A  compound has 40% Carbon 6.6% Hydrogen & 53.3% Oxygen. Calculate the Empirical Formula of the compound.

Sure, here are the steps to calculate the empirical formula of the compound:

1. Assume that you have 100g of the compound. This means that you have 40g of Carbon (C), 6.6g of Hydrogen (H), and 53.3g of Oxygen (O).

2. Convert the mass of each element to moles. You can do this by dividing the mass of each element by its atomic mass. The atomic mass of C is approximately 12g/mol, H is approximately 1g/mol, and O is approximately 16g/mol.

   So, you have:

   • Carbon: 40g / 12g/mol = 3.33 mol
   • Hydrogen: 6.6g / 1g/mol = 6.6 mol
   • Oxygen: 53.3g / 16g/mol = 3.33 mol

3. To find the empirical formula, you need to write the ratio of the moles of each element in the compound. In this case, the ratio is C: 3.33, H: 6.6, O: 3.33.

4. To simplify this ratio, divide each number by the smallest number of moles calculated in step 2. In this case, the smallest number is 3.33.

   So, you have:

   • Carbon: 3.33 / 3.33 = 1
   • Hydrogen: 6.6 / 3.33 = 2
   • Oxygen: 3.33 / 3.33 = 1

5. Therefore, the empirical formula of the compound is CH2O.

To calculate the empirical formula of a compound, follow these steps:

1. Assume that you have 100g of the compound. This means that you have 40g of Carbon (C), 6.6g of Hydrogen (H), and 53.3g of Oxygen (O).

2. Convert the mass of each element to moles. You do this by dividing the mass of each element by its atomic mass. The atomic mass of C is approximately 12g/mol, H is approximately 1g/mol, and O is approximately 16g/mol.

   So, you have:

   • Carbon: 40g / 12g/mol = 3.33 mol
   • Hydrogen: 6.6g / 1g/mol = 6.6 mol
   • Oxygen: 53.3g / 16g/mol = 3.33 mol

3. Write the ratio of the moles of each element to the smallest number of moles calculated. In this case, the smallest number of moles is 3.33 mol.

   So, you have:

   • Carbon: 3.33 mol / 3.33 = 1
   • Hydrogen: 6.6 mol / 3.33 = 2
   • Oxygen: 3.33 mol / 3.33 = 1

4. The empirical formula of the compound is therefore CH2O.