0.5mols of a mono alkanoicacid weighs 44 grams, determine the molecular formula and name the acid. (show the workings)
To determine the molecular formula and name of the acid, we can start by finding the molar mass of the acid.
Given:
Number of moles (n) = 0.5 mols
Mass (m) = 44 grams
The molar mass of the acid can be calculated using the formula:
Molar mass (M) = Mass (m) / Number of moles (n)
Substituting the given values:
M = 44 grams / 0.5 mols
M = 88 grams/mols
Now, let's find the empirical formula of the acid. The empirical formula represents the simplest whole number ratio of the atoms in a compound.
To do this, we need to know the percentage composition of the acid. Unfortunately, it is not provided in the given question. Therefore, we cannot determine the molecular formula and name of the acid without additional information.
To determine the molecular formula and name of the acid, we need to use the given information that 0.5 moles of the acid weighs 44 grams.
First, we need to find the molar mass of the acid by dividing its weight (44 grams) by the number of moles (0.5 moles):
Molar mass = Weight / Moles
Molar mass = 44 g / 0.5 mol
Molar mass = 88 g/mol
So, the molar mass of the acid is 88 g/mol.
Next, we need to determine the empirical formula of the acid. The empirical formula gives the simplest whole number ratio of atoms present in a compound. To find it, we will divide the molecular formula by the greatest common divisor (GCD) of its subscripts.
Since we are given the molar mass, we need to deduce the empirical formula from it. We know that the molecular formula will be a multiple of the empirical formula.
Let's assume the empirical formula to be "CₓHᵧO".
The molar mass for the empirical formula (CₓHᵧO) can be calculated using the atomic masses of carbon, hydrogen, and oxygen. The atomic masses are approximately 12 g/mol for carbon, 1 g/mol for hydrogen, and 16 g/mol for oxygen.
Molar mass of empirical formula (CₓHᵧO) = (12 * ₓ) + (1 * ᵧ) + (16 * 1)
Molar mass of empirical formula = 12ₓ + ᵧ + 16
Given that the molar mass of the acid is 88 g/mol, we can set up an equation to solve for the values of ₓ and ᵧ:
88 = 12ₓ + ᵧ + 16
Since the question states that it is a mono alkanoic acid, it means that only one molecule of carboxyl group (-COOH) is present in the acid. Hence, ₓ will be equal to 1.
Substituting ₓ = 1, the equation becomes:
88 = 12 + ᵧ + 16
ᵧ = 60
Therefore, the empirical formula of the acid is C₁H₆O₃.
Next, we need to find the molecular formula, which represents the actual number of atoms in the compound.
To find the molecular formula, we need to know the molar mass of the acid.
The molar mass of the acid is 88 g/mol, which is also the molar mass of the empirical formula (C₁H₆O₃).
Since the empirical formula mass is equal to the molar mass, the empirical formula is the same as the molecular formula. Therefore, the molecular formula and the name of the acid is:
Molecular formula: C₁H₆O₃
Name of the acid: Methanoic acid (also known as formic acid)