a neurotransmitter consists of carbon, hydrogen, nitrogen, and oxygen. 24.972 grams of carbon dioxide, 6.135…

a neurotransmitter consists of carbon, hydrogen, nitrogen, and oxygen. 24.972 grams of carbon dioxide, 6.135 grams of water, and 1.590 grams of nitrogen (n₂) are recovered from the combustion of a 10.00 gram sample. determine the empirical formula of the substance. what is the subscript for c? c?hno hint: be careful to get single element mole values for nitrogen and oxygen.

a neurotransmitter consists of carbon, hydrogen, nitrogen, and oxygen. 24.972 grams of carbon dioxide, 6.135 grams of water, and 1.590 grams of nitrogen (n₂) are recovered from the combustion of a 10.00 gram sample. determine the empirical formula of the substance. what is the subscript for c? c?hno hint: be careful to get single element mole values for nitrogen and oxygen.

Answer

Explanation:

Step1: Calculate moles of C from CO₂

Molar mass of (CO_2) is (44.01\ g/mol). Moles of (CO_2=\frac{24.972\ g}{44.01\ g/mol}\approx0.5674\ mol). Since 1 mole of (CO_2) has 1 mole of C, moles of C = 0.5674 mol.

Step2: Calculate moles of H from H₂O

Molar mass of (H_2O) is (18.02\ g/mol). Moles of (H_2O=\frac{6.135\ g}{18.02\ g/mol}\approx0.3404\ mol). Each (H_2O) has 2 H, so moles of H = (2\times0.3404 = 0.6808) mol.

Step3: Calculate moles of N from N₂

Molar mass of (N_2) is (28.02\ g/mol). Moles of (N_2=\frac{1.590\ g}{28.02\ g/mol}\approx0.05674\ mol). Each (N_2) has 2 N, so moles of N = (2\times0.05674 = 0.1135) mol.

Step4: Calculate mass of O in sample

Mass of C: (0.5674\ mol\times12.01\ g/mol\approx6.814\ g)
Mass of H: (0.6808\ mol\times1.008\ g/mol\approx0.686\ g)
Mass of N: (1.590\ g) (given as (N_2) mass, so mass of N is 1.590 g)
Total mass of C + H + N = (6.814 + 0.686 + 1.590 = 9.09\ g)
Mass of O = sample mass - (C + H + N) = (10.00 - 9.09 = 0.91\ g)
Molar mass of O is (16.00\ g/mol), moles of O = (\frac{0.91\ g}{16.00\ g/mol}\approx0.0569\ mol)

Step5: Find the smallest mole value (O: ~0.0569, N: ~0.1135, H: ~0.6808, C: ~0.5674)

Divide each mole by 0.0569:
C: (\frac{0.5674}{0.0569}\approx9.97\approx10)
H: (\frac{0.6808}{0.0569}\approx11.96\approx12)
N: (\frac{0.1135}{0.0569}\approx1.99\approx2)
O: (\frac{0.0569}{0.0569}=1)

So the empirical formula is (C_{10}H_{12}N_2O), subscript for C is 10.

Answer:

10