This a tough one folks here's the entire prompt:
Nicotine, the addictive ingredient in tobacco, contains the elements carbon, hydrogen, and nitrogen. When a 1.567g sample of nicotine is burned completely in oxygen, it produces 4.251g CO2 and 1.216g H20. What is the empirical formula for nicotine and the percent by mass of each element in the sample?
Check back tomorrow. I'll take care of this tomorrow morning. I promise.
First things first. Nicotine is composed of C, H, and N. You have data for C and H but nothing for N. But that can be determined.
Convert CO2 to g C, H2O to grams H. I will round my numbers but you should go through with a little more care. Note that mm stands for molar mass.
g C = 4.251 g C x (mm C/mmCO2) = 4.251 x (12/44) = about 1.159
g H = 1.216 g H2O x (2H/H2O) = 1.216 x (2/18) = about 0.1351
g N = g sample - g C - g H - 1.567- gC - gH = about 0.273
Now convert those g C, H, N to mols.
mols C = 1.159/12 = about 0.0965
mols H = 0.135/1 = 0.135
mols N = 0.273/14 = about 0.0195
Divide mols C, mols H, and mols N by the smallest number. I get
C = 4.9 which rounds to 5
H = 6.9 which rounds to 7
N = 1
So C5H7N is the empirical formula.
To find percent of each element. For example for C it is
%C = 100*(atomic mass C*12/mm C5H7N)
same process for H and N.
Post your work if you get stuck.
To determine the empirical formula of nicotine and the percent by mass of each element in the sample, we need to go through several steps. Here's how you can solve this problem:
Step 1: Convert the given mass of each compound to moles.
First, we need to convert the masses of CO2 and H2O into moles using their respective molar masses.
The molar mass of CO2 is calculated as follows:
12.01 g/mol (C) + 2 * 16.00 g/mol (O) = 44.01 g/mol
The molar mass of H2O is calculated as follows:
2 * 1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol
We can now calculate the moles of CO2 and H2O produced:
moles of CO2 = 4.251 g / 44.01 g/mol = 0.0964 mol
moles of H2O = 1.216 g / 18.02 g/mol = 0.0676 mol
Step 2: Calculate the moles of carbon (C), hydrogen (H), and nitrogen (N).
The nicotine molecule contains C, H, and N atoms. We need to find the moles of each element in the given sample.
From the balanced chemical equation (combustion of nicotine), we know that each mole of nicotine produces one mole of CO2 and one mole of H2O.
moles of C = moles of CO2 = 0.0964 mol
moles of H = (moles of H2O) * 2 = 0.0676 mol * 2 = 0.1352 mol
moles of N = moles of nicotine - (moles of C + moles of H) = 1 - (0.0964 + 0.1352) = 0.7684 mol
Step 3: Find the simplest whole number ratio.
To determine the empirical formula, we need to express the moles as a ratio of small whole numbers.
Divide each of the moles by the smallest of them (moles of C in this case).
moles of C = 0.0964 mol / 0.0964 mol = 1
moles of H = 0.1352 mol / 0.0964 mol = 1.4
moles of N = 0.7684 mol / 0.0964 mol = 7.96
The ratio of C : H : N is approximately 1 : 1.4 : 7.96.
To simplify this ratio, we can multiply each number by 5 to convert 1.4 to a whole number:
5 * C : 5 * H : 5 * N = 5 : 7 : 39
Therefore, the empirical formula of nicotine is C5H7N.
Step 4: Calculate the percent by mass of each element in the sample.
To calculate the percent by mass, we need to determine the mass of each element in the sample and divide by the total molar mass of nicotine.
The molar mass of nicotine is calculated as follows:
(5 * 12.01 g/mol) + (7 * 1.01 g/mol) + (1 * 14.01 g/mol) = 162.19 g/mol
The percent by mass of each element can be determined as follows:
Percent of C = (5 * 12.01 g/mol) / 162.19 g/mol * 100% = 36.99%
Percent of H = (7 * 1.01 g/mol) / 162.19 g/mol * 100% = 4.36%
Percent of N = (1 * 14.01 g/mol) / 162.19 g/mol * 100% = 8.63%
Therefore, the percent by mass of each element in the nicotine sample is approximately 36.99% carbon, 4.36% hydrogen, and 8.63% nitrogen.