Question

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

Answer 1

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)))||`

Answer 2

`"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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