Question

How do I verify the percent composition of `"MgSO"_4cdotx"H"_2"O"` in this Epsom Salt package and solve for `x`?

Answer 1

COMPARING THE `bb"Mg/S"` RATIOS

The percentages are stated as

`9.87%` `"Mg"`
`12.98%` `"S"`

in `"MgSO"_4`, specifically, but it can also be in the entire compound since water does not contain `"Mg"` or `"S"`. This gives a `"Mg/S"` ratio of:

`0.0987/0.1298 = color(green)(0.7604)`

The molar mass of this compound is:

`M_(MgSO_4) = stackrel("Mg")overbrace"24.305 g/mol" + stackrel("S")overbrace"32.065 g/mol" + 4 xx stackrel("O")overbrace"15.999 g/mol"`

`=` `"120.366 g/mol"`

The theoretical percentages of `"Mg"` and `"S"` are found by simply dividing their molar masses by the molecular molar mass:

`%"Mg" = "24.305 g/mol"/"120.366 g/mol" = 0.2019 = 20.19%`

`%"S" = "32.065 g/mol"/"120.366 g/mol" = 0.2664 = 26.64%`

Even though the percentages are different, the theoretical ratio of `"Mg/S"` is actually what we're looking for:

`0.2019/0.2664 = color(green)(0.7580)`

The ratios are very close: `0.7604` (actual) vs. `0.7580` (theoretical). If we wish, we can find the percent difference:

`(0.7604 - 0.7580)/(0.7580) xx 100% = 1.28%`,

which is small. Therefore, it is fairly likely that `"MgSO"_4` is indeed in the Epsom Salt package.

SOLVING FOR NUMBER OF WATERS

In the "Fertilizer uses" box, we see it says `9.84%` "water soluble" magnesium. I think we can assume it means `9.84%` is the percentage of `"Mg"` in `"MgSO"_4cdotx"H"_2"O"`. So:

(1) `" "` `0.0984 + 0.1298 + chi_(O1) + chi_(O2) + chi_H = 1`,

where `chi_(O1) + chi_(O2)` is the fraction of oxygen atom mass in the hydrated compound, but `chi_(O1)` is ONLY in the sulfate.

One way of proceeding is:

(2) `" "` `M_(MgSO_4cdotx"H"_2"O") = stackrel("Mg")overbrace"24.305 g/mol" + stackrel("S")overbrace"32.065 g/mol" + 4 xx stackrel("O")overbrace"15.999 g/mol" + x*stackrel("H"_2"O")overbrace("18.015 g/mol")`

`= stackrel("MgSO"_4)overbrace("120.366 g/mol") + x*stackrel("H"_2"O")overbrace("18.015 g/mol")`

As-written, it is not possible to solve since we have two unknowns and one equation, but we can solve for the percentage of `"O"` in `"SO"_4` from the percentage of `"S"` and using the fact that the ratio of molar masses allows one to convert to the identity of another species.

`chi_(O1) = chi_S xx (4 xx "15.999 g/mol")/("32.065 g/mol") = 0.2591`

This means from (1), we have:

`0.0984 + 0.1298 + 0.2591 + stackrel("water only")overbrace(chi_(O2) + chi_H)`

`=> 0.4873 + chi_(H_2O) = 1`,

`=> chi_(H_2O) = 0.5127`

Thus, we can write another equation:

(3) `" "` `0.5127M_(MgSO_4cdotx"H"_2"O") = x*"18.015 g/mol"`

Now we can solve for the molar mass. Plug (3) into (2):

`M_(MgSO_4cdotx"H"_2"O") = stackrel("MgSO"_4)overbrace("120.366 g/mol") + 0.5127M_(MgSO_4cdotx"H"_2"O")`

`M_(MgSO_4cdotx"H"_2"O") - 0.5127M_(MgSO_4cdotx"H"_2"O") = stackrel("MgSO"_4)overbrace("120.366 g/mol")`

`0.4873M_(MgSO_4cdotx"H"_2"O") = stackrel("MgSO"_4)overbrace("120.366 g/mol")`

`=> M_(MgSO_4cdotx"H"_2"O") = (stackrel("MgSO"_4)overbrace("120.366 g/mol"))/0.4873`

`=` `"247.03 g/mol"`

As a result, from (3) we get:

`color(blue)(x) = (0.5127M_(MgSO_4cdotx"H"_2"O"))/"18.015 g/mol" = (0.5127*"247.03 g/mol")/"18.015 g/mol" ~~ color(blue)(7)`

WRITING THE CHEMICAL FORMULA

Therefore, the formula for the hydrate is predicted to be:

`color(blue)("MgSO"_4cdot7"H"_2"O")`