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. What is the molecular formula of this compound? (C:12 g/mol, H:1 g/mol, O:16g/mol)
41.4 % carbon, 3.47 % hydrogen, and 55.1 % oxygen. A 0.050 mol sample of this compound weighs 5.80 g
Take a 100 g sample which gives you 41.4 g C, 3.47 g H, 55.1 g O.
mols C = 41.4/12 = 3.45
mols H = 3.47/1 = 3.47
mols O = 55.1/16 = 3.44
So the ratio (rounded) is C 3.5; H 3.5; O 3.5. or C1H1O1 and the empirical mass is 12 + 1 + 16 = 29
mols = g/molar mass or molar mass = g/mol = 5.80/0.05 = 116
empirical mass x some whole number = molar mass
29 x ? = 116 and ? = 116/29 = 4 so the molecular formula is (C1H1O1)4 or C4H4O4.
Well, let's break this down! First, we need to find the molar mass of the compound.
The molar mass of carbon is 12 g/mol, hydrogen is 1 g/mol, and oxygen is 16 g/mol.
Now, let's calculate the masses of each element in the 0.050 mol sample:
Carbon: (41.4/100) * 5.80 g = 2.4012 g
Hydrogen: (3.47/100) * 5.80 g = 0.20126 g
Oxygen: (55.1/100) * 5.80 g = 3.1948 g
Next, we need to find the number of moles of each element in the sample:
Carbon: 2.4012 g / 12 g/mol = 0.2001 mol
Hydrogen: 0.20126 g / 1 g/mol = 0.20126 mol
Oxygen: 3.1948 g / 16 g/mol = 0.1997 mol
Now, let's find the simplest whole number ratio by dividing all the mole values by the smallest mole value (0.1997 mol):
Carbon: 0.2001 mol / 0.1997 mol = 1.001
Hydrogen: 0.20126 mol / 0.1997 mol = 1.007
Oxygen: 0.1997 mol / 0.1997 mol = 1.000
The ratios are very close to 1, which suggests that the empirical formula could be CH1O1. However, we can simplify this further by dividing all the subscripts by their greatest common divisor, which is 1:
Empirical formula: CH1O1
So, the molecular formula of this compound is CH1O1. It looks like we have found the elusive "One-Hydrogen-One-Oxygen" compound! Congratulations!
To determine the molecular formula of fumaric acid, you need to find the empirical formula first. The empirical formula gives the simplest whole number ratio of the atoms present in a compound.
1. Calculate the molar mass of each element:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Oxygen (O): 16 g/mol
2. Convert the mass percentages to grams:
- Carbon: (41.4% * 5.80 g) = 2.403 g
- Hydrogen: (3.47% * 5.80 g) = 0.201 g
- Oxygen: (55.1% * 5.80 g) = 3.194 g
3. Calculate the number of moles for each element:
- Moles of Carbon: (2.403 g / 12 g/mol) = 0.2003 mol
- Moles of Hydrogen: (0.201 g / 1 g/mol) = 0.201 mol
- Moles of Oxygen: (3.194 g / 16 g/mol) = 0.1996 mol
4. Divide the moles of each element by the smallest number of moles to obtain the simplest whole number ratio:
- Carbon: 0.2003 mol / 0.1996 mol ≈ 1
- Hydrogen: 0.201 mol / 0.1996 mol ≈ 1
- Oxygen: 0.1996 mol / 0.1996 mol = 1
5. The empirical formula for fumaric acid is CH2O.
To find the molecular formula, you need to determine the ratio between the empirical formula mass and the molar mass of the compound.
6. Calculate the empirical formula mass:
- C: 1 atom * 12 g/mol = 12 g/mol
- H: 2 atoms * 1 g/mol = 2 g/mol
- O: 1 atom * 16 g/mol = 16 g/mol
Empirical formula mass = 12 g/mol + 2 g/mol + 16 g/mol = 30 g/mol
7. Calculate the ratio:
- Molecular mass of the compound: 5.80 g / 0.050 mol = 116 g/mol
8. Divide the molecular mass by the empirical formula mass:
- Molecular formula ratio: 116 g/mol / 30 g/mol ≈ 3.87
9. Round the ratio to the nearest whole number to obtain the molecular formula:
- Molecular formula = 3 * CH2O = C3H6O3
Therefore, the molecular formula of this compound is C3H6O3.
To identify the molecular formula of the compound, we need to determine the empirical formula and then find the molecular formula based on its molar mass.
1. Start by assuming we have 100g of the compound, which makes it easier to deal with percentages.
- From this assumption, we can calculate that the compound contains:
- 41.4g of carbon (41.4% of 100g)
- 3.47g of hydrogen (3.47% of 100g)
- 55.1g of oxygen (55.1% of 100g)
2. Next, convert the masses to moles by dividing each value by its respective atomic mass:
- For carbon (C), the atomic mass is 12 g/mol:
- Moles of carbon = 41.4g / 12 g/mol ≈ 3.45 mol
- For hydrogen (H), the atomic mass is 1 g/mol:
- Moles of hydrogen = 3.47g / 1 g/mol = 3.47 mol
- For oxygen (O), the atomic mass is 16 g/mol:
- Moles of oxygen = 55.1g / 16 g/mol ≈ 3.44 mol
3. Now, divide each mole value by the smallest number of moles to find the simplest whole number ratio. In this case, the smallest value is approximately 3.44 mol. Divide each value by 3.44:
- Divide moles of carbon by 3.44: 3.45 mol / 3.44 mol ≈ 1 mol
- Divide moles of hydrogen by 3.44: 3.47 mol / 3.44 mol ≈ 1 mol
- Divide moles of oxygen by 3.44: 3.44 mol / 3.44 mol = 1 mol
4. The resulting ratio of 1:1:1 gives us the empirical formula CH₃O.
5. Finally, 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 5.80g, we can calculate the molar mass:
- Molar mass = Mass (g) / Moles
- Molar mass = 5.80g / 0.050 mol = 116 g/mol
6. To determine the molecular formula from the empirical formula, we need to find the ratio of the molar mass to the empirical formula mass:
- Molecular formula ratio = Molar mass / Empirical formula mass
- Molecular formula ratio = 116 g/mol / 31 g/mol (12 g/mol for C + 1 g/mol for H + 16 g/mol for O)
- Molecular formula ratio ≈ 3.74
7. Since the molecular formula ratio is approximately 3.74, we need to multiply the empirical formula by this ratio to obtain the molecular formula:
- Molecular formula = (CH₃O)ₓ
- Molecular formula = (CH₃O)₃.₇₄
- Rounded to the nearest whole number, the molecular formula is C₄H₁₀O₃.
Therefore, the molecular formula of the compound is C₄H₁₀O₃.