Answer:
The empirical formula is CH6N2
Explanation:
A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound
Step 1: Data given
Mass of the compound = 35.0 mg = 0.035 grams
Mass of CO2 = 33.5 mg = 0.0335 grams
Mass of H2O = 41.1 mg = 0.0411 grams
Molar mass CO2 = 44.01 g/mol
Molar mass H2O = 18.02 g/mol
Molar mass C = 12.01 g/mol
Molar mass O = 16.0 g/mol
Molar mass H = 1.01 g/mol
Step 2: Calculate moles CO2
Moles CO2 = 0.0335 grams / 44.01 g/mol
Moles CO2 = 7.61 *10^-4 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C
Step 4: Calculate mass C
Mass C = 7.61 *10^-4 moles * 12.01 g/mol
Mass C = 0.00914 grams = 9.14 mg
Step 5: Calculate moles H2O
Moles H2O = 0.0411 grams / 18.02 g/mol
Moles H2O = 0.00228 moles
Step 6: Calculate moles H
For 1 mol H2O we have 2 moles H
For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H
Step 7: Calculate mass H
Mass H = 0.00456 moles * 1.01 g/mol
Mass H = 0.00461 grams = 4.61 mg
Step 8: Calculate mass N
Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams
Step 9: Calculate moles N
Moles N = 0.02125 grams / 14.0 g/mol
Moles N = 0.00152 moles
Step 10: Calculate the mol ratio
We divide by the smallest amount of moes
C: 0.000761 moles / 0.000761 moles= 1
H: 0.00456 moles / 0.000761 moles = 6
N: 0.00152 moles / 0.000761 moles = 2
For every C atom we have 6 H atoms and 2 N atoms
The empirical formula is CH6N2
= 6.02x10⁻⁶