A compound containing Na, S, H, and O where, Na =14.28 %, S = 9.92%, H = 6.2% and the rest are O. The molecular mass of the compound is 322. Calculate the molecular formula of compound considering all H atoms are present as water of crystallization?

The last part which explains that H is present as the water of crystallization is a bit confusing for me.

1 Answer
Jun 1, 2018

Answer:

Warning! Long Answer. The molecular formula is #"Na"_2"SO"_4·10"H"_2"O"#.

Explanation:

Step 1. Calculate the empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of #"Na"# to #"S"# to #"H"# to #"O"#.

Your compound contains 14.28 % #"Na"#, 9.92 % #"S"#, and 6.2 % #"H"#.

Assume that you have 100 g of sample.

Then it contains 14.28 g of #"Na"#, 9.92 g of #"S"#, and 6.2 g of #"H"#.

#"Mass of O = (100 - 14.28 - 9.92 - 6.2) g = 69.6 g"#

#"Moles of Na" = 14.28 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "1.189 mol B"#

#"Moles of S" = 9.92 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(32.06 color(red)(cancel(color(black)( "g C")))) = "0.3094 mol B"#

#"Moles of H" = 6.2 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "6.15 mol H"#

#"Moles of O" = 69.6 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "4.350 mol O"#

From this point on, I like to summarize the calculations in a table.

#ulbb("Element"color(white)(m) "Mass/g"color(white)(Xl) "Moles"color(white)(Xll) "Ratio"color(white)(m)"Integers")#
#color(white)(m)"Na" color(white)(XXXml)14.28 color(white)(Xml)0.6211 color(white)(mm)2.007color(white)(mmmm)2#
#color(white)(m)"S" color(white)(XXXmmll)9.92 color(white)(Xml)0.3094color(white)(mml)1color(white)(mmmmmll)1#
#color(white)(m)"H" color(white)(XXXXmll)6.2 color(white)(mmll)6.15 color(white)(Xmm)19.9 color(white)(mmmml)20#
#color(white)(m)"O"color(white)(XXXmm)69.6 color(white)(Xmll)4.350color(white)(mml)14.06color(white)(mmmm)14#

The empirical formula is #"Na"_2"SH"_20"O"_14#.

Step 2. Calculate the molecular formula of the compound

The empirical formula mass of #"Na"_2"SH"_20"O"_14# is 322.19 u.

The molecular mass is 322 u.

The molecular mass must be an integral multiple of the empirical formula mass.

#"MM"/"EFM" = (322color(red)(cancel(color(black)("u"))))/(322.19 color(red)(cancel(color(black)("u")))) = 0.999 ≈ 1#

The molecular formula must be the same as the empirical formula.

#"MF" = "EF" = "Na"_2"SH"_20"O"_14#

Step 3. Calculate the water of crystallization

If all the #"H"# atoms are present as water of crystallization,

#"20H = 10H"_2"O"#

The molecular formula is #"Na"_2"SO"_4·10"H"_2"O"#.