An organic compound containing carbon, hydrogen and oxygen only was found to contain 64.9% carbon and 13.5% hydrogen. If the molecular mass of the compound is 74g/mol.
I) calculate the percentage of oxygen in the compound
Ii) determine the empirical formula and molecular formula
III) write the structural formula of the three possible isomers of the compound
iv) if the compound in (Ii) above gives a fragrant smell on heating with ethanoic acid in the presence of concentrated H2SO4, name the functional group present in the compound
I) To find the percentage of oxygen in the compound, we can start by subtracting the percentages of carbon and hydrogen from 100%:
100% - 64.9% (carbon) - 13.5% (hydrogen) = 21.6% (oxygen)
Therefore, the compound contains 21.6% oxygen.
II) To determine the empirical formula, we need to find the ratio of the different elements present in the compound.
First, we assume that we have a 100g sample of the compound. From the given percentages, we can determine the mass of each element:
64.9g carbon and 13.5g hydrogen (since C = 64.9% of 100g and H = 13.5% of 100g)
Next, we need to convert the mass of each element to moles:
1 mole of carbon has a molar mass of 12g/mol, so 64.9g of carbon is equal to 5.41 moles of carbon
1 mole of hydrogen has a molar mass of 1g/mol, so 13.5g of hydrogen is equal to 13.5 moles of hydrogen
Now, we need to find the ratio of the moles of each element. Dividing the number of moles by the smallest number of moles (5.41), we get:
Carbon: 5.41/5.41 ≈ 1
Hydrogen: 13.5/5.41 ≈ 2.5
Multiplying these ratios by 2 to get whole numbers, we get:
Carbon: 2
Hydrogen: 5
Therefore, the empirical formula of the compound is CH2O.
To determine the molecular formula, we need to know the molecular mass of the compound (given as 74g/mol). The empirical formula mass of CH2O is:
(1 * 12) + (2 * 1) + (1 * 16) = 12 + 2 + 16 = 30g/mol
Dividing the molecular mass by the empirical formula mass, we get:
74g/mol / 30g/mol ≈ 2.47
This means that the molecular formula is approximately 2.47 times the empirical formula. So, the molecular formula of the compound is C2.47H4.94O.
III) To determine the possible isomers of the compound, we need to consider the different ways the atoms can be arranged in the molecular formula: C2.47H4.94O.
The possible isomers are:
1. Dimethyl ether: CH3-O-CH3
2. Formaldehyde: O=CH2
3. Methyl alcohol (methanol): CH3-OH
IV) If the compound gives off a fragrant smell when heated with ethanoic acid in the presence of concentrated H2SO4, it suggests the presence of an alcohol functional group (OH).
Therefore, the functional group present in the compound is the alcohol functional group (OH).
I) To calculate the percentage of oxygen in the compound, we can subtract the percentages of carbon and hydrogen from 100%:
Percentage of oxygen = 100% - 64.9% - 13.5%
Percentage of oxygen = 21.6%
II) The empirical formula can be determined by finding the ratio of the number of atoms of each element present in the compound. First, we need to find the number of moles of each element:
Number of moles of carbon = (64.9% / atomic mass of carbon) * (molecular mass of the compound / 100)
Number of moles of carbon = (64.9 / 12) * (74 / 100)
Number of moles of carbon = 0.54042
Number of moles of hydrogen = (13.5% / atomic mass of hydrogen) * (molecular mass of the compound / 100)
Number of moles of hydrogen = (13.5 / 1) * (74 / 100)
Number of moles of hydrogen = 0.945
Number of moles of oxygen = (21.6% / atomic mass of oxygen) * (molecular mass of the compound / 100)
Number of moles of oxygen = (21.6 / 16) * (74 / 100)
Number of moles of oxygen = 0.08325
Next, we divide each number of moles by the smallest number of moles to get the simplest, whole-number ratio.
Dividing by 0.08325, we get:
Number of moles of carbon = 0.54042 / 0.08325 ≈ 6.5
Number of moles of hydrogen = 0.945 / 0.08325 ≈ 11.35
Number of moles of oxygen = 0.08325 / 0.08325 = 1
Since we cannot have fractions in the empirical formula, we need to multiply each number by 2:
Number of moles of carbon = 6.5 * 2 = 13
Number of moles of hydrogen = 11.35 * 2 ≈ 23
Number of moles of oxygen = 1 * 2 = 2
Therefore, the empirical formula is C13H23O2.
