A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 , 1.09 g h 2 o, and 1.70 g n 2 . if the molar mass of the compound is 129.1 g/mol, what is the chemical formula of the compound?

Question

Raymond Ward

Chemistry

19 March, 01:59

A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 , 1.09 g h 2 o, and 1.70 g n 2 . if the molar mass of the compound is 129.1 g/mol, what is the chemical formula of the compound?

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Answers (1)

Know the Answer?

Dean Bright · Answer 1

Answer is: C₃H₃N₃O₃.

Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.

m (CₐHₓNₓ) = 5,214 g.

m (CO₂) = 5,34 g.

m (H₂) = 1,09 g.

m (N₂) = 1,70 g.

n (CO₂) = n (C) = 5,34 g : 44 g/mol = 0,121 mol.

n (H₂O) = 1,09 g : 18 g/mol = 0,06 mol.

n (H) = 2 · 0,0605 mol = 0,121 mol.

n (N₂) = 1,7 g : 28 g/mol = 0,0607 mol.

n (N) = 0,0607 mol · 2 = 0,121 mol.

n (C) : n (H) : n (N) = 0,121 mol : 0,121 mol : 0,121 mol / : 0,121

n (C) : n (H) : n (N) = 1 : 1 : 1.

M (CHN) = 27 g/mol.

m (O₂) = 8,13 g - 5,214 g = 2,914 g.

n (O₂) = 2,914 g : 32 g/mol = 0,09 mol.

n (CₓHₓNₓOₓ) = 5,214 g : 129,1 g/mol = 0,0404 mol.

n (CₓHₓNₓOₓ) : n (CO₂) = 1 : 3.