You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2 and 0.030 g H2O (we'll assume that these are the products, along with nitrogen). You know that TNT is made up solely of C, H, O, and N, and that it contains 18.5% nitrogen by mass.

Question

Chemistry

Deandre Vargas

Today, 14:33

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2 and 0.030 g H2O (we'll assume that these are the products, along with nitrogen). You know that TNT is made up solely of C, H, O, and N, and that it contains 18.5% nitrogen by mass.

+2

Answers (1)

Know the Answer?

Davian · Answer 1

Answer is: the formula is C₇N₃O₆H₅.

1) M (TNT) = 0.150 g; mass of the trinitrotoluene.

ω (N) = 18.5% : 100%.

ω (N) = 0.185; mass percentage of the nitrogen.

m (N) = 0.150 g · 0.185.

m (N) = 0.02775 ·; mass of the nitrogen.

n (N) = 0.02775 g : 14 g/mol.

n (N) = 0.002 mol; amount of the nitrogen.

n (CO₂) = 0.204 g : 44 g/mol.

n (CO₂) = 0.0046 mol.

n (C) = n (CO₂) = 0.0046 mol; amount of the carbon.

m (C) = 0.0046 mol · 12 g/mol.

m (C) = 0.0552 g; mass of the carbon.

n (H₂O) = 0.030 g : 18 g/mol.

n (H₂O) = 0.00166 mol.

n (H) = 2 · n (H₂O) = 0.0033 mol; amount of the hydrogen.

m (H) = 0.0033 mol · 1 g/mol.

m (H) = 0.0033 g; mass of the hydrogen.

2) m (O) = m (TNT) - m (N) - m (C) - m (H).

m (O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m (O) = 0.06375 g.

n (O) = 0.06375 g : 16 g/mol.

n (O) = 0.004 mol; amount of oxygen.

n (C) : n (N) : n (O) : n (H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n (C) : n (N) : n (O) : n (H) = 2.33 : 1 : 2 : 1.66 / *3.

n (C) : n (N) : n (O) : n (H) = 7 : 3 : 6 : 5.