# Question abc31

May 12, 2017

${C}_{6} {H}_{8} {O}_{8}$

#### Explanation:

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

$\text{C" = 34.6" g}$
$\text{H" = 3.85" g}$
$\text{O" = 61.55" g}$

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

$\text{C" = (34.6 cancel"g")/1 *(1" mol")/(12 cancel"g") = 2.88" mol}$

$\text{H" = (3.85 cancel"g")/1 *(1" mol")/(1.00 cancel"g") = 3.85" mol}$

$\text{O" = (61.55 cancel"g")/1 *(1" mol")/(16 cancel"g") = 3.85" mol}$

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

"C" = (2.88cancel"mol")/(2.88 cancel"mol") = 1

"H" = (3.85cancel"mol")/(2.88 cancel"mol") = 1.33

"O" = (3.85 cancel"mol")/(2.88 cancel"mol") = 1.33

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

$\text{C} = 1 \cdot \left(3\right) = 3$

$\text{H} = 1.33 \cdot \left(3\right) = 3.99 \approx 4$

$\text{O} = 1.33 \cdot \left(3\right) = 3.99 \approx 4$

$\textcolor{b l u e}{\text{Step 5: Combine elements and attach their corresponding mole ratios. Then, find the empirical formula mass}}$

C_(3)H_(4)O_(4)->(104" g")/"mol" = "empirical formula mass"

$\textcolor{b l u e}{\text{Step 6: Divide given molecular formula mass by the empirical formula mass to get a factor.}}$
This factor will tell you how many times to multiply the subscripts to get your mole ratios for the molecular formula

("Molecular formula mass")/("Empirical formula mass") ->[(208 cancel"g")/(cancel"mol")]/[(104 cancel"g")/(cancel"mol")]= 2#

${C}_{3} {H}_{4} {O}_{4} \to \textcolor{\mathmr{and} a n \ge}{{C}_{6} {H}_{8} {O}_{8} = \text{Molecular Formula}}$

$\text{Answer} : {C}_{6} {H}_{8} {O}_{8}$