help please ......

Question

In the compound MgBr_2, what mass of bromine is present for each 46.0 g of magnesium?

Answer 1

This problem just requires a little algebra.

Mg = 24.305g

Br = 79.904g

First, you need to calculate the percent composition of Magnesium in one molecule of MgBr₂. One molecule weighs 184.113g. 24.305g/184.113g = 13.2% of the molecule's weight. This percentage will always be consistent no matter how much (or how little) MgBr₂ there is.

When the problem says there is 46.0g of Magnesium, that 46.0g is still equal to 13.2% of the MgBr₂, by mass. Therefore, you can set this up as an algebra problem:

46.0g = .132 (X) --> in which "X" equals the total weight of the MgBr₂ sample

(divide both sides of the equation by .132, and you get...)

348.4848... = X

I would round this up to 348.485g.

If the total mass is 348.485g and 46.0g of that is Bromine (because that information is given in the problem), then the remainder has to be Magnesium.

348.485 - 46.0 = 302.485g.