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

May 24, 2017

Empirical formula of sorbitol: ${C}_{3} {H}_{7} {O}_{3}$

Molecular formula of sorbitol ${C}_{6} {H}_{14} {O}_{6}$

#### Explanation:

$\textcolor{b l u e}{\text{Step 1: Assume 100 g sample to find mass of each element}}$

$\text{C" = 39.56" g}$
$\text{H" = 7.74" g}$
$\text{O" = 52.70" g}$

$\textcolor{w h i t e}{a a a a}$

$\textcolor{b l u e}{\text{Step 2: Find number of moles from the mass of each element.}}$ Use the periodic table to find molar masses of each element

$\text{C" = (39.56 cancel"g")/1 *(1" mol")/(12 cancel"g") = 3.29" mol}$

$\text{H" = (7.74 cancel"g")/1 *(1" mol")/(1.00 cancel"g") = 7.74" mol}$

$\text{O" = (52.70 cancel"g")/1 *(1" mol")/(16 cancel"g") = 3.29" mol}$

$\textcolor{w h i t e}{a a a a}$

$\textcolor{b l u e}{\text{Step 3: Divide all the mole numbers by the smallest mole value of them all}}$

"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

$\textcolor{w h i t e}{a a a a}$

$\textcolor{b l u e}{\text{Step 4: Multiply the numbers by a factor which will give out a whole number ratio.}}$

$\text{C} = 1.00 \cdot \left(3\right) \to 3$

$\text{H} = 2.35 \cdot \left(3\right) \to 7.05 \approx 7$

$\text{O} = 1.00 \cdot \left(3\right) \to 3$

$\textcolor{w h i t e}{a a a a}$

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" )||

$\textcolor{w h i t e}{a a a a}$

$\textcolor{b l u e}{\text{Step 6: Divide given molecular formula mass by empirical formula mass}}$
$\textcolor{w h i t e}{- - -} \textcolor{b l u e}{\text{to get a factor.}}$

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

$\textcolor{w h i t e}{a a a a}$

$\textcolor{b l u e}{\text{Step 7: Take this factor and multiply it to the empirical formula}}$
$\textcolor{w h i t e}{- - -} \textcolor{b l u e}{\text{to get the molecular formula of sorbitol}}$

$\textcolor{w h i t e}{- - - - -} \stackrel{\text{empirical formula"||ul(C_(3)H_(7)O_(3))||xx2 -> stackrel"molecular formula}}{|} | \underline{\textcolor{\mathmr{and} a n \ge}{{C}_{6} {H}_{14} {O}_{6}}} | |$

May 25, 2017

${\text{C"_3"H"_7"O}}_{3}$

${\text{C"_6"H"_14"O}}_{6}$

#### Explanation:

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

$\text{C" -> 39.56" g}$
$\text{H" -> 7.74" g}$
$\text{O" -> 52.70" g}$

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

$\text{C" = 39.56/12.011 = 3.29" mol}$

$\text{H" = 7.74/1.0079 = 7.68" mol}$

$\text{O" = (52.70)/(15.9994) = 3.29" mol}$

Step 3. Divide by the lowest.

$\text{C} = \frac{3.29}{3.29} = 1$

$\text{H} = \frac{7.68}{3.29} = 2.33$

$\text{O} = \frac{3.29}{3.29} = 1$

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

$\text{C} = 1 \times 3 = 3$

$\text{H} = 2.33 \times 3 = 6.99 \approx 7$

$\text{O} = 1 \times 3 = 3$

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

${\text{C"_3"H"_7"O}}_{3}$

Using average atomic masses we get mass of empirical formula as

$3 \times 12.011 + 7 \times 1.0079 + 3 \times 15.9994 = 91.0865$
Step 6. Divide the given molecular formula mass by empirical formula mass to get integral multiplying factor.

$\left(\text{Molecular formula mass")/("Empirical formula mass}\right) = \frac{182}{91.0865} \approx 2$

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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