If the weight of metal chloride is #x# grams containing #y# grams of metal, then what is the equivalent weight of the metal in terms of #x#, #y#, and #35.5#?

1 Answer
Stefan V.
Aug 17, 2017

#35.5 * y/(x-y)# #"g"#

Explanation:

For this compound, the equivalent mass of the metal represents the mass of metal that combines with exactly #"35.5 g"# of chlorine.

You know that the total mass of the oxide is equal to #x# #"g"# and that this sample contains #y# #"g"# of metal, so you can say that the mass of chlorine present in the sample is equal to

#overbrace(xcolor(white)(.)"g")^(color(blue)("mass of metal chloride")) - overbrace(ycolor(white)(.)"g")^(color(blue)("mass of metal")) = overbrace((x-y)color(white)(.)"g")^(color(blue)("mass of chlorine"))#

So, you know that #y# #"g"# of metal combine with #(x-y)# #"g"# of chlorine, so you can say that #"35.5 g"# of chlorine will combine with

#35.5 color(red)(cancel(color(black)("g Cl"))) * (y color(white)(.)"g metal")/((x-y)color(red)(cancel(color(black)("g Cl")))) = (35.5 * y/(x-y))color(white)(.)"g metal"#

Therefore, the equivalent mass of the metal in this compound is equal to

#color(darkgreen)(ul(color(black)("equivalent mass" = (35.5 * y/(x-y))color(white)(.)"g")))#

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