An analysis of nicotine (MW = 162 g/mol)
gives 74.0% carbon, 8.65% hydrogen, and
17.3% nitrogen. What is the TRUE
(MOLECULAR) formula for nicotine?
All I need are the formulas!!
C10H14N2
To determine the molecular formula of nicotine, we need to find the empirical formula first.
1. Assume we have a 100g sample of nicotine. This means we have:
- 74.0g of carbon (74.0% of 100g)
- 8.65g of hydrogen (8.65% of 100g)
- 17.3g of nitrogen (17.3% of 100g)
2. Convert the masses of each element to moles:
- Moles of carbon = mass of carbon / molar mass of carbon = 74.0g / 12.01 g/mol = 6.164 moles of carbon
- Moles of hydrogen = mass of hydrogen / molar mass of hydrogen = 8.65g / 1.008 g/mol = 8.581 moles of hydrogen
- Moles of nitrogen = mass of nitrogen / molar mass of nitrogen = 17.3g / 14.01 g/mol = 1.235 moles of nitrogen
3. Divide each mole value by the smallest mole value:
- Divide carbon moles by 1.235: 6.164 moles / 1.235 moles = 4.993
- Divide hydrogen moles by 1.235: 8.581 moles / 1.235 moles = 6.950
- Divide nitrogen moles by 1.235: 1.235 moles / 1.235 moles = 1.000
4. Round each value to the nearest whole number:
- Carbon: 5
- Hydrogen: 7
- Nitrogen: 1
The empirical formula of nicotine is C5H7N.
Now, to find the true molecular formula, we need to know the molecular mass of nicotine. Given that the molecular weight of nicotine is 162 g/mol, we can compare it to the empirical formula.
The empirical formula mass is:
(5 * molar mass of carbon) + (7 * molar mass of hydrogen) + (1 * molar mass of nitrogen) = (5 * 12.01) + (7 * 1.008) + (1 * 14.01) = 82.14 g/mol.
To find the true molecular formula, we divide the molecular weight of nicotine by the empirical formula mass:
162 g/mol / 82.14 g/mol = 1.97
Since 1.97 is close to 2, it means that the true molecular formula is approximately twice the empirical formula.
Therefore, the true molecular formula of nicotine is C10H14N2.
To determine the true molecular formula of nicotine, we need to calculate the empirical formula first. The empirical formula represents the simplest whole-number ratio of the elements in a compound.
To calculate the empirical formula, we can assume we have 100 grams of nicotine to make the percentage calculations easier.
1. Calculate the mass of carbon (C):
Mass of C = 100 g * 0.74 = 74 g
2. Calculate the mass of hydrogen (H):
Mass of H = 100 g * 0.0865 = 8.65 g
3. Calculate the mass of nitrogen (N):
Mass of N = 100 g * 0.173 = 17.3 g
4. Convert the masses of each element to moles by dividing by their respective molar masses:
Moles of C = 74 g / 12.01 g/mol = 6.16 mol
Moles of H = 8.65 g / 1.008 g/mol = 8.59 mol
Moles of N = 17.3 g / 14.01 g/mol = 1.23 mol
5. Find the smallest mol value, which in this case is 1.23 mol (for N).
Divide the other moles by this small value:
Moles of C (rounded) = 6.16 mol / 1.23 mol = 5
Moles of H (rounded) = 8.59 mol / 1.23 mol = 7
6. The empirical formula is then determined by combining the elements with their respective ratios:
Empirical formula = C5H7N
Now that we have the empirical formula, we need to determine the molecular formula. The molecular formula represents the actual number of atoms of each element in a molecule.
To find the molecular formula, we need the molar mass of nicotine, which is given as 162 g/mol. We need to compare this with the empirical formula mass.
7. Calculate the empirical formula mass:
Empirical formula mass = (5 * molar mass of C) + (7 * molar mass of H) + (1 * molar mass of N)
Empirical formula mass = (5 * 12.01 g/mol) + (7 * 1.008 g/mol) + (1 * 14.01 g/mol)
Empirical formula mass = 92.067 g/mol
8. Determine the ratio of the molecular mass to the empirical formula mass to find the integer multiple:
Molecular mass / Empirical formula mass = 162 g/mol / 92.067 g/mol ≈ 1.759
9. Round the ratio to the nearest whole number (since we cannot have a fraction of atoms):
Ratio (rounded) = 2
10. Multiply the subscripts of the empirical formula by the ratio:
Molecular formula = (C5H7N) * 2 = C10H14N
Therefore, the TRUE (MOLECULAR) formula for nicotine is C10H14N.
Note this doesn't add to 100% but I'll assume it does. I don't think that will change the formula.
Take 100 g sample to give you
74.0 g C
8.65 g H
17.3 g N
Convert to mols.
74/12 = ?
8.65/1 = ?
17.3/14 = ?
Find the ratio of these numbers with the smallest being 1.00. The easy way to do that is to divide the smallest number by itself (which automatically gives you 1.00), then divide the other two numbers by the smallest number. This gives you the empirical formula.
Then determine the weight of the empirical formula.
molecular formula = (CxHyNz)n.
Your job then is to determine n. Here is how you get that.
(162/empirical formula) = ? and round to a whole number.