Question

Calculate the percentage of sodium in sodium aluminum fluoride.
([Na₃AlF₆] correct to the nearest whole number).
(F=$19$; Na = $23$; AI =$27$)

Answer 1

Firstly, we calculate the molecular mass of Na₃[AIF₆] = $3\times 23+27+6\times 19$

$$\begin{gathered} =69+27+114 \\ =210 \end{gathered}$$

%age of sodium(Na) in [Na₃AIF₆]$=\frac{69}{210}\times 100$

$$\begin{gathered} =\frac{69\times 10}{21} \\ =\frac{230}{21} \\ =32.9 \end{gathered}$$

The percentage of sodium in sodium aluminum fluoride is $32.9\%$.