Question

A 2.70 g of a C.H.O. compound was combusted in air to make 6.14 g of CO2 and 2.51 g H:0. What is the empirical formula? Balance the chemical equation C. 12.0ige Gilgg cho = 1.675 goed 0.13% (1.67670M)-2.70 -0.884 gind H. 44,0lgustus Hs holt x 2.5lg kyo = 0.141 glad 941 0.055 01396 med 2254

Answer 1

Answer:

The empirical formula is = $C_3H_6O$

Balanced reaction is:

$C_3H_6O+4O_2\rightarrow 3CO_2+3H_2O$

Explanation:

Mass of water obtained = 2.51 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 2.51 g /18 g/mol = 0.1394 moles

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

Moles of H = 2 x 0.1394 = 0.2789 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.2789 x 1.008 = 0.2811 g

Mass of carbon dioxide obtained = 6.14 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 6.14 g  /44.01 g/mol = 0.1395 moles

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

Moles of C = 0.1395 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.1395 x 12.0107 = 1.6755 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 = 2.70 g

Mass of O in sample = 2.70 - 1.6755 - 0.2811 = 0.7434 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 2.3028 / 15.999  = 0.0464 moles

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

0.2789 : 0.0464 : 0.1395

= 6 : 1 : 3

The empirical formula is = $C_3H_6O$

Balanced reaction is:

$C_3H_6O+4O_2\rightarrow 3CO_2+3H_2O$