Unknown compound has only C,H, and O (CxHyOz). Combustion of 6.50 g of this compound produced 9.53 g of carbon dioxide and 3.90 g of water.?
How many moles of carbon, C, were in the original sample?
How many moles of hydrogen, H, were in the original sample?
If 6.50 g of the unknown compound contained 0.217 mol of C and 0.433 mol of H, how many moles of oxygen, O, were in the sample?
- hcbiochemLv 71 month agoFavorite Answer
moles and mass C:
9.53 g CO2 / 44.0 g/mol X 1 mol C/1 mol CO2 = 0.217 mol C
0.217 mol C X 12 g/mol = 2.60 g C
Moles and mass H:
3.90 g H2O / 18.0 g/mol X (2 mol H/1 mol H2O) = 0.433 mol H
0.433 mol H X 1.008 g/mol = 0.437 g H
Completing the rest of the likely question:
mass and moles O:
mass O = 6.50 g - (2.60 + 0.44) = 3.46 g O
moles O = 3.46 g / 16 g/mol = 0.216
empirical formula = CH2O
- King LeoLv 71 month ago
— — How many moles of carbon, C, were in the original sample?
➤ moles of carbon equal moles of carbon dioxide produced
n( C ) = n( CO₂ )
= ( 9.53g / 44gmol⁻¹ )
= 0.217 mol
— — How many moles of hydrogen, H, were in the original sample?
➤ moles of hydrogen equal twice the moles of water produced
n( H ) = 2 * n( H₂O )
= 2 ( 3.90g / 18gmol⁻¹ )
= 0.433 mol