in a chemical reaction you cannot create or destroy matter, so there must be the same number of C, H, and N atoms present in the products as there were in the reactants. All of the C from the nicotine reacts to produce CO2. So you can work out how many moles and what mass of C was in the original sample from the mass of CO2 produced. moles = mass / molar mass moles CO2 = 0.014242 g / 44.01 g/mol moles CO2 = 3.236x10^-4 moles CO2 Each CO2 has 1 C in it Therefore moles C = moles CO2 moles C = 3.236x10^-4 moles C This is how many moles of C was in the 0.000525 g of nicotine mass = moles x molar mass mass C = 3.236x10^-4 x 12.01 g/mol mass C = 0.003887 g C in the nicotine All the H from the nicotine ends up in the H2O moles H2O = mass / molar mass moles H2O = 0.004083 g / 18.016 g/mol = 2.2663x10^-4 moles H2O Each H2O has 2 moles H in it Therefore moles H = 2 x 2.2663x10^-4 moles moles H = 4.533x10^-4 moles This is how many moles of H were in the nicotine mass H = molar mass x moles mass H = 1.008 g/mol x 4.533x10^-4 moles mass H = 0.000457 g The mass of N in the nicotine = total mass - mass C - mass H mass N = 0.00525 g - 0.003887 g - 0.000457 g mass N = 0.000906 g moles N = mass / molar mass moles N = 0.000906 g / 14.01 g/mol = 6.467x10^-5 moles Now you know all number of moles of each atom, write them out in ratio format moles C : H : N = 3.236x10^-4 moles C : 4.533x10^-4 moles H : 6.467x10^-5 moles N divide each number by the lowest number in the ratio C : H : N (3.236x10^-4 / 6.467x10^-5) : (4.533x10^-4 / 6.467x10^-5) : (6.467x10^-5 /6.467x10^-5) = 5 : 7 : 1 empirical formula is thus C5H7N Now, you can work out the molecular formula by determining how many times the empirical formula fits into the molecular weight by dividing the molecular weight by the mass of the empirical formula Emirical mass = [5 x 12.1] + [7 x 1.008] + 14.1 = 81.1 g/mol molecular mass / emirical mass = 160 / 81 = 2 So the emirical formula fits in twice multily the emirical formula by to to give the molecular formula 2 x [C5H7N] gives molecular formula C10H14N2

Answer 4

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RE:
What is the empirical formula for nicotine?
Nicotine, a component of tobacco, is composed of C, H, and N. A 5.775 mg sample of nicotine was combusted, producing 15.666 mg of CO2 and 4.491 mg of H2O. What is the empirical formula for nicotine?

Answer 5

Nicotine has an empirical formula of C5H7N and a molecular formula of C10H14N2.

Answer 6

What is the empirical formula of nicotine? Change % to grams, convert to moles, divide by smallest

C 74.03 g/(12.011 g/mol) = 6.16 mol; 6.16/1.233 = 5
H 8.70 g/(1.008 g/mol) = 8.63 mol; 8.63/1.233 = 7
N 17.27 g/(14.01 g/mol) = 1.233 mol; 1.233/1.233 = 1

C5H7N ANSWER