Question

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?h n o hint: be careful to get single element mole values for nitrogen and oxygen.

Explanation:

Step1: Calculate moles of C from CO₂

Molar mass of CO₂ = 12.01 + 2×16.00 = 44.01 g/mol.
Moles of CO₂ = $\frac{24.972\ \text{g}}{44.01\ \text{g/mol}}$ = 0.5674 mol.
Since 1 mol CO₂ has 1 mol C, moles of C = 0.5674 mol.

Step2: Calculate moles of H from H₂O

Molar mass of H₂O = 2×1.008 + 16.00 = 18.016 g/mol.
Moles of H₂O = $\frac{6.135\ \text{g}}{18.016\ \text{g/mol}}$ ≈ 0.3405 mol.
Since 1 mol H₂O has 2 mol H, moles of H = 2×0.3405 = 0.6810 mol.

Step3: Calculate moles of N from N₂

Molar mass of N₂ = 2×14.01 = 28.02 g/mol.
Moles of N₂ = $\frac{1.590\ \text{g}}{28.02\ \text{g/mol}}$ ≈ 0.05675 mol.
Since 1 mol N₂ has 2 mol N, moles of N = 2×0.05675 = 0.1135 mol.

Step4: Calculate mass of O in sample

Mass of C = 0.5674 mol × 12.01 g/mol ≈ 6.814 g.
Mass of H = 0.6810 mol × 1.008 g/mol ≈ 0.686 g.
Mass of N = 1.590 g (given as N₂, so mass of N is 1.590 g).
Mass of O = 10.00 g - (6.814 + 0.686 + 1.590) g = 10.00 - 9.09 = 0.91 g.

Step5: Calculate moles of O

Molar mass of O = 16.00 g/mol.
Moles of O = $\frac{0.91\ \text{g}}{16.00\ \text{g/mol}}$ ≈ 0.056875 mol.

Step6: Find the smallest mole value

Smallest moles: N (0.1135) or O (0.056875)? Wait, no—wait, moles of O is ~0.0569, N is ~0.1135, C is 0.5674, H is 0.6810. Wait, no, wait: Wait, moles of O calculation: Wait, mass of O: 10.00 - (C + H + N) = 10 - (6.814 + 0.686 + 1.590) = 10 - 9.09 = 0.91 g. Then moles O = 0.91 / 16 ≈ 0.056875.

Wait, but let's recheck: Wait, moles of N: N₂ is 1.590 g, molar mass 28.02, so moles N₂ = 1.590 / 28.02 ≈ 0.05675, so moles N = 2×0.05675 = 0.1135. Ah, I see, earlier mistake: moles of N₂ is 0.05675, so N is 0.1135 mol.

Now, find the smallest mole value among C (0.5674), H (0.6810), N (0.1135), O (0.056875). The smallest is O: ~0.056875. Wait, no, O is 0.056875, N is 0.1135 (which is ~twice O). Wait, maybe I messed up O calculation. Wait, mass of sample is 10.00 g. Mass of C: 0.5674 mol ×12.01 = ~6.814 g. Mass of H: 0.6810 mol ×1.008 = ~0.686 g. Mass of N: 1.590 g (since N₂ is 1.590 g, so N is 1.590 g). So total mass of C + H + N = 6.814 + 0.686 + 1.590 = 9.09 g. Therefore, mass of O = 10.00 - 9.09 = 0.91 g. Moles of O = 0.91 / 16 = 0.056875 mol.

Now, divide each mole by the smallest mole (O: 0.056875) to get ratios:

C: 0.5674 / 0.056875 ≈ 9.977 ≈ 10.
H: 0.6810 / 0.056875 ≈ 11.97 ≈ 12.
N: 0.1135 / 0.056875 ≈ 1.996 ≈ 2.
O: 0.056875 / 0.056875 = 1.

Wait, but wait, maybe the smallest is N? Wait, no, N is 0.1135, O is 0.056875 (half of N). Wait, maybe I made a mistake in O mass. Wait, let's recalculate mass of O:

Mass of C: from CO₂: 24.972 g CO₂ × (12.01 / 44.01) = 24.972 × 0.2729 ≈ 6.814 g (correct).

Mass of H: from H₂O: 6.135 g H₂O × (2×1.008 / 18.016) = 6.135 × 0.1119 ≈ 0.686 g (correct).

Mass of N: 1.590 g (correct, since N₂ is 1.590 g, so N is 1.590 g).

Total: 6.814 + 0.686 + 1.590 = 9.09 g. So O is 10.00 - 9.09 = 0.91 g (correct). Moles O: 0.91 / 16 = 0.056875 (correct).

Now, moles:

C: 0.5674, H: 0.6810, N: 0.1135, O: 0.056875.

Divide each by O's moles (0.056875):

C: 0.5674 / 0.056875 ≈ 9.977 ≈ 10.

Ah, so the subscript for C is 10.

Wait, let's check again. Moles of C: 24.972 g CO₂. Molar mass CO₂ 44.01, so moles CO₂ = 24.972 / 44.01 = 0.5674 mol. So C is 0.5674 mol.

Moles of O from CO₂: 2×0.5674 = 1.1348 mol.

Moles of O from H₂O: 0.3405 mol (since H₂O is 0.3405 mol, O is 0.3405 mol).

Moles of O from N₂? No, N₂ has no O. So total O in products: 1.1348 + 0.3405 = 1.4753 mol.

But O in sample: O in sample = total O in products - O from O₂ (but wait, combustion: sample + O₂ → CO₂ + H₂O + N₂. So O in sample = (O in CO₂ + O in H₂O) - O in O₂. But we don't know O₂. Wait, earlier method was wrong! Oh no, I made a mistake. The correct way to find O in the sample is: mass of sample - mass of C - mass of H - mass of N. Which is what I did before: 10 - 6.814 - 0.686 - 1.590 = 0.91 g. That is correct, because mass is conserved. So O in sample is 0.91 g, moles O = 0.91 / 16 = 0.056875 mol. That is correct.

So going back, moles:

C: 0.5674

H: 0.6810

N: 0.1135 (from N₂: 1.590 g N₂, moles N₂ = 1.590 / 28.02 = 0.05675, so N is 2×0.05675 = 0.1135)

O: 0.056875 (from sample: 10 - 6.814 - 0.686 - 1.590 = 0.91 g, moles 0.91 / 16)

Now, to find the empirical formula, we divide each mole by the smallest mole. The smallest mole is O: 0.056875.

C: 0.5674 / 0.056875 ≈ 9.977 ≈ 10

H: 0.6810 / 0.056875 ≈ 11.97 ≈ 12

N: 0.1135 / 0.056875 ≈ 1.996 ≈ 2

O: 0.056875 / 0.056875 = 1

So the ratios are approximately C:10, H:12, N:2, O:1. So empirical formula is C₁₀H₁₂N₂O. Wait, but let's check the moles again. Wait, N: 0.1135 / 0.056875 ≈ 2, which is correct. C: 0.5674 / 0.056875 ≈ 10, correct. So the subscript for C is 10.

Answer:

10