Final answer:
The percent composition of oxygen in lithium oxide (Li2O) is approximately 53.55%, which is calculated by the formula (mass of O / molar mass of Li2O) x 100%.
Step-by-step explanation:
The question is asking to calculate the percent composition of oxygen in lithium oxide (Li2O). To find this, we need the molar masses of lithium and oxygen. Lithium has an atomic mass of approximately 6.94 g/mol, and oxygen has an atomic mass of about 16.00 g/mol.
In the formula Li2O, there are two lithium atoms and one oxygen atom. Therefore, the molar mass of lithium oxide would be (2 × 6.94) + 16.00 = 29.88 g/mol.
The mass of oxygen in the compound is taken from the molecular formula, which is 16.00 g/mol. The percent composition is found by dividing the mass of oxygen by the molar mass of the compound and multiplying by 100%.
Percent Composition of Oxygen = (mass of O / molar mass of Li2O) × 100%
= (16.00 g/mol / 29.88 g/mol) × 100%
= 53.55%
Therefore, oxygen makes up approximately 53.55% of the mass of lithium oxide.