Question 27 Unsaved
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<h2> 162g/mol</h2>Explanation:
The question is incomplete. The complete question includes the information to find the empirical formula of nicotine:
<em>Nicotine has the formula </em><em> . To determine its composition, a sample is burned in excess oxygen, producing the following results:</em>
- <em>1.0 mol of CO₂</em>
- <em>0.70 mol of H₂O</em>
- <em>0.20 mol of NO₂</em>
<em>Assume that all the atoms in nicotine are present as products </em>
<h2>Solution</h2>To find the empirical formula you need to find the moles of C, H, and N in each of the compound.
- 1.0 mol of CO₂ has 1.0 mol of C
- 0.70 mol of H₂O has 1.4 mol of H
- 0.20 mol of NO₂ has 0.20 mol of N
Thus, the ratio of moles is:
- C: 1.0
- H: 1.4
- N: 0.20
Divide all by the smallest number: 0.20
- C: 1.0 / 0.20 = 5
- H: 1.4 / 0.20 = 7
- N: 0.20 / 0.20 = 1
Hence, the empirical formula is C₅H₇N
Find the mass of 1 mole of units of the empirical formula:
- C: 5mol × 12g/mol = 60g
- H: 7mol × 1g/mol = 7 g
- N: 1 mol × 14g/mol = 14g
Total mass = 60g + 7g + 14g = 81g
Two moles of units of the empirical formula weighs 2 × 81g = 162g and three units weighs 3 × 81g = 243 g.
Thus, since the molar mass is between 150 and 180 g/mol, the correct molar mass is 162g/mol and the molecular formula is twice the empirical formula: C₁₀H₁₄N₂.