What is C7H6O3 in empirical form and how do you get there?
The empirical formula is the simplest formula that can be written for a compound; in this case what you have written is already in the lowest form since 7 is a prime number and can't be divided by anything except 7.IF it were (and it isn't) C6H6O3 it could be divided by 3 and written as C2H2O as the empirical formula.
C7H6O3 in empirical form would be CH2O. How did we get there, you ask? Well, it's like trying to simplify a complicated recipe. We need to find the simplest ratio of atoms present in the compound.
First, we need to calculate the molar mass of each element: Carbon (C) has a molar mass of approximately 12.01 g/mol, Hydrogen (H) has a molar mass of 1.01 g/mol, and Oxygen (O) has a molar mass of about 16.00 g/mol.
The molecular mass of C7H6O3 is then calculated as follows:
(7 * 12.01) + (6 * 1.01) + (3 * 16.00) = 91.08 g/mol
Now, to find the empirical formula, we divide each element's number of moles by the molar mass of the compound:
C: (7 * 12.01) / 91.08 = 0.93 mol
H: (6 * 1.01) / 91.08 = 0.07 mol
O: (3 * 16.00) / 91.08 = 0.52 mol
To simplify, we make sure the mole ratios are in their simplest form:
C: 0.93 / 0.07 ≈ 13.3 ≈ 13
H: 0.07 / 0.07 = 1
O: 0.52 / 0.07 ≈ 7.4 ≈ 7
Therefore, the empirical formula becomes CH2O. Voila!
To determine the empirical formula of C7H6O3, we need to find the simplest whole number ratio of the atoms present.
1. Write down the given chemical formula: C7H6O3.
2. Determine the molar mass of each element by looking up their atomic masses on the periodic table. The atomic masses are approximately: C = 12.01 g/mol, H = 1.01 g/mol, and O = 16.00 g/mol.
3. Calculate the total molar mass of the compound by adding up the molar masses of each individual element.
Total molar mass = (7 × C) + (6 × H) + (3 × O)
= (7 × 12.01 g/mol) + (6 × 1.01 g/mol) + (3 × 16.00 g/mol)
= 84.07 g/mol + 6.06 g/mol + 48.00 g/mol
= 138.13 g/mol
4. Divide the molar mass of each element by the smallest molar mass among them (H in this case) to find the ratio of atoms.
C: H: O = (84.07 g/mol / 1.01 g/mol): (6.06 g/mol / 1.01 g/mol) : (48.00 g/mol / 1.01 g/mol)
= 83.24 : 6 : 47.52
5. Round off the ratio to the nearest whole number.
C: H: O = 83 : 6 : 48
6. The empirical formula is obtained by writing the elements and their corresponding subscripts in the simplest ratio: C83H6O48.
So, the empirical formula of C7H6O3 is C83H6O48.
To find the empirical formula of a compound, you need to determine the simplest, whole-number ratio of atoms present in the compound. We can start by finding the molar masses of the elements in the compound:
Carbon (C) has a molar mass of 12.01 g/mol.
Hydrogen (H) has a molar mass of 1.01 g/mol.
Oxygen (O) has a molar mass of 16.00 g/mol.
Given the molecular formula C7H6O3, we can determine the molar mass:
(7 * 12.01) + (6 * 1.01) + (3 * 16.00) = 98.07 g/mol
Now, we will divide the molar mass of each element by the molar mass of the entire compound and multiply by a common factor to obtain whole numbers.
For carbon (C):
(7 * 12.01 g/mol) / 98.07 g/mol ≈ 0.854
For hydrogen (H):
(6 * 1.01 g/mol) / 98.07 g/mol ≈ 0.061
For oxygen (O):
(3 * 16.00 g/mol) / 98.07 g/mol ≈ 0.489
To get whole numbers, we can multiply each of these values by a common factor of approximately 8. Thus, the empirical formula of C7H6O3 in its simplest, whole-number ratio is C6H5O2.