A sample of a substance (containing only C, H, and N) is burned in oxygen.

7.961 g of CO2, 1.164 g of H2O and 1.551 g of NO are the sole products of combustion.
What is the empirical formula of the compound?

Cindy, Tiffany, et al. See your other posts. We prefer you stick to one screen name. Posting with more than one name often delays help.

To determine the empirical formula of a compound, we need to find the simplest whole-number ratio of the elements present in the compound. In this case, we have a sample that contains carbon (C), hydrogen (H), and nitrogen (N), and the products of combustion are carbon dioxide (CO2), water (H2O), and nitrogen monoxide (NO).

1. Start by calculating the number of moles of each compound formed:
- Moles of CO2 = mass of CO2 / molar mass of CO2
- Moles of H2O = mass of H2O / molar mass of H2O
- Moles of NO = mass of NO / molar mass of NO

2. Determine the mole ratio between the different elements:
- Divide the moles of each compound by the smallest number of moles obtained from step 1.

3. Use the mole ratio from step 2 to write the empirical formula of the compound.
- The subscripts in the empirical formula represent the number of atoms of each element.

Let's calculate the moles of CO2, H2O, and NO:

1. Moles of CO2:
- Mass of CO2 = 7.961 g
- Molar mass of CO2 = 12.01 g/mol (mass of C) + (2 * 16.00 g/mol) (mass of O) = 44.01 g/mol
- Moles of CO2 = 7.961 g / 44.01 g/mol ≈ 0.181 mol

2. Moles of H2O:
- Mass of H2O = 1.164 g
- Molar mass of H2O = 2 * 1.01 g/mol (mass of H) + 16.00 g/mol (mass of O) = 18.02 g/mol
- Moles of H2O = 1.164 g / 18.02 g/mol ≈ 0.065 mol

3. Moles of NO:
- Mass of NO = 1.551 g
- Molar mass of NO = 14.01 g/mol (mass of N) + 16.00 g/mol (mass of O) = 30.01 g/mol
- Moles of NO = 1.551 g / 30.01 g/mol ≈ 0.052 mol

Next, we find the mole ratio:

- The lowest number of moles obtained is 0.052 mol (from NO).
- Divide the moles of each compound by the smallest number of moles:
- Moles of CO2 / 0.052 ≈ 3.48
- Moles of H2O / 0.052 ≈ 1.25
- Moles of NO / 0.052 ≈ 1.00

The approximate whole-number mole ratio is 3.48:1.25:1.00, or approximately 7:3:2.

Therefore, the empirical formula of the compound is C7H3N2.