Question

which formula correctly shows how to calculate the percent composition of oxygen in magnesium nitrite?
periodic table
$\frac{95.9964amu}{148.3148amu}\times100$
$\frac{31.9988amu}{70.3105amu}\times100$
$\frac{15.9994amu}{116.316amu}\times100$
$\frac{63.9976amu}{116.316amu}\times100$

Explanation:

Step1: Write the formula of magnesium nitrite

The formula of magnesium nitrite is $Mg(NO_2)_2$.

Step2: Calculate molar - mass of magnesium nitrite

The molar - mass of $Mg$ is approximately $24.31\ g/mol$, the molar - mass of $N$ is approximately $14.01\ g/mol$, and the molar - mass of $O$ is approximately $15.9994\ g/mol$.
The molar - mass of $Mg(NO_2)_2$ is $M = 24.31+2\times(14.01 + 2\times15.9994)=24.31+2\times(14.01 + 31.9988)=24.31+2\times46.0088=24.31 + 92.0176=116.316\ g/mol$.

Step3: Calculate the total molar - mass of oxygen in magnesium nitrite

There are 4 oxygen atoms in $Mg(NO_2)_2$. The total molar - mass of oxygen is $m_O=4\times15.9994 = 63.9976\ g/mol$.

Step4: Calculate the percent composition of oxygen

The percent composition of oxygen is given by the formula $\text{Percent composition of }O=\frac{\text{Total molar - mass of }O}{\text{Molar - mass of }Mg(NO_2)_2}\times100=\frac{63.9976\ amu}{116.316\ amu}\times100$.

Answer:

$\frac{63.9976\ amu}{116.316\ amu}\times100$ (the fourth option)