Question

1. A compound containing only C, H and O was subjected to combustion analysis. A sample of 6.130×10¹ g produced 1.625×10² g of CO₂ and 7.603×10¹ g of H₂O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.
C? H? O?

2.If the molar mass of the compound is 464.816 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.

C? H? O?

3.When a 0.339 g sample of Cu(ClO₃)₂·xH₂O is heated to remove all water, a mass of 0.230 g remains. Determine the value of x and report it in the answer box below. Only answers that are integers are allowed.

Answer 1

1)

let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 162.5/44

= 3.693

Number of moles of H2O = mass of H2O / molar mass H2O

= 76.03/18

= 4.224

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 3.693

so, x = 3.693

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*4.224 = 8.448

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 61.3 - 3.693*12 - 8.448*1

= 8.534

number of mol of O = mass of O / molar mass of O

= 8.534/16.0

= 0.5334

so, z = 0.5334

Divide by smallest to get simplest whole number ratio:

C: 3.693/0.5334 = 7

H: 8.448/0.5334 = 16

O: 0.5334/0.5334 = 1

So empirical formula is:C7H16O

Answer: C7H16O

2)

Molar mass of C7H16O,

MM = 7*MM(C) + 16*MM(H) + 1*MM(O)

= 7*12.01 + 16*1.008 + 1*16.0

= 116.198 g/mol

Now we have:

Molar mass = 464.816 g/mol

Empirical formula mass = 116.198 g/mol

Multiplying factor = molar mass / empirical formula mass

= 464.816/116.198

= 4

So molecular formula is:C28H64O4

Answer: C28H64O4

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