Question

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 6.00 g of this compound produced 8.80 g of carbon dioxide and 3.60 g of water.

Part A:How many moles of carbon,C , were in the original sample?Express your answer numerically in moles.Part B:How many moles of hydrogen,H , were in the original sample?Express your answer numerically in moles.

Answer 1

Answer:

There were 0.20 moles Carbon in the original sample

There were 0.40 moles hydrogen in the original sample

Explanation:

Step 1:  The balanced equation

CxHyOz + O2 → CO2 + H2O

Step 2: Data given

mass of the unknown compound = 6.00 g

mass of carbon dioxide produced = 8.80 g

mass of water produced = 3.60 g

Molar mass of Carbon = 12.01 g/mole

Molar mass of Hydrygen = 1.01 g/mole  ; 2.02 for H2

Molar mass of Oxygen = 16 g/mole

Molar mass of CO2 = 44.01 g/mole

Molar mass of H2O = 18.02 g/mole

Step 3: Calculate mass

8.80 g CO2 * 12.01 g/mole / 44.01 g/mole = 2.40 g carbon

3.60 g H2O * 2.02 g/mole / 18.02 g/mole = 0.404 g hydrogen

6.00 g total - 2.40 g carbon - 0.404 g hydrogen = 3.196 g oxygen

Step 4: Calculate number of moles

moles = mass / Molar mass

Mass of  Carbon = 2.40g / 12.01 g/mole = 0.20 mole Carbon

Mass of hydrogen = 0.40 g / 1.01 g/mole = 0.40 mole Hydrogen