Question

A 0.165 R of a C.H.O. compound was combusted in air to make 0.403 gofC0, and 0.165 H:0. What is the empirical formula? Balance the chemical equation. Answers: (1) CH.O, 1,4,3,3.(2) CH3, 4, 11, 8, 6,(3) CH.O,1,3,2,3,(4) CH,0, 2, 11. 8,8

Answer 1

Answer:

The empirical formula is = $C_4H_8O$

Balanced reaction is:

$2C_4H_8O+11O_2\rightarrow 8CO_2+8H_2O$

Explanation:

Mass of water obtained = 0.165 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.165 g /18 g/mol = 0.009167 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.009167 = 0.01833 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01833 x 1.008 = 0.01848 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g  /44.01 g/mol = 0.009157 moles

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

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the compound only contains hydrogen, oxygen and carbon. So,

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

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.01848 = 0.03652 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.036412  / 15.999  = 0.002278 moles

Taking the simplest ratio for H, O and C as:

0.01833 : 0.002278 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

Balanced reaction is:

$2C_4H_8O+11O_2\rightarrow 8CO_2+8H_2O$