Final answer:
To determine the quantity in moles of oxygen in a 175.7 g sample of manganese(IV) perchlorate, calculate the number of moles of oxygen using the molar mass of the compound (Mn(ClO₄)₄). There are approximately 1.547 moles of oxygen in 175.7 g of manganese(IV) perchlorate.
Step-by-step explanation:
To determine the quantity in moles of oxygen in a 175.7 g sample of manganese(IV) perchlorate, we need to calculate the number of moles of oxygen using the molar mass of the compound. The molar mass of manganese(IV) perchlorate (Mn(ClO₄)₄) is 452.74 g/mol. From the formula Mn(ClO₄)₄, we can see that there are four oxygen atoms in one molecule of manganese(IV) perchlorate. Therefore, the number of moles of oxygen can be calculated as follows:
175.7 g Mn(ClO₄)₄ x (1 mol Mn(ClO₄)₄ / 452.74 g Mn(ClO₄)₄) x (4 mol O / 1 mol Mn(ClO₄)₄) = 1.547 mol O
Therefore, there are approximately 1.547 moles of oxygen in 175.7 g of manganese(IV) perchlorate.