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A 34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon dioxide and 76.5 grams of water. What is a possible molecular formula of the hydro-carbon?

Respuesta :

joseaaronlara

Answer:

The answer to your question is  CH₄

Explanation:

Process

1.- Write the possible chemical reaction

                   CxHy + O₂   ⇒    CO₂  +   H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of COâ‚‚ = 12 + 32 = 44 g

                         44 g of CO₂ ------------------- 12 g of carbon

                         93.5 g of CO₂ ----------------  x

                          x = (93.5 x 12) / 44

                          x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

                         18 g of water -------------   2 g of hydrogen

                          76.5 g of water --------    x g of hydrogen

                           x = (76.5 x 2) / 18

                           x = 8.5 g of hydrogen

4.- Calculate the moles of carbon and hydrogen in the sample

                           12g of carbon ------------- 1 mol

                           25.5g of carbon ---------- x

                             x = (25.5 x 1) / 12

                              x = 2.12 moles of carbon

                            1 g of hydrogen ----------- 1 mol

                            8.5 g of hydrogen -------- x

                             x = (8.5 x 1) / 1

                             x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

                                                    CH₄                                        

Answer:

[tex]CH_4[/tex]

Explanation:

Given that:-

Mass of carbon dioxide obtained = 93.5 grams

Molar mass of carbon dioxide = 44.01 g/mol

Moles of [tex]CO_2[/tex] = 93.5 g  /44.01 g/mol = 2.12451 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 2.12451 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 2.12451 x 12.0107 = 25.51685 g

Given that the compound is a hydrocarbon which only contains hydrogen and carbon. So,

Mass of H in the sample = Total mass - Mass of C

Mass of the sample = 34.0 g

Mass of H in sample = 34.0 - 25.51685 g = 8.48315 g  

Molar mass of H = 1.0078 g/mol

Moles of H  = 8.48315 / 1.0078  = 8.41749 moles

Taking the simplest ratio for H and C as:

8.41749 : 2.12451  = 4 : 1

The possible molecular formula of the hydrocarbon formula is = [tex]CH_4[/tex]

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