A sample has a molar mass of 162g/mol. When a sample of mass 0.385g was burned , 1.072g of CO2 ,0.307g H2O and 0.068g of nitrogen were produced. What is empirial and molecular formular of a sample . Write equation for combustion
Could you please recheck your given CO2, H20 & N2 gram weights. From the values given, the sum of %C + %H + %N from original sample > 100%. Which means %Oxy is a negative number. Not workable for empirical formula analysis.
To determine the empirical and molecular formula of the sample, we need to follow a series of steps:
Step 1: Calculate the number of moles of CO2 produced.
To do this, we divide the mass of CO2 produced (1.072 g) by its molar mass (44.01 g/mol):
Number of moles of CO2 = 1.072 g / 44.01 g/mol ≈ 0.0243 mol
Step 2: Calculate the number of moles of H2O produced.
Similarly, divide the mass of H2O produced (0.307 g) by its molar mass (18.015 g/mol):
Number of moles of H2O = 0.307 g / 18.015 g/mol ≈ 0.017 mol
Step 3: Calculate the number of moles of nitrogen produced.
Again, divide the mass of nitrogen produced (0.068 g) by its molar mass (28.013 g/mol):
Number of moles of nitrogen = 0.068 g / 28.013 g/mol ≈ 0.0024 mol
Step 4: Calculate the molar ratio of each element.
Divide the number of moles of each element obtained in steps 1-3 by the smallest number of moles. In this case, the smallest number of moles is 0.0024 mol:
CO2: 0.0243 mol / 0.0024 mol ≈ 10
H2O: 0.017 mol / 0.0024 mol ≈ 7
Nitrogen: 0.0024 mol / 0.0024 mol = 1
Step 5: Find the empirical formula.
The empirical formula represents the simplest whole number ratio of atoms in a compound. Based on the molar ratios calculated in step 4, we can write the empirical formula as:
C10H7N
Step 6: Determine the molecular formula.
The molecular formula represents the actual number of atoms of each element present in a compound. To determine the molecular formula, we need to know the molar mass of the compound. In this case, the molar mass is given as 162 g/mol.
The molar mass of the empirical formula (C10H7N) can be calculated as follows:
(10 * 12.01 g/mol) + (7 * 1.008 g/mol) + (1 * 14.01 g/mol) = 162.15 g/mol
Since the molar mass of the empirical formula is very close to the given molar mass (162 g/mol), the empirical formula is also the molecular formula.
So, the empirical and molecular formula of the sample is C10H7N.
Finally, let's write the balanced equation for the combustion of this compound:
C10H7N + 10.4O2 -> 10CO2 + 4.5H2O + 0.1N2