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Report (any type)/Brief report
22/09/2021
The molecular formula is ##”C”_6″H”_8″O”_6″##.
We need to determine the empirical formula and the empirical formula mass. Then divide the molecular mass by the empirical mass, then multiply the empirical formula times the result.
EMPIRICAL FORMULA
Determine the masses
The percentages given add up to ##”100.00%”##. In a ##”100 g”## sample, the masses would be the percentages in grams.
##”C”=”40.92 g”##
##”H”=”4.58 g”##
##”O”=”54.50 g”##
Determine the molar masses of the (atomic weight in g/mol).
##”C”=”12.0107 g/mol”##
##”H”=”1.00794 g/mol”##
##”O”=”15.999 g/mol”##
Determine the number of moles of each element from the given mass and the molar mass.
##”C”:####40.92 cancel”g C”xx(1 “mol C”)/(12.010 cancel”g C”)=”3.407 mol C”##
##”H”:## ##4.58 cancel”g H”xx(1 “mol H”)/(1.00794 cancel”g H”)=”4.54 mol H”##
##”O”:## ##54.50 cancel”g O”xx(1 “mol O”)/(15.999 cancel”g O”)=”3.406 mol O”##
Determine by dividing the moles for each element by the smallest number of moles.
##”C”:## ##(3.407 “mol”)/(3.406 “mol”)=1.000##
##”H”:## ##(4.54 “mol”)/(3.406 “mol”)=1.33##
##”H”:## ##(3.406 “mol”)/(3.406 “mol”)=1.000##
Multiply the ratios times ##3## to get all whole numbers.
##”C”:## ##1.000xx3=3.000##
##”H”:## ##1.33xx3=3.99##
##”O”:## ##1.000xx=3.000##
The empirical formula is ##”C”_3″H”_4″O”_3##.
The empirical formula mass = ##(3xx12.0107″g/mol”)+(4xx1.00794″g/mol”)+(3xx15.999″g/mol”)=88.061″g/mol”##
MOLECULAR FORMULA
Divide molecular mass by empirical mass. Multiply the empirical formula times the result.
##”Molecular mass”https://studydaddy.com/”empirical mass” = “176.1 g/mol”https://studydaddy.com/”88.061 g/mol”=2.000##
Molecular formula ##=## ##”C”_3″H”_4″O”_3″## times 2 ##=####”C”_6″H”_8″O”_6##
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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