a compound was analyzed and found to contain C,H,N,O, and Cl. When o.150g sample of this compound was burned, it produced 0.138 g of CO2 and 0.0566 g of H2O. All of the Nitrogen in another sample of mass 0.200 g was converted to 0.0238 g of NH3. Finally, the chlorine in a 0.500 g sample of the compound was converted to Cl- an precipitated as AgCl. the precipitate had a mass of 1..004 g. Find the empirical formula of the compound.
when going from CO2 to C or from H20 to H and so on. Will we have to balance the equation or just 1 mole of co2 = 1 mole of C?!! Help plz
1 mole of CO2 = 1 mole C.
2 moles H = 1 mole H2O
etc.
ok i got the answer but seem different i get C3H6Cl2 N O
To determine the empirical formula of the compound, we need to find the simplest, whole-number ratio of the elements present.
Let's start by calculating the number of moles of each element in the given samples:
1. Carbon (C):
- Mass of CO2 produced: 0.138 g
- Molar mass of CO2: 44.01 g/mol
- Number of moles of CO2: 0.138 g / 44.01 g/mol = 0.00314 mol
- Since the stoichiometry of the balanced equation shows that 1 mole of CO2 contains 1 mole of carbon, we have 0.00314 moles of carbon.
2. Hydrogen (H):
- Mass of H2O produced: 0.0566 g
- Molar mass of H2O: 18.02 g/mol
- Number of moles of H2O: 0.0566 g / 18.02 g/mol = 0.00314 mol
- Since the stoichiometry of the balanced equation shows that 1 mole of H2O contains 2 moles of hydrogen, we have 0.00314 mol x 2 = 0.00629 moles of hydrogen.
3. Nitrogen (N):
- Mass of NH3 produced: 0.0238 g
- Molar mass of NH3: 17.03 g/mol
- Number of moles of NH3: 0.0238 g / 17.03 g/mol = 0.00140 mol
- Since the stoichiometry of the balanced equation shows that 1 mole of NH3 contains 1 mole of nitrogen, we have 0.00140 moles of nitrogen.
4. Chlorine (Cl):
- Mass of AgCl precipitate: 1.004 g
- Molar mass of AgCl: 143.32 g/mol (from periodic table)
- Number of moles of AgCl: 1.004 g / 143.32 g/mol = 0.00701 mol
- Since the stoichiometry of the balanced equation shows that 1 mole of AgCl contains 1 mole of chlorine, we have 0.00701 moles of chlorine.
Now, we need to find the simplest, whole-number ratio of the elements:
- Carbon (C): 0.00314 mol
- Hydrogen (H): 0.00629 mol
- Nitrogen (N): 0.00140 mol
- Chlorine (Cl): 0.00701 mol
To obtain the empirical formula, divide the number of moles of each element by the smallest number of moles:
- Carbon: 0.00314 mol / 0.00140 mol = 2.24
- Hydrogen: 0.00629 mol / 0.00140 mol = 4.49
- Nitrogen: 0.00140 mol / 0.00140 mol = 1.00
- Chlorine: 0.00701 mol / 0.00140 mol = 5
The numbers obtained are approximately 2.24, 4.49, 1.00, and 5. To simplify, we multiply all numbers by 4 to get whole numbers:
- Carbon: 2.24 x 4 = 8.96 ≈ 9
- Hydrogen: 4.49 x 4 = 17.96 ≈ 18
- Nitrogen: 1.00 x 4 = 4
- Chlorine: 5 x 4 = 20
Therefore, the empirical formula of the compound is C9H18N4Cl5.