The Mole and Stoichiometry: Question 1

Worked solution

Step 1: Recall the formula of barium nitrate

Barium nitrate, $\text{Ba(NO}_3\text{)}_2$, contains, in every formula unit:

• $1$ barium atom
• $2$ nitrogen atoms (one from each of the two $\text{NO}_3$ groups)
• $6$ oxygen atoms (three from each of the two $\text{NO}_3$ groups)

Step 2: Add up the relative atomic masses

$M_r = A_r(\text{Ba}) + 2\times A_r(\text{N}) + 6\times A_r(\text{O})$

$M_r = 137 + (2\times14) + (6\times16)$

$M_r = 137 + 28 + 96 = 261$

Why the other options are wrong

• A ($124$): this is $2\times(14+3\times16) = 2\times62=124$, which comes from forgetting to add the mass of barium at all.
• B ($199$): this is $137+14+48=199$, which comes from forgetting that the outer subscript $2$ also multiplies the nitrogen and oxygen atoms, so only one $\text{NO}_3$ group was counted.
• C ($213$): this is $137+28+48=213$, which comes from correctly doubling the nitrogen atoms but using only $3$ oxygen atoms instead of $6$.

Final answer

• $M_r(\text{Ba(NO}_3\text{)}_2) = \boxed{261}$, option D.