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

Apr 1, 2017

COMPARING THE $\boldsymbol{\text{Mg/S}}$ RATIOS

The percentages are stated as

9.87% $\text{Mg}$
12.98% $\text{S}$

in ${\text{MgSO}}_{4}$, specifically, but it can also be in the entire compound since water does not contain $\text{Mg}$ or $\text{S}$. This gives a $\text{Mg/S}$ ratio of:

$\frac{0.0987}{0.1298} = \textcolor{g r e e n}{0.7604}$

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"

$=$ $\text{120.366 g/mol}$

The theoretical percentages of $\text{Mg}$ and $\text{S}$ 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 $\text{Mg/S}$ is actually what we're looking for:

$\frac{0.2019}{0.2664} = \textcolor{g r e e n}{0.7580}$

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 ${\text{MgSO}}_{4}$ 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 $\text{Mg}$ in $\text{MgSO"_4cdotx"H"_2"O}$. So:

(1) $\text{ }$$0.0984 + 0.1298 + {\chi}_{O 1} + {\chi}_{O 2} + {\chi}_{H} = 1$,

where ${\chi}_{O 1} + {\chi}_{O 2}$ is the fraction of oxygen atom mass in the hydrated compound, but ${\chi}_{O 1}$ is ONLY in the sulfate.

One way of proceeding is:

(2) $\text{ }$${M}_{M g S {O}_{4} \cdot x \text{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 $\text{O}$ in ${\text{SO}}_{4}$ from the percentage of $\text{S}$ and using the fact that the ratio of molar masses allows one to convert to the identity of another species.

${\chi}_{O 1} = {\chi}_{S} \times \left(4 \times \text{15.999 g/mol")/("32.065 g/mol}\right) = 0.2591$

This means from (1), we have:

$0.0984 + 0.1298 + 0.2591 + \stackrel{\text{water only}}{\overbrace{{\chi}_{O 2} + {\chi}_{H}}}$

$\implies 0.4873 + {\chi}_{{H}_{2} O} = 1$,

$\implies {\chi}_{{H}_{2} O} = 0.5127$

Thus, we can write another equation:

(3) $\text{ }$0.5127M_(MgSO_4cdotx"H"_2"O") = x*"18.015 g/mol"

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

${M}_{M g S {O}_{4} \cdot x \text{H"_2"O") = stackrel("MgSO"_4)overbrace("120.366 g/mol") + 0.5127M_(MgSO_4cdotx"H"_2"O}}$

${M}_{M g S {O}_{4} \cdot x \text{H"_2"O") - 0.5127M_(MgSO_4cdotx"H"_2"O") = stackrel("MgSO"_4)overbrace("120.366 g/mol}}$

$0.4873 {M}_{M g S {O}_{4} \cdot x \text{H"_2"O") = stackrel("MgSO"_4)overbrace("120.366 g/mol}}$

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

$=$ $\text{247.03 g/mol}$

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:

$\textcolor{b l u e}{\text{MgSO"_4cdot7"H"_2"O}}$