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Serjik [45]
3 years ago
15

You are asked to identify compound X (a white, crystalline solid), which was extracted from a plant seized by customs inspectors

. An elemental analysis of X shows that the mass percentage composition of the compound is 26.86% C and 2.239% H, with the remainder being oxygen. A mass spectrum of X yields a molar mass of 90.0 g/mol. Write the molecular formula of X.
Chemistry
1 answer:
igor_vitrenko [27]3 years ago
4 0

Answer:

C2H2O4

Explanation:

To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :

C = 26.86%

H = 2.239%

O = 100 - ( 26.86 + 2.239) = 70.901%

We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u

The division is as follows:

C = 26.86/12 = 2.2383

H = 2.239/1 = 2.239

O = 70.901/16 = 4.4313

We now divide each by the smallest number I.e 2.2383

C = 2.2383/2.2383 = 1

H = 2.239/2.2383 = 1

O = 4.4313/2.2383 = 1.98 = 2

Thus, the empirical formula is CHO2.

To get the molecular formula, we use the molar mass .

(CHO2)n = 90

We add the atomic masses multiplied by n.

(12 + 1 + 2(16))n = 90

45n = 90

n = 90/45 = 2.

Thus , the molecular formula is C2H2O4

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Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/
Ivenika [448]

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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