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

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")#