A compound contains 40.45% C, 7.87%H, 15.73% N and 35.96%O. How many atoms of H are present in the empirical formula of the compound?Group of answer choices1327

To find the number of atoms of H in the empirical formula of the compound, we need to convert the percentages to moles.

1. First, assume that you have 100g of the compound. This means that you have 40.45g of C, 7.87g of H, 15.73g of N, and 35.96g of O.

2. Next, convert these amounts to moles. The atomic masses of C, H, N, and O are approximately 12.01g/mol, 1.008g/mol, 14.01g/mol, and 16.00g/mol, respectively.

   Moles of C = 40.45g / 12.01g/mol = 3.37 mol Moles of H = 7.87g / 1.008g/mol = 7.81 mol Moles of N = 15.73g / 14.01g/mol = 1.12 mol Moles of O = 35.96g / 16.00g/mol = 2.25 mol

3. To find the empirical formula, divide each of these amounts by the smallest amount, which is 1.12 mol (N).

   Ratio of C = 3.37 mol / 1.12 mol = 3 Ratio of H = 7.81 mol / 1.12 mol = 7 Ratio of N = 1.12 mol / 1.12 mol = 1 Ratio of O = 2.25 mol / 1.12 mol = 2

So, the empirical formula of the compound is C3H7N1O2. This means that there are 7 atoms of H in the empirical formula of the compound.