This is a multi-step problem. The solution involves the following steps:
STEP 1: To determine the mass of the butane in the lighter, recall the definition of density:
[tex]density = \frac{mass}{volume}[/tex]
Since the density and the volume of butane are given, the mass can be calculated as follows:
[tex]mass = density \times volume\\mass \ of \ C_4H_{10} \ = 0.601 \frac{g}{mL} \times 7.25 \ mL\\\boxed {mass \ of \ C_4H_{10} \ = 4.35725 \ g}[/tex]
STEP 2: Use the molar mass of butane (58.124 g/mol) to determine the moles of the sample.
[tex]moles \ of \ C_4H_{10} \ = \frac{4.35725 \ g}{58.124 \frac{g}{mol}}\\ \\\boxed {moles \ of \ C_4H_{10} \ = 0.074965 \ mol}\\[/tex]
STEP 3: Determine how many moles of C are in the given moles of the butane sample. Use the mole ratio indicated in the chemical formula: 1 mol of C₄H₁₀ = 4 mol of C
[tex]moles \ of \ C \ = 0.074965 \ mol \ C_4H_{10} \times \frac{4 \ mol \ C}{1 \ mol \ C_4H_{10}}\\\\\boxed {moles \ of \ C \ = 0.29986 \ mol}[/tex]
STEP 4: Use the molar mass of carbon to get the mass of carbon in the sample.
[tex]mass \ of \ C \ = 0.29986 \ mol \ C \times \frac{12.011 \ g \ C}{1 \ mol \ C}\\\\\boxed {mass \ of \ C \ = 3.60162 \ g}[/tex]
Expressed with 3 significant figures, the final answer should be:
[tex]\boxed {\boxed {mass \ of \ C \ = 3.60 \ g}}[/tex]
Alternatively, all four steps may be expressed in the same equation:
[tex]mass \ of \ C = 7.25 \ mL \ C_4H_{10} \times \frac{0.601 \ g}{1 \ mL \ C_4H_{10}} \times \frac{1 \ mol \ C_4H_{10}}{58.124 \ g C_4H_{10}} \times \frac{4 \ mol \ C}{1 \ mol \ C_4H_{10}} \times \frac{12.011 \ g \ C}{1 \ mol \ C}\\\\mass \ of \ C \ = \frac{209.3397 \ g}{58.124}\\\\mass \ of \ C \ = 3.601606 \ g\\\\\boxed {\boxed {mass \ of \ C \ = 3.60 \ g}}[/tex]
