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?

Update:

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?

2 Answers

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  • 1 month ago
    Favorite 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

  • 1 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

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