To determine the molecular formula, we need to know the molecular mass of the empirical formula. The empirical formula mass can be calculated by summing the atomic masses of each element in the empirical formula:
Empirical formula mass = (13 * atomic mass of carbon) + (23 * atomic mass of hydrogen) + (2 * atomic mass of oxygen)
Empirical formula mass = (13 * 12) + (23 * 1) + (2 * 16)
Empirical formula mass ≈ 180
The molecular formula can now be found by dividing the molecular mass of the compound by the empirical formula mass:
Molecular formula = Molecular mass of the compound / Empirical formula mass
Molecular formula = 74 / 180
Since the result is less than 1, it means we have one unit of the empirical formula, so the molecular formula is the same as the empirical formula.
Therefore, the empirical formula and molecular formula of the compound are C13H23O2.
III) Writing the structural formulas of the three possible isomers of the compound is difficult without additional information about the arrangement of atoms within the molecule. Can you provide any more information or details?
IV) If the compound from II above gives a fragrant smell on heating with ethanoic acid in the presence of concentrated H2SO4, it suggests the presence of an ester functional group. Esters are known for their pleasant, fragrant smells.
To solve this problem, we'll need to break it down into several steps. Let's go through each part of the question.
I) To calculate the percentage of oxygen in the compound, we need to subtract the sum of the percentages of carbon and hydrogen from 100%. Since the compound contains only carbon, hydrogen, and oxygen, we can assume that the remaining percentage is due to oxygen.
Percentage of oxygen = 100% - (Percentage of carbon + Percentage of hydrogen)
= 100% - (64.9% + 13.5%)
= 21.6%
Therefore, the compound contains 21.6% oxygen.
Ii) To determine the empirical formula and molecular formula, we need to convert the percentages of each element into moles. The empirical formula represents the simplest whole number ratio of the elements, while the molecular formula gives the actual number of atoms of each element in a molecule.
1. Convert the percentages of carbon, hydrogen, and oxygen into moles:
- Moles of carbon = (Percentage of carbon / atomic mass of carbon) * molecular mass
= (64.9 / 12.01) * 74
≈ 406.79
- Moles of hydrogen = (Percentage of hydrogen / atomic mass of hydrogen) * molecular mass
= (13.5 / 1.01) * 74
≈ 982.18
- Moles of oxygen = (Percentage of oxygen / atomic mass of oxygen) * molecular mass
= (21.6 / 16.00) * 74
≈ 99.82
2. Determine the empirical formula:
The empirical formula is the simplest ratio of atoms. We divide the moles of each element by the smallest value to find their ratio.
- Moles of carbon in empirical formula = 406.79 / 99.82 ≈ 4.08
- Moles of hydrogen in empirical formula = 982.18 / 99.82 ≈ 9.84
- Moles of oxygen in empirical formula = 99.82 / 99.82 = 1
The empirical formula is C4H10O.
3. Determine the molecular formula:
To determine the molecular formula, we need to know the molecular mass of the empirical formula. The empirical formula mass can be calculated by adding up the atomic masses of all the atoms in the empirical formula.
- Empirical formula mass = (4 * atomic mass of carbon) + (10 * atomic mass of hydrogen) + (1 * atomic mass of oxygen)
= (4 * 12.01) + (10 * 1.01) + (1 * 16.00)
= 58.12
The molecular formula will be a multiple of the empirical formula. Divide the molecular mass by the empirical formula mass to find the multiple.
- Multiple = molecular mass / empirical formula mass
= 74 / 58.12 ≈ 1.27 (rounded to two decimal places)
Multiply the subscripts in the empirical formula by the multiple to get the molecular formula:
- Molecular formula = (C4H10O)1.27 (rounded to the nearest whole number)
= C5H13O
Therefore, the empirical formula is C4H10O, and the molecular formula is C5H13O.
III) To write the structural formula of the three possible isomers of the compound, we need to consider the different ways the atoms can be arranged.
1. Isomer 1: n-pentanol (C5H13O)
Structural formula: CH3-CH2-CH2-CH2-OH
2. Isomer 2: 2-methylbutanol (C5H13O)
Structural formula: CH3-CH(CH3)-CH2-CH2-OH
3. Isomer 3: 2,2-dimethylpropanol (C5H13O)
Structural formula: CH3-C(CH3)3-OH
These are three possible isomers of the compound with the molecular formula C5H13O.
iv) If the compound gives a fragrant smell on heating with ethanoic acid in the presence of concentrated H2SO4, the functional group present in the compound is an alcohol (-OH).
I hope this explanation helps you understand how to approach and solve this problem.