Question #b9857
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"How can molarity be used as a conversion factor?"
For each element, you divide the total mass of its atoms by the empirical formula mass and multiply by 100.
EXAMPLE
What is the percent composition of a compound with the empirical formula #"CH"_2"O"#?
Solution
Step 1: Find the masses of the individual atoms.
#"C = 12.01 u"#
#"H = 1.008 u"#
#"O = 16.00 u"#
Step 2: Find the total mass of each atom in the formula.
#"1 C = 12.01 u"#
#"2 H = 2.016 u"#
#"1 O = 16.00 u"#
Step 3: Find the total mass of all the atoms in the formula.
#"CH"_2"O = (12.01 + 2.016 + 16.00) u = 30.03 u"#
Step 4: Find the mass percent of each atom.
#"mass %" = "mass of atom"/"total mass" × "100 %#
#"mass % C" = (12.01 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) × "100 %" = "40.00 %"#
#"mass % H" = (2.016 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) × "100 %" = "6.714 %"#
#"mass % O" = (16.00 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) × "100 %" = "53.29 %"#
Step 5: Check that your answers add up to 100 %.
#40.00 + 6.714 + 53.29 = 100.00#
