Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of a 2.18 g sample burns in excess oxygen yields 3.94 g of CO2 and 1.89 g of H2O. A separate experiment shows that a 1.23 g sample contains 0.235 g of N. Calculate the moles of C in the sample.

Answer 1

Answer:

1.074 g

Explanation:

The compound contains Carbon, hydrogen, nitrogen and oxygen.

Molar mass of carbon dioxide = 44.009 g/mol

mole of carbon dioxide = mass of carbon dioxide / molar mass = 3.94 g / 44.009 g/mol = 0.0895 mol

mass of carbon = 0.0895 mol  × 12.011 g/mol = 1.074 g

mole of H₂O = 1.89g × 2 / 18.02 g/mol = 0.210

mass  of Hydrogen = 0.210 × 1.008 mole = 0.211 g

since the mass of 1.23g yield 0.235 g

2.18 g will yield 2.18 g ×0.235 g / 1.23 = 0.417 g

mole of nitrogen = 0.417 g / 14.0067 g/mol = 0.0298

Mass of oxygen in the sample = 2.18 - ( 0.417 g + 0.211 g + 1.074 g) = 0.478 g

mole of oxygen present = 0.478 g / 15.999 g/mol = 0.0299 mol

the ratio of C : H : N : O

0.0895 mol  : 0.210 : 0.0298 : 0.0299 mol

C₃H₇NO is the emperical formula

Answer 2

Answer:

1.074 g

Explanation:

Molar mass of CO2 = 12 + (16×2)

= 44 g/mol

Number of moles of CO2 = mass/molar mass = 3.94/44

= 0.0895 mol

mass of carbon = 0.0895 × 12.011

= 1.074 g

Number of mole of H₂O = 1.89 × 2 / 18

= 0.210 mol

mass  of Hydrogen = 0.210 × 1

= 0.210 g

since the mass of 1.23g yield 0.235 g

2.18 g will yield 2.18 g ×0.235 g / 1.23 = 0.417 g

mole of nitrogen = 0.417 g / 14

= 0.0298 mol

Mass of oxygen in the sample = 2.18 - ( 0.417 g + 0.211 g + 1.074 g)

= 0.478 g

mole of oxygen present = 0.478/16 = 0.03 mol.

0.0895  : 0.210 : 0.0298 : 0.0299