Final answer:
To find the mass of lead (IV) ions in 15 grams of lead (IV) hydroxide, you calculate the molar mass of Pb(OH)4, determine the number of moles of Pb(OH)4, and use this to find the mass of Pb4+ ions, resulting in approximately 12.99 g.
Step-by-step explanation:
To calculate the mass of lead (IV) ions in 15 grams of lead (IV) hydroxide, one must first determine the formula for lead (IV) hydroxide, which is Pb(OH)4. The next step is to calculate the molar mass of Pb(OH)4 consisting of one lead atom (Pb), which has an atomic mass of 207.2 g/mol, and four hydroxide ions (OH), each with a molar mass of approximately 17 g/mol (16.0 g/mol for oxygen and 1.008 g/mol for hydrogen). The molar mass of Pb(OH)4 is then:
(207.2 g/mol) + 4 × (15.999 g/mol + 1.008 g/mol) = 239.2 g/mol.
Next, we calculate the number of moles of lead (IV) hydroxide in 15 grams:
Number of moles = mass / molar mass = 15 g / 239.2 g/mol ≈ 0.0627 moles.
Since there is one lead (IV) ion for each molecule of lead (IV) hydroxide, the number of moles of lead (IV) ions is the same as the number of moles of lead (IV) hydroxide. Finally, we multiply the number of moles of lead (IV) ions by the molar mass of lead (IV) to obtain the mass:
Mass of Pb4+ ions = number of moles × atomic mass of lead (IV) = 0.0627 moles × 207.2 g/mol ≈ 12.99 g.