Answer:
The empirical formula is C7H5N3O6
Explanation:
Step 1: Data given
Mass of sample = 0.150 grams
Mass of CO2 = 0.204 grams
Molar mass CO2 = 44.01 g/mol
Mass of H2O = 0.030 grams
Molar mass H2O = 18.02 g/mol
Molar mass C = 12.01 g/mol
Molar mass H = 1.01 g/mol
Molar mass O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 0.204 grams / 44.01 g/mol
Moles CO2 = 0.00464 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.00464 moles we have 0.00464 moles C
Step 4: Calculate mass C
Mass C = 0.00464 moles * 12.01 g/mol
Mass C = 0.0557 grams
Step 5: Calculate moles H2O
Moles H2O = 0.030 grams / 18.02 g/mol
Moles H2O = 0.00166 moles
Step 6: Calculate moles H
For 1 mol H2O we have 2 moles H
For 0.00166 moles H2O we have 2* 0.00166 = 0.00332 moles H
Step 7: Calculate mass H
Mass H = 0.00332 moles * 1.01 g/mol
Mass H = 0.00335 grams
Step 8: Calculate mass N
Mass N = 0.185 * 0.150 grams
Mass N = 0.02775 grams
Step 9: Calculate moles N
Moles N = 0.02775 grams / 14.0 g/mol
Moles N = 0.00198 moles
Step 10: Calculate mass O
Mass O = 0.150 grams - 0.02775 - 0.00335 - 0.0557
Mass O = 0.0632 grams
Step 11: Calculate moles O
Moles O = 0.0632 grams / 16.0 g/mol
Moles O = 0.00395 moles
Step 11: Calculate mol ratio
We divide by the smallest amount of moles
C: 0.00464 moles / 0.00198 moles =2.33
H: 0.00332 moles / 0.00198 moles = 1.66
N: 0.00198 moles / 0.00198 moles = 1
O: 0.00395 moles / 0.00198 moles = 2
For 1 mol N we have 2.33 moles C, 1.66 moles H and 2 moles O
OR
For 3 moles N we have 7 moles C, 5 moles H and 6 moles O
The empirical formula is C7H5N3O6
