a sample of compound C,H,N and O with a mass of 0.6216 was found to contain 0.1735g C,0.01455g H,0.2024g N.Its molecular mass 129.calculate its epirical and molecular formulas.
To determine the empirical formula of the compound, we first need to calculate the number of moles of each element present in the compound.
1. Calculate the number of moles for each element:
- Moles of carbon (C) = mass of C / molar mass of C = 0.1735 g / 12.01 g/mol = 0.01445 mol
- Moles of hydrogen (H) = mass of H / molar mass of H = 0.01455 g / 1.008 g/mol = 0.01444 mol
- Moles of nitrogen (N) = mass of N / molar mass of N = 0.2024 g / 14.01 g/mol = 0.01445 mol
2. Divide the number of moles of each element by the smallest number of moles:
- C: 0.01445 mol / 0.01444 mol = 1
- H: 0.01444 mol / 0.01444 mol = 1
- N: 0.01445 mol / 0.01444 mol = 1
The empirical formula of the compound is C₁H₁N₁.
To determine the molecular formula, we need to know the molecular mass of the compound. In this case, it is given as 129 g/mol.
3. Calculate the empirical formula mass:
- Empirical formula mass = (atomic mass of C × number of carbon atoms) + (atomic mass of H × number of hydrogen atoms) + (atomic mass of N × number of nitrogen atoms)
- Empirical formula mass = (12.01 g/mol × 1) + (1.008 g/mol × 1) + (14.01 g/mol × 1) = 27.028 g/mol
4. Calculate the ratio between the molecular mass and the empirical formula mass:
- Ratio = molecular mass / empirical formula mass = 129 g/mol / 27.028 g/mol ≈ 4.77
5. Round the ratio to the nearest whole number to find the number of empirical formula units in the molecular formula:
- Rounded ratio = 5
6. Multiply the subscripts in the empirical formula by the rounded ratio:
- Molecular formula = (C₁H₁N₁) × 5 = C₅H₅N₅
The molecular formula of the compound is C₅H₅N₅.
To calculate the empirical and molecular formulas, we need to follow a step-by-step process. Let's start with the empirical formula.
1. Determine the number of moles of each element:
To find the moles of each element, divide the mass of each element by its molar mass.
Moles of C = 0.1735 g / molar mass of C (12.01 g/mol)
Moles of H = 0.01455 g / molar mass of H (1.008 g/mol)
Moles of N = 0.2024 g / molar mass of N (14.01 g/mol)
Moles of O = (Total mass - sum of masses of C, H, and N) / molar mass of O (16.00 g/mol)
2. Find the mole ratios:
Divide the moles of each element by the smallest number of moles calculated in step 1. This provides the simplest whole-number ratios.
Divide the moles of each element by the smallest value obtained:
C: moles of C / smallest mole value
H: moles of H / smallest mole value
N: moles of N / smallest mole value
O: moles of O / smallest mole value
3. Determine the empirical formula:
The resulting mole ratios from step 2 give us the subscripts for each element to determine the empirical formula.
The empirical formula is represented by the whole number ratios of the elements.
Now, let's go through the calculations.
1. Moles of C:
Moles of C = 0.1735 g / 12.01 g/mol = 0.0145 mol
2. Moles of H:
Moles of H = 0.01455 g / 1.008 g/mol = 0.0144 mol
3. Moles of N:
Moles of N = 0.2024 g / 14.01 g/mol = 0.0145 mol
4. Moles of O:
Moles of O = (0.6216 g - (0.1735 g + 0.01455 g + 0.2024 g)) / 16.00 g/mol
Moles of O = 0.2312 g / 16.00 g/mol = 0.0145 mol
5. Mole Ratios:
C: 0.0145 mol / 0.0144 mol ≈ 1
H: 0.0144 mol / 0.0144 mol = 1
N: 0.0145 mol / 0.0144 mol ≈ 1
O: 0.0145 mol / 0.0144 mol ≈ 1
Based on the above ratios, the empirical formula is C1H1N1O1.
6. Calculate the molecular formula:
To find the molecular formula, we need the molar mass of the compound.
Given: Molecular mass = 129 g/mol
Divide the molecular mass by the empirical formula mass to find the multiplier:
Multiplier = Molecular mass / Empirical formula mass
Empirical formula mass = (molar mass of C × subscripts of C) + (molar mass of H × subscripts of H) + (molar mass of N × subscripts of N) + (molar mass of O × subscripts of O)
= (12.01 g/mol × 1) + (1.008 g/mol × 1) + (14.01 g/mol × 1) + (16.00 g/mol × 1)
= 43.03 g/mol
Multiplier = 129 g/mol / 43.03 g/mol ≈ 3
Multiply the subscripts in the empirical formula by the multiplier:
C1H1N1O1 × 3 = C3H3N3O3
Therefore, the empirical formula is C1H1N1O1, and the molecular formula is C3H3N3O3.
mol = g/molar mass.
Calculate mols of each element; e.g.,
0.1735/12 = 0.0145 = mols C.
Do the same for H and N.
grams O = 0.6216-(C+H+N)
mols O = g/molar mass.
Find the ratio of the elements to each other. That gives you the empirical formula. Now find the empirical formula mass and determine molecular formula this way.
empirical formula mass x factor = molar mass.
Solve for factor(call that n) and round to a whole number (if it isn't whole to begin with), then the molecular formula is
(CxHyNzOw)n