Final Answer:
The metal in the oxide containing 16.99% O by mass is Aluminum (Al). (Option c)
Step-by-step explanation:
Identify the unknowns:
Metal (M) in the oxide MO.
Percentage mass of the metal (%M) in the oxide.
Given information:
Percentage mass of oxygen (%O) in the oxide = 16.99%
Relationship between percentages:
%O + %M = 100%
Calculate the percentage mass of the metal:
%M = 100% - %O = 100% - 16.99% = 83.01%
Consider the possible metals and their oxide formulas:
a. Sodium (Na): Na₂O, molecular mass = 62 g/mol
b. Magnesium (Mg): MgO, molecular mass = 40 g/mol
c. Aluminum (Al): Al₂O₃, molecular mass = 102 g/mol
d. Silicon (Si): SiO₂ (silicon dioxide), molecular mass = 60 g/mol
Calculate the mass of oxygen in each oxide:
Na₂O: 16 g O / 62 g oxide * 100% = 25.8%
MgO: 16 g O / 40 g oxide * 100% = 40%
Al₂O₃: 48 g O / 102 g oxide * 100% = 47.06%
SiO₂: 32 g O / 60 g oxide * 100% = 53.33%
Compare the calculated oxygen percentages with the given value:
Only the oxygen percentage in Al₂O₃ (47.06%) is close to the given 16.99%. Therefore, the metal in the oxide is Aluminum (Al).