Calculate the empirical formula of a compound containing 52.14% C, 13.12% H, and 34.73% O.
I don't understand how they got the answer. please help
calculate the empirical formula of compound containing 52.14%of carbon,13.12% of hydrogen and 34.73% of oxygen
To determine the empirical formula of a compound, you need to find the simplest whole-number ratio of the elements present in the compound. Here's how you can calculate the empirical formula using the given percentages:
Step 1: Convert the percentage to grams.
Assume we have 100 grams of the compound. Therefore,
- The mass of Carbon (C) = 52.14 grams
- The mass of Hydrogen (H) = 13.12 grams
- The mass of Oxygen (O) = 34.73 grams
Step 2: Determine the number of moles of each element.
To find the number of moles, divide the mass of each element by its molar mass. The molar mass of C is approximately 12.01 g/mol, H is approximately 1.01 g/mol, and O is approximately 16.00 g/mol.
- Moles of C = 52.14 g / 12.01 g/mol ≈ 4.34 mol
- Moles of H = 13.12 g / 1.01 g/mol ≈ 12.99 mol
- Moles of O = 34.73 g / 16.00 g/mol ≈ 2.17 mol
Step 3: Divide each mole value by the smallest mole value.
By dividing each mole value by the smallest mole value, you will obtain the simplest whole-number ratio.
- Divide Moles of C by 2.17 ≈ 2
- Divide Moles of H by 2.17 ≈ 6
- Divide Moles of O by 2.17 ≈ 1
Step 4: Write the empirical formula.
Using the whole-number ratio obtained in Step 3, we can now write the empirical formula.
- Empirical formula = C2H6O
So, the empirical formula of the compound containing 52.14% C, 13.12% H, and 34.73% O is C2H6O.
To calculate the empirical formula of a compound, you need to determine the simplest whole-number ratio of the elements present in the compound. Here's how you can calculate the empirical formula for a compound containing 52.14% C, 13.12% H, and 34.73% O:
1. Assume you have 100g of the compound. This assumption allows you to work with grams, which can be easily converted to moles for further calculations.
2. Convert each element's percentage into grams. For example:
- Carbon (C): 52.14% of 100g = 52.14g
- Hydrogen (H): 13.12% of 100g = 13.12g
- Oxygen (O): 34.73% of 100g = 34.73g
3. Convert each element's mass into moles using their molar mass:
- Carbon (C): The molar mass of carbon (C) is 12.01 g/mol. Divide the mass of carbon by its molar mass:
moles of carbon = mass of carbon / molar mass of carbon = 52.14g / 12.01 g/mol = 4.34 mol
- Hydrogen (H): The molar mass of hydrogen (H) is 1.01 g/mol. Divide the mass of hydrogen by its molar mass:
moles of hydrogen = mass of hydrogen / molar mass of hydrogen = 13.12g / 1.01 g/mol = 12.99 mol
- Oxygen (O): The molar mass of oxygen (O) is 16.00 g/mol. Divide the mass of oxygen by its molar mass:
moles of oxygen = mass of oxygen / molar mass of oxygen = 34.73g / 16.00 g/mol = 2.17 mol
4. Find the simplest whole-number ratio of moles by dividing each mole value by the smallest mole value. In this case, the smallest mole value is 2.17 mol:
- moles of carbon / 2.17 mol = 4.34 mol / 2.17 mol = 2
- moles of hydrogen / 2.17 mol = 12.99 mol / 2.17 mol = 6
- moles of oxygen / 2.17 mol = 2.17 mol / 2.17 mol = 1
5. Write the empirical formula using the whole-number ratio obtained in the previous step:
The empirical formula of the compound is C2H6O.
Therefore, the empirical formula of the compound containing 52.14% C, 13.12% H, and 34.73% O is C2H6O.
look at example 2
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm