/> The moles of CO₂ produced can be calculated by dividing the mass of CO₂ produced by its molar mass (44.01 g/mol):

$$\text{moles of CO}_{2} = \frac{0.220 \,\mathrm{g}}{44.01 \,\mathrm{g/mol}} = 0.005 \, \mathrm{mol}$$

Since each mole of CO₂ contains one mole of carbon, there are 0.005 moles of carbon in the compound.

Now, let's calculate the moles of H₂O produced:

$$\text{moles of H}_{2}\text{O} = \frac{0.126 \,\mathrm{g}}{18.02 \,\mathrm{g/mol}} = 0.007 \, \mathrm{mol}$$

Since each mole of H₂O contains two moles of hydrogen, there are 0.014 moles of hydrogen in the compound.

Now, let's find the masses of carbon and hydrogen in the compound:

Mass of carbon = moles of carbon × molar mass of carbon

$$\text{Mass of carbon} = 0.005 \, \mathrm{mol} \times 12.01 \,\mathrm{g/mol} = 0.060 \,\mathrm{g}$$

Mass of hydrogen = moles of hydrogen × molar mass of hydrogen

$$\text{Mass of hydrogen} = 0.014 \, \mathrm{mol} \times 1.008 \,\mathrm{g/mol} = 0.014 \,\mathrm{g}$$

Now, we are given that the percentage of carbon is 24%. Let's find the total mass of the compound:

$$\text{Total mass of compound} = \frac{\text{Mass of carbon}}{\% \text{ of carbon}} = \frac{0.060 \,\mathrm{g}}{0.24} = 0.250 \,\mathrm{g}$$

Finally, let's find the percentage of hydrogen:

$$\% \text{ of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Total mass of compound}} \times 100 = \frac{0.014 \,\mathrm{g}}{0.250 \,\mathrm{g}} \times 100 = 5.6$$

So, the percentage of hydrogen is 5.6% or 56 × 10^-1.

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