17 mg of a hydrocarbon (M.F. C₁₀H₁₆) takes up 8.40 mL of the H₂ gas measured at 0°C and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields
CH₃-C-CH₃, H-C-H, H-C-CH₂-CH₂-C-C-H
The number of double bond/s present in the hydrocarbon is ___________.

Solution

Chemistry ma’am · 4 steps

Each C=C takes up one H2, so find how many moles of H2 one mole of hydrocarbon absorbs.

1. Moles of hydrocarbon. The molar mass of C₁₀H₁₆ is 120 + 16 = 136 g mol⁻¹. So 17 mg is 0.017/136 = 1.25 × 10⁻⁴ mol.
2. Moles of H2. At STP one mole of gas occupies 22400 mL. So 8.40 mL of H₂ is 8.40/22400 = 3.75 × 10⁻⁴ mol.

   n(H₂)/n(C₁₀H₁₆) = (3.75 × 10⁻⁴)/(1.25 × 10⁻⁴) = 3

3. Check with ozonolysis. C₁₀H₁₆ has (2(10) + 2 - 16)/2 = 3 degrees of unsaturation, and all 3 are used up by H₂, so there is no ring. The ozonolysis products contain only C=O groups, so these are C=C bonds, not triple bonds. Getting 10 carbons needs HCHO twice, which fits myrcene, (CH₃)₂C=CH-CH₂-CH₂-C(=CH₂)-CH=CH₂.
4. So. One mole of the hydrocarbon takes up 3 moles of H₂, so it has 3 double bonds.

The idea to keep. Moles of H2 absorbed per mole of compound counts the C=C bonds. Then let ozonolysis tell you where those bonds sit, with each C=C giving two C=O ends.

Answer from the NTA final answer key. Solution by PrepWiser.