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7 November, 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.

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  1. 7 November, 16:03
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    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.
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