you are given an unknown gaseous binary compound. when 10.0g of the compound isburned in excess oxygen, 16.3g of water is produced. the compound has 1.38 times that of oxygen gas at the same conditions of temperature and pressure. give a possible identity for the compound.

you are given an unknown gaseous binary compound. when 10.0g of the compound isburned in excess oxygen, 16.3g of water is produced. the compound has a density 1.38 times that of oxygen gas at the same conditions of temperature and pressure. give a possible identity for the compound

molar mass oxygen = 32/22.4 = about 1.43 and that time 1.38 = about 1.97.

So the molar mass of the unknown gas must be 22.4 x 1.97 = about 44 and it must contain H. I ran through the acids (HF, HCl, H2S, NH3, PH3, etc and there is no density that will match 1.97 or even that close and none with a molar mass of 44. But propane, a gas, C3H8 does. See if that will produce 16.3 g H2O and check is density.

To identify the compound, we need to determine its empirical formula. The empirical formula represents the simplest whole-number ratio of atoms in a compound.

Let's start by calculating the molar mass of water (H₂O):
- The molar mass of two hydrogen atoms (H₂) is 2 * 1.01 = 2.02 g/mol.
- The molar mass of one oxygen atom (O) is 1 * 16.00 = 16.00 g/mol.
Therefore, the molar mass of water (H₂O) is 2.02 + 16.00 = 18.02 g/mol.

Based on the given information, 10.0 g of the unknown compound produces 16.3 g of water. We need to convert the mass of water to moles using its molar mass:
16.3 g of water * (1 mol/18.02 g) = 0.905 mol of water.

Since water consists of two hydrogen atoms and one oxygen atom, the molar ratio of hydrogen to oxygen in water is 2:1. Therefore, the number of moles of oxygen in water is half the number of moles of water:
0.905 mol of water * (1/2) = 0.4525 mol of oxygen.

According to the information given, the molar ratio of the unknown compound to oxygen is 1.38:1. Therefore, we can calculate the number of moles of the unknown compound:
0.4525 mol of oxygen * (1/1.38) ≈ 0.328 mol of the unknown compound.

Now, divide the mass of the compound (10.0 g) by the number of moles calculated (0.328 mol):
10.0 g / 0.328 mol ≈ 30.5 g/mol.

The molar mass of the compound is approximately 30.5 g/mol.

Next, let's determine the possible empirical formula by finding the simplest ratio of atoms. Divide the molar mass of the compound by the molar mass of oxygen (16.00 g/mol):
30.5 g/mol / 16.00 g/mol ≈ 1.91.

Since we need a whole-number ratio, we can multiply all the subscripts by 2 to get the empirical formula:
2 * 1.91 = 3.82.

The empirical formula of the compound is approximately O₃.82.

Based on the empirical formula, a possible identity for the compound could be ozone (O₃), which is a gaseous binary compound composed of oxygen atoms.