An ammonium salt of an alkanoic acid gain 60.5% of carbon 6.5% of hydrogen gas. If 0.309g of the salt yields 0.0313g of nitrogen. Determine the empirical formula of the salt
To determine the empirical formula of the salt, we need to calculate the ratio of elements present in the compound based on the given percentages.
Given:
- Carbon (C) = 60.5%
- Hydrogen (H) = 6.5%
- Nitrogen (N) = 0.0313g
Step 1: Convert percentage to mass:
Calculate the mass of each element based on the given percentages.
Mass of carbon (C):
0.605 * 0.309g = 0.187045g
Mass of hydrogen (H):
0.065 * 0.309g = 0.020085g
Step 2: Convert mass to moles:
Calculate the number of moles of each element using their molar masses.
Molar mass of carbon (C) = 12.01 g/mol
Number of moles of carbon (C) = mass / molar mass
= 0.187045g / 12.01 g/mol
≈ 0.01556 mol
Molar mass of hydrogen (H) = 1.01 g/mol
Number of moles of hydrogen (H) = mass / molar mass
= 0.020085g / 1.01 g/mol
≈ 0.01986 mol
Molar mass of nitrogen (N) = 14.01 g/mol
Number of moles of nitrogen (N) = mass / molar mass
= 0.0313g / 14.01 g/mol
≈ 0.00223 mol
Step 3: Find the simplest ratio of the elements:
Divide each element's mole value by the smallest mole value obtained.
C: 0.01556 mol / 0.00223 mol ≈ 6.98
H: 0.01986 mol / 0.00223 mol ≈ 8.9
N: 0.00223 mol / 0.00223 mol = 1
Step 4: Determine the empirical formula:
The ratio of elements obtained from the previous step represents the empirical formula.
Therefore, the empirical formula of the salt is C7H9N.
To determine the empirical formula of the salt, we need to find the relative proportions of the elements present in it.
First, let's calculate the mass of each element in the given sample of the salt.
Given:
Mass of C = 60.5% of 0.309g = 0.60 * 0.309g = 0.1854g
Mass of H = 6.5% of 0.309g = 0.065 * 0.309g = 0.020085g
Mass of N = 0.0313g
Next, we need to convert the masses of C, H, and N into moles using their respective atomic masses from the periodic table.
The atomic masses (rounded to two decimal places) are:
C (Carbon) = 12.01 g/mol
H (Hydrogen) = 1.01 g/mol
N (Nitrogen) = 14.01 g/mol
Number of moles of C = mass of C / atomic mass of C = 0.1854g / 12.01 g/mol = 0.01544 mol
Number of moles of H = mass of H / atomic mass of H = 0.020085g / 1.01 g/mol = 0.01988 mol
Number of moles of N = mass of N / atomic mass of N = 0.0313g / 14.01 g/mol = 0.00223 mol
Now, let's find the mole ratios of C, H, and N by dividing the number of moles of each element by the smallest number of moles obtained.
Dividing by 0.00223 mol (the smallest number of moles):
Ratio of C = 0.01544 mol / 0.00223 mol ≈ 6.92
Ratio of H = 0.01988 mol / 0.00223 mol ≈ 8.91
Ratio of N = 0.00223 mol / 0.00223 mol = 1.00
Based on the obtained ratios, the empirical formula of the salt is C6.92H8.91N. However, we should always simplify the subscripts to the nearest whole numbers.
Rounding the subscripts to the nearest whole numbers, the empirical formula of the salt is C7H9N.