Fumaric acid, which occurs in many plants, contains, by mass, 41.4% carbon, 3.47% hydrogen, and 55.1%
oxygen. A 0.050-mol sample of this compound weighs 5.80 g. The molecular formula of this compound is
Take a 100 g sample. That gives you
41.4 g C
3.47 g H
55.1 g O
Convert each of the grams to moles. moles = g/molar mass.
Then determine the ratio of the elements to the smallest (found by dividing all of the numbers by the smallest number of moles). That gives the empirical formula.
Since moles = g/molar mass, the second part of the problem allows you to determine the molar mass.
Divide molar mass/empirical formula mass to determine the number of empirical formula units in the mole.
To find the molecular formula of the compound, we need to determine the empirical formula first. The empirical formula gives the simplest whole number ratio of atoms in a compound.
1. Convert the percentages to grams:
- Carbon (C): 41.4% of 5.80 g = 2.4012 g
- Hydrogen (H): 3.47% of 5.80 g = 0.2009 g
- Oxygen (O): 55.1% of 5.80 g = 3.1968 g
2. Determine the number of moles for each element:
- Carbon (C): 2.4012 g / molar mass of carbon (12.01 g/mol) = 0.1999 mol
- Hydrogen (H): 0.2009 g / molar mass of hydrogen (1.01 g/mol) = 0.1997 mol
- Oxygen (O): 3.1968 g / molar mass of oxygen (16.00 g/mol) = 0.1998 mol
3. Calculate the mole ratio of the elements by dividing each number of moles by the smallest value:
- Carbon (C): 0.1999 mol / 0.1997 mol = 1.0006
- Hydrogen (H): 0.1997 mol / 0.1997 mol = 1.0000
- Oxygen (O): 0.1998 mol / 0.1997 mol = 1.0005
4. Round the mole ratios to the nearest whole number:
- Carbon (C): 1
- Hydrogen (H): 1
- Oxygen (O): 1
Therefore, the empirical formula of the compound is CH₄O.
Next, to find the molecular formula, we need to compare the empirical formula mass (52.05 g/mol) with the molar mass of the compound (unknown).
5. Calculate the empirical formula mass:
- Carbon (C): 1 x molar mass of carbon (12.01 g/mol) = 12.01 g/mol
- Hydrogen (H): 4 x molar mass of hydrogen (1.01 g/mol) = 4.04 g/mol
- Oxygen (O): 1 x molar mass of oxygen (16.00 g/mol) = 16.00 g/mol
Empirical formula mass = 12.01 g/mol + 4.04 g/mol + 16.00 g/mol = 32.05 g/mol
6. Divide the molar mass of the compound (unknown) by the empirical formula mass to determine the molecular formula factor:
Molar mass of the compound (unknown) / Empirical formula mass = Molecular formula factor
Molar mass of the compound (unknown) = (0.050 mol x 5.80 g/mol) = 0.29 g
0.29 g / 32.05 g/mol = 0.009 mole
7. Multiply the empirical formula by the molecular formula factor to find the molecular formula:
Empirical formula (CH₄O) x Molecular formula factor (0.009) = Molecular formula
Molecular formula = C₀.₀₁₈H₀.₀₉₈O₀.₀₉₈
After rounding to the nearest whole number, the molecular formula of the compound is C₁H₂O₁, which can be further simplified as CH₂O.
To determine the molecular formula of the compound, we need to find the empirical formula first. The empirical formula gives the simplest whole-number ratio of atoms in a compound.
Given that the compound contains 41.4% carbon, 3.47% hydrogen, and 55.1% oxygen, we can assume that we have 100g of the compound. By calculating the mass of each element in the 100g sample, we can then convert these masses to moles.
1. Carbon:
Mass of carbon = (41.4 / 100) * 100g = 41.4g
Molar mass of carbon = 12.01 g/mol
Moles of carbon = 41.4g / 12.01 g/mol = 3.45 mol
2. Hydrogen:
Mass of hydrogen = (3.47 / 100) * 100g = 3.47g
Molar mass of hydrogen = 1.01 g/mol
Moles of hydrogen = 3.47g / 1.01 g/mol = 3.44 mol
3. Oxygen:
Mass of oxygen = (55.1 / 100) * 100g = 55.1g
Molar mass of oxygen = 16.00 g/mol
Moles of oxygen = 55.1g / 16.00 g/mol = 3.44 mol
Now we need to find the simplest whole-number ratio of moles for each element. In this case, carbon, hydrogen, and oxygen all have approximately the same number of moles (3.45 mol, 3.44 mol, and 3.44 mol, respectively). So, the empirical formula is approximately CH2O.
To find the molecular formula, we need to know the molar mass of the compound.
Given that a 0.050-mol sample of the compound weighs 5.80 g, we can calculate the molar mass as follows:
Molar mass of compound = (mass of sample) / (moles of sample)
Molar mass of compound = 5.80 g / 0.050 mol = 116 g/mol
Now we can calculate the ratio between the molar mass of the empirical formula (CH2O) and the molar mass of the compound:
Molar mass ratio = (molar mass of compound) / (molar mass of empirical formula)
Molar mass ratio = 116 g/mol / (12.01 g/mol + 2 * 1.01 g/mol + 16.00 g/mol)
Molar mass ratio = 116 g/mol / 30.03 g/mol
Simplifying the ratio:
Molar mass ratio ≈ 3.863
The molecular formula is obtained by multiplying the empirical formula by the molar mass ratio:
Molecular formula = empirical formula * molar mass ratio
Molecular formula = CH2O * 3.863
Molecular formula ≈ C4H6O4
Therefore, the molecular formula of the compound is C4H6O4.