A0.2121g sample of an organic compound was burned in a stream of oxygen and the carbon dioxide produced was collected in a stream of barium hydroxide . calculate the percentage carbon in the sample if 0.6006g barium carbonate was formed

The answer to your question is: 17.26% of carbon

Explanation:

Data

CxHy = 0.2121 g

BaCO₃ = 0.6006 g

Molecular mass BaCO₃ = 137 + 12 + 48 = 197 g

Reaction

                    CO₂  +  Ba(OH)₂  ⇒   BaCO₃  +  H₂O

Process

1.- Find the amount of carbon in BaCO₃

                              197 g of BaCO₃    12 g of Carbon

                               0.6006 g              x

                               x = (0.6006 x 12) / 197

                               x = 0.0366 g of carbon

2.- Calculate the percentage of carbon in the organic compound

           0.2121 g of organic compound    100%

           0.0366g                                          x

                              x = (0.0366 x 100) / 0.2121

                              x = 17.26%

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explanation:

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm