Question

A fertilizer is found to have the following mass percentage composition: 12.2% N, 5.26% H, 26.9% P, and 55.6% O. What is the empirical formula of the compound?

A. Convert mass percentages to moles.
B. Determine the mole ratio of the elements.
C. Find the empirical formula using the smallest whole numbers.
D. Identify the elements present in the fertilizer.

Answer 1

To determine the empirical formula of the compound, convert the mass percentages to moles, determine the mole ratio of the elements, find the empirical formula using the smallest whole numbers, and identify the elements present in the fertilizer.

Step-by-step explanation:

The empirical formula of the compound can be determined by following these steps:

1. Convert the mass percentages to moles for each element. For example, if you have a 100g sample, then you would have 12.2g of N, 5.26g of H, 26.9g of P, and 55.6g of O.
2. Determine the mole ratio of the elements by dividing the moles of each element by the smallest number of moles. In this case, let's find the moles for N, H, P, and O: N - 12.2g/14.01 g/mol = 0.873 mol; H - 5.26g/1.008 g/mol = 5.22 mol; P - 26.9g/30.97 g/mol = 0.868 mol; O - 55.6g/16.00 g/mol = 3.475 mol. The smallest number of moles is 0.868 mol, so divide all the moles by 0.868 mol to find the mole ratio: N - 0.873 mol/0.868 mol = 1, H - 5.22 mol/0.868 mol = 6, P - 0.868 mol/0.868 mol = 1, O - 3.475 mol/0.868 mol = 4.
3. Find the empirical formula using the smallest whole numbers obtained in the previous step. The empirical formula for the compound is N1H6P1O4.
4. Identify the elements present in the fertilizer, which are N (nitrogen), H (hydrogen), P (phosphorus), and O (oxygen).