Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 gCO_(2) and 1.3102 gH_(2)O. Find the empirical formula of the compound.The answer I got was CH_(16)O, but I'm not sure if I did it right or not. Can someone confirm this for me?

Answers

Answer 1
Answer:

Answer:

The empirical formula of the compound is C_4H_8O_1

Explanation:

Moles of carbon dioxide = (3.2007 g)/(44 g/mol)=0.07274 mol

Moles of carbon in sample = 1* 0.07274 mol=0.07274 mol

Moles of water = (1.3101 g)/(18 g/mol)=0.07279 mol

Moles of hydrogen in sample = 2* 0.07279 mol=0.14558 mol

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of oxygen

Mass of oxygen =

1.3109 g - (12 g/mol × 0.07274 mol) - (1 g/mol× 0.14558 mol)

Mass of oxygen = 0.29244 g

Moles of oxygen in sample = (0.29244 g)/(16 g/mol)=0.01828 mol

For empirical formula divide the moles of element which are is less amount with all the moles of every element.

Carbon =(0.07274 mol)/(0.01828 mol)=3.9\approx 4

Hydrogen =(0.14558 mol)/(0.01828 mol)=7.9\approx 8

oxygen=(0.01828 mol)/(0.01828 mol)=1

The empirical formula of the compound is C_4H_8O_1

Answer 2
Answer:

Answer:

The answer to your question is:   C₃H₃O  This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

                                     44 g -------------------- 12 g

                                     3.2007 g ------------    x

                                      x = (3.2007 x 12) / 44

                                      x = 0.8729 g of Carbon

                                     12 g of C --------------  1 mol

                                     0.8729 g --------------  x

                                     x = (0.8729 x 1) / 12

                                     x = 0.0727 mol of Carbon

For Hydrogen

                                 18 g ---------------------- 1 g

                              1.3102 g -------------------  x

                                   x = (1.3102 x 1) / 18

                                  x = 0.0727 g of Hydrogen                                      

                                  1 g ------------------------ 1 mol

                                  0.0727g ----------------  x

                                  x = (0.0727 x 1)/1

                                  x = 0.0727 mol of Hydrogen

For oxygen

                    g of Oxygen = g of sample - g of Carbon - g of hydrogen

                    g of Oxygen = 1.3109 - 0.8709 - 0.0727

                    g of Oxygen = 0.3673

                                  16 g of Oxygen ------------- 1 mol of O

                                  0.3673 g ---------------------   x

                                   x = (0.3673 x 1)/ 16

                                   x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon              0.0727 / 0.023  = 3.1  ≈ 3

Hydrogen         0.0727 / 0.023 = 3.1  ≈ 3

Oxygen             0.0230 / 0.023 = 1

                                        C₃H₃O


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