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Report (any type)/Brief report
22/09/2021
The molecular formula is ##”H”_2″O”_2″##.
Since the percentages add up to 100%, we can assume that we have a 100-g sample, which will allow us to convert the percentages into grams.
##”H”:####5.94%=>”5.94 g”##
##”O”:####94.06%=>”94.06 g”##
Determine Moles of Each Element
First we need to determine moles of H and O by dividing their given masses by their molar masses (atomic weight on the periodic table) in g/mol.
##”H”:####5.94cancel”g H”xx(1″mol H”)/(1.00794cancel”g H”)=”5.89 mol H”##
##”O”:####94.06″g O”xx(1″mol O”)/(15.999″g O”)=”5.88 mol O”##
Determine and Empirical Formula
Since the number of moles for H and O are equal, ratios are 1.
The empirical formula is ##”HO”##
Determine the molecular formula.
The empirical formula mass is ##(1xx1.00794″g/mol”)+(1xx15.999″g/mol”)=”17.007 g/mol”##
The molecular formula mass is ##”34.01 g/mol”##. To get the multiplication factor, divide the molecular formula mass by the empirical formula mass.
##”Multiplication factor”=(34.01″g/mol”)/(17.007″g/mol”)=”2.000″##
To get the molecular formula, multiply the subscripts of the empirical formula times 2.
The molecular formula is ##”H”_2″O”_2″##.
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