Find the molecular and empirical formula from the given information below?

2 Answers
May 24, 2017

Answer:

Empirical formula of sorbitol: #C_(3)H_(7)O_(3)#

Molecular formula of sorbitol #C_(6)H_(14)O_(6)#

Explanation:

#color(blue)("Step 1: Assume 100 g sample to find mass of each element")#

#"C" = 39.56" g"#
#"H" = 7.74" g"#
#"O" = 52.70" g"#

#color(white)(aaaa)#

#color(blue)("Step 2: Find number of moles from the mass of each element.")# Use the periodic table to find molar masses of each element

#"C" = (39.56 cancel"g")/1 *(1" mol")/(12 cancel"g") = 3.29" mol"#

#"H" = (7.74 cancel"g")/1 *(1" mol")/(1.00 cancel"g") = 7.74" mol"#

#"O" = (52.70 cancel"g")/1 *(1" mol")/(16 cancel"g") = 3.29" mol"#

#color(white)(aaaa)#

#color(blue)("Step 3: Divide all the mole numbers by the smallest mole value of them all")#

#"C" = (3.29cancel"mol")/(3.29 cancel"mol") = 1.00#

#"H" = (7.74cancel"mol")/(3.29 cancel"mol") = 2.35#

#"O" = (3.29 cancel"mol")/(3.29 cancel"mol") = 1.00#

#color(white)(aaaa)#

#color(blue)("Step 4: Multiply the numbers by a factor which will give out a whole number ratio.")#

#"C" = 1.00 * (3) -> 3#

#"H" = 2.35 * (3) -> 7.05~~7#

#"O" = 1.00 * (3) -> 3#

#color(white)(aaaa)#

#color(blue)("Step 5: Combine elements and attach corresponding mole values."#
#color(white)(---)color(blue)("Then, find empirical formula mass using the periodic table"#

#color(white)(----)stackrel"empirical formula"||ul(C_(3)H_(7)O_(3))|| -> stackrel"empirical formula mass"||ul((91" g")/"mol" )||#

#color(white)(aaaa)#

#color(blue)("Step 6: Divide given molecular formula mass by empirical formula mass")#
#color(white)(---)color(blue)("to get a factor.")#

#color(white)(aaaaaaaaa)("Molecular formula mass")/("Empirical formula mass") ->[(182 cancel"g")/(cancel"mol")]/[(91 cancel"g")/(cancel"mol")]= 2#

#color(white)(aaaa)#

#color(blue)("Step 7: Take this factor and multiply it to the empirical formula")#
#color(white)(---)color(blue)"to get the molecular formula of sorbitol"#

#color(white)(-----)stackrel"empirical formula"||ul(C_(3)H_(7)O_(3))||xx2 -> stackrel"molecular formula"||ul(color(orange)(C_(6)H_(14)O_(6)))||#

May 25, 2017

Answer:

#"C"_3"H"_7"O"_3#

# "C"_6"H"_14"O"_6#

Explanation:

Step 1. Assume #100 g# of the compound, sorbitol, is present. This is to change the percentage to grams.

#"C" -> 39.56" g"#
#"H" -> 7.74" g"#
#"O" -> 52.70" g"#

Step 2. Convert the masses to moles. We need to use Average Atomic masses of elements.

#"C" = 39.56/12.011 = 3.29" mol"#

#"H" = 7.74/1.0079 = 7.68" mol"#

#"O" = (52.70)/(15.9994) = 3.29" mol"#

Step 3. Divide by the lowest.

#"C" = 3.29/3.29 = 1#

#"H" = 7.68/3.29= 2.33#

#"O" = 3.29/3.29 = 1#

Step 4. Multiply to get smallest whole-number ratio. Key here is #2.33#. if you multiply it by #3# we get very close to a whole number

#"C" = 1 xx 3 = 3#

#"H" = 2.33 xx3 = 6.99~~7#

#"O" = 1 xx 3 =3#

Step 5. We get the empirical formula from these whole numbers

#"C"_3"H"_7"O"_3#

Using average atomic masses we get mass of empirical formula as

#3xx12.011+7xx1.0079+3xx15.9994=91.0865#
Step 6. Divide the given molecular formula mass by empirical formula mass to get integral multiplying factor.

#("Molecular formula mass")/("Empirical formula mass") =182/91.0865approx 2#

Step 7. Multiply empirical formula with this integer to get the molecular formula of sorbitol.

#("C"_3"H"_7"O"_3)xx2 -> "C"_6"H"_14"O"_6#

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