Question

2. (2 Points) Cyanidin, contains carbon, hydrogen, and oxygen. The complete combustion of a 2.0000 g sample in excess O2 produces 4.5964 g CO2 and 0.6901 g H2O. Cyanidin's molecular mass is approximately 290 g/mol. What is the molecular formula of cyanidin?

Answer 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

= 4.5964/44

= 0.1045

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

= 0.6901/18

= 3.834*10^-2

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.1045

so, x = 0.1045

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*3.834*10^-2 = 7.668*10^-2

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 2.0 - 0.1045*12 - 7.668*10^-2*1

= 0.6698

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

= 0.6698/16.0

= 4.186*10^-2

so, z = 4.186*10^-2

Divide by smallest:

C: 0.1045/4.186*10^-2 = 2.5

H: 7.668*10^-2/4.186*10^-2 = 1.83 = 11/6

O: 4.186*10^-2/4.186*10^-2 = 1

Multiply by 6 to get simplest whole number ratio:

C: 2.5*6 = 15

H: 11/6*6 = 11

O: 1*6 = 6

So empirical formula is:C15H11O6

Molar mass of C15H11O6,

MM = 15*MM(C) + 11*MM(H) + 6*MM(O)

= 15*12.01 + 11*1.008 + 6*16.0

= 287.238 g/mol

Now we have:

Molar mass = 290.0 g/mol

Empirical formula mass = 287.238 g/mol

Multiplying factor = molar mass / empirical formula mass

= 290.0/287.238

= 1

So molecular formula is:C15H11O6

Answer: C15H11O6