If a binary metal halide whose $% C l$ $%Cl$ is $60.68 %$ $60.68%$ , what is the likely salt?

Question

1671 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-08T04:58:11

We are given $M C l$ $MCl$ , where chloride content is $60.68 %$ $60.68%$ by mass....we gots sodium metal.....

So we solve for the atomic mass of $M$ $M$ .....

$60.68 % = \frac{35.5 \cdot g}{Formula mass of salt, g} \times 100 %$ $60.68%=(35.5*g)/("Formula mass of salt, g")xx100%$

$Formula mass of salt = \frac{35.5 \cdot g}{0.6068} = 58.5 \cdot g$ $"Formula mass of salt"=(35.5*g)/(0.6068)=58.5*g$

But because there is ONE CHLORIDE, we know that the formula mass of the metal is.....

$M_{mass} = (58.5 - 35.45) \cdot g = 23.0 \cdot g$ $M_"mass"=(58.5-35.45)*g=23.0*g$

And a quick glance at the Periodic Table tells us that the metal is likely to be sodium, i.e. $22.99 \cdot g \cdot m o l^{- 1}$ $22.99*g*mol^-1$ .

Capisce?

If a binary metal halide whose %Cl%Cl is 60.68%60.68%, what is the likely salt?