Question

a metal (m) forms an oxide with the formula mo. if the oxide contains 28.53 % o by mass, what is the identity of the metal? express your answer as a chemical symbol.

Explanation:

Step1: Determine molar mass of oxygen

The molar mass of oxygen ($O$) is $M_O = 16\ g/mol$.

Step2: Set up mass - percentage equation

Let the molar mass of the metal be $M_M$. The molar mass of the oxide $MO$ is $M = M_M+M_O$. Given that the mass - percentage of $O$ in $MO$ is $28.53\%$, we have the equation $\frac{M_O}{M_M + M_O}\times100\%=28.53\%$. Substituting $M_O = 16\ g/mol$ into the equation, we get $\frac{16}{M_M + 16}=0.2853$.

Step3: Solve for the molar mass of the metal

Cross - multiply the equation $\frac{16}{M_M + 16}=0.2853$:
$16=0.2853\times(M_M + 16)$.
$16 = 0.2853M_M+0.2853\times16$.
$16 = 0.2853M_M + 4.5648$.
$0.2853M_M=16 - 4.5648$.
$0.2853M_M = 11.4352$.
$M_M=\frac{11.4352}{0.2853}\approx40\ g/mol$.
The metal with a molar mass close to $40\ g/mol$ is calcium ($Ca$).

Answer:

$Ca$