An unknown substance contains 71 percent carbon and 16 percent ...?

Method 1:

Mole ratio C : H : N
= (71/12) : [(100 - 71 - 16)/1] : (16/14) :
= 5.92 : 13 : 1.14
= (5.92/1.14) : (13/1.14) : (1.14/1.14)
≈ 5 : 11: 1

Let molecular formula = (C₅H₁₁N)ₙ

Molecular mass:
(12×5 + 1×11 + 14) × n = 85
85n = 85
Molecular formula = C₅H₁₁N

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Method 2:

In 1 mole of the compound:
Moles of C = (85 × 71%) / 12 ≈ 5
Moles of H = [85 × (100 - 71 - 16)%] / 1 ≈ 11
Moles of N = (85 × 16%) / 14 ≈ 1

Molecular formula = C₅H₁₁N

since you have molar mass, you can convert %'s directly to the formula

.. .. .. 71g C.. ... ... 1 mol C.. .. ... 85g CxHyNz
 ----- ----- ---- ---- x ----- ---- ---- x ----- ---- ---- ---- = 5 mol C / mol CxHyNz
.. 100g CxHyNz ... 12.011g C... 1 mol CxHyNz

.. .. .. 16g N.. ... ... 1 mol N.. ..... 85g CxHyNz
 ----- ----- ---- ---- x ---- ---- ---- x ----- ---- ---- ---- = 1 mol N / mol CxHyNz
.. 100g CxHyNz ... 14.01g N.... 1 mol CxHyNz

.. .. .. 13g H.. ... ... 1 mol H.. ..... 85g CxHyNz
 ----- ----- ---- ---- x ---- ---- ---- x ----- ---- ---- ---- = 11 mol H / mol CxHyNz
.. 100g CxHyNz ... 1.008g H.... 1 mol CxHyNz

formula is C5H11N