A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.123g carbon dioxide and 0.0310g water. In another experiment, it is found that 0.103g of the compound produces 0.0230g ammonia. What is the expirical formula of the compound?
Well, first tell me what you think it is. Then I will gladly correct you.
c=0.03g, O=0.09
To determine the empirical formula of the compound, we need to find the ratio of the atoms present in it.
Let's start by calculating the number of moles for each component:
1. Carbon dioxide (CO2):
- Molar mass of carbon dioxide (CO2) = 44.01 g/mol
- Moles of carbon dioxide produced = mass / molar mass = 0.123g / 44.01 g/mol
2. Water (H2O):
- Molar mass of water (H2O) = 18.02 g/mol
- Moles of water produced = mass / molar mass = 0.0310g / 18.02 g/mol
3. Ammonia (NH3):
- Molar mass of ammonia (NH3) = 17.03 g/mol
- Moles of ammonia produced = mass / molar mass = 0.0230g / 17.03 g/mol
Next, let's convert the moles of each component into the simplest whole number ratio by dividing each by the smallest value obtained.
For carbon dioxide:
Moles of carbon dioxide = 0.123g / 44.01 g/mol = 0.002793 mol
For water:
Moles of water = 0.0310g / 18.02 g/mol = 0.001719 mol
For ammonia:
Moles of ammonia = 0.0230g / 17.03 g/mol = 0.001350 mol
Now, divide each of the above values by the smallest value (0.001350 mol) to find the ratio in terms of whole numbers:
Carbon dioxide: 0.002793 mol / 0.001350 mol = 2.068 ≈ 2
Water: 0.001719 mol / 0.001350 mol = 1.275 ≈ 1
Ammonia: 0.001350 mol / 0.001350 mol = 1
The empirical formula is determined by the whole number ratio of atoms. In this case, the ratio is approximately C2H2NO. Therefore, the empirical formula of the compound is C2H2NO.