Question

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOzCxHyOz). Combustion of 5.50 gg of this compound produced 8.07 gg of carbon dioxide and 3.30 gg of water. How many moles of carbon, C, were in the original sample? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

Moles of Carbon in the product = 0.183 mol

Explanation:

Complete combustion of an organic compound in the presence of excess oxygen will give carbon dioxide (CO2) and water vapour(H2O).

Equation of reaction

CxHyOz(s) + (2x + y/2 - z)/2 O2(g) --> x CO2(g) + y/2 H2O(l)

Moles of products

CO2

Molar mass of CO2 = 12 + (16*2)

= 44 g/mol

n(CO2) = 8.07/44.0

= 0.183 mol

One compound of CO2 has 1 Carbon atom and 2 Oxygen atom.

So if we have,

0.183 moles of carbon dioxide then

0.183 moles of carbon.

The mass of carbon in the product, m = 0.183*12

= 2.20 g

H2O:

Molar mass of H2O = (1*2) + 16

= 18 g/mol

Number of moles of H2O =

= 3.3/18

= 0.183 mol

One compound of H2O has 2 Hydrogen atom and 1 Oxygen atom.

0.183 moles of water then we also have

= 2*(0.183)

= 0.367 moles of hydrogen in the sample.

The mass of hydrogen in the compound, m

= 0.367*1

= 0.367 g

Adding these two values together will give us the mass of our compound that C and H;

2.2 + 0.367

= 2.567 g

So for the Oxygen,

5.5 g - 2.567

= 2.933 g

Moles of Oxygen;

Molar mass of O = 16 g/mol

= 2.933/16

= 0.183 mol

Answer 2

Answer:

Explanation:

8.07gCO2/44 g/mol= 0.183 mol

Therefor 0.183 mol C