# If 250g of sugar is completely fermented to ethanol, what is the theoretical yield of ethyl alcohol in: ?

## i. grams or milliliters (density of ethanol is 0.79g/mL) ii. what is the final %v/v of alcohol if that 250 sugar was used to make 1L of wine? (assume no volume changed during fermentation) please, show me the working out. Thanks

Aug 22, 2017

Here's what I got.

#### Explanation:

Start by writing the chemical equation that describes this reaction

${\text{C"_ 6"H"_ 12"O"_ (6(aq)) -> 2"C"_ 2"H"_ 5"OH"_ ((aq)) + 2"CO}}_{2 \left(g\right)} \uparrow$

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$
SIDE NOTE As you can see, I am assuming that by $\text{250 g}$ of sugar you mean $\text{250 g}$ of glucose, not sucrose, ${\text{C"_12"H"_22"O}}_{11}$.

If you were supposed to work with sucrose instead of glucose, add this step

${\text{C"_ 12"H"_ 22"O"_ (11(aq)) + "H"_ 2"O"_ ((l)) -> 2"C"_ 6"H"_ 12 "O}}_{6 \left(a q\right)}$

and redo the calculations with the mass of glucose that you'll get from this reaction.
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

Now, notice that you have a $1 : 2$ mole ratio between glucose and ethanol, so convert the mass of glucose to moles by using the compound's molar mass

250 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "1.388 moles glucose"

This means that the reaction will produce

1.388 color(red)(cancel(color(black)("moles glucose"))) * "2 moles ethanol"/(1color(red)(cancel(color(black)("mole glucose")))) = "2.776 moles ethanol"

To convert this to grams, use the molar mass of ethanol

2.776 color(red)(cancel(color(black)("moles ethanol"))) * "46.07 g"/(1color(red)(cancel(color(black)("mole ethanol")))) = "127.9 g"

Rounded to two sig figs, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{theoretical yield = 130 g}}}}$

To convert this to milliliters, use the density of ethanol

$127.9 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "1 mL"/(0.79color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("160 mL}}}}$

To find the volume by volume percent concentration of the wine, $\text{% v/v}$, you should assume that you're working with the volume of ethanol produced by the reaction in a total volume of $\text{1 L} = {10}^{3}$ $\text{mL}$ of wine.

Now, the solution's volume by volume percent concentration tells you the volume of solute present for every $\text{100 mL}$ of solution.

In your case, $\text{100 mL}$ of your solution will contain

100 color(red)(cancel(color(black)("mL solution"))) * "160 mL ethanol"/(10^3color(red)(cancel(color(black)("mL solution")))) = "16 mL ethanol"

Therefore, the concentration by volume will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{% v/v = 16% ethanol}}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the volume of wine.

Aug 22, 2017

i. The theoretical yield of ethanol is 135 g or 170 mL; ii.the final concentration is 17 % v/v.

#### Explanation:

We know that we will need a balanced equation with masses and moles, so let's gather all the information in one place.

I assume that by "sugar" you mean sucrose, ${\text{C"_12"H"_22"O}}_{11}$.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m} 342.30 \textcolor{w h i t e}{m m m m m m m m l l} 46.07$
$\textcolor{w h i t e}{m m m m} {\text{C"_12"H"_22"O"_11 + "H"_2"O" → 4"CH"_3"CH"_2"OH" + 4"CO}}_{2}$
$\text{Mass/g:} \textcolor{w h i t e}{m l l} 250$

Step 1. Calculate the moles of sugar

$\text{Moles of sugar" = 250 color(red)(cancel(color(black)("g sugar"))) × "1 mol sugar"/(342.30 color(red)(cancel(color(black)("g sugar")))) = "0.7304 mol sugar}$

Step 2. Calculate the moles of ethanol

$\text{Moles of ethanol" = 0.7304 color(red)(cancel(color(black)("mol sugar"))) × "4 mol ethanol"/(1 color(red)(cancel(color(black)("mol sugar")))) = "2.921 mol ethanol}$

Step 3. Calculate the mass of ethanol

$\text{Mass of ethanol" = 2.921 color(red)(cancel(color(black)("mol ethanol"))) × "46.07 g ethanol"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "135 g ethanol}$

Step 4. Calculate the volume of ethanol

$\text{Volume of ethanol" = 135 color(red)(cancel(color(black)("g ethanol"))) × "1 mL ethanol"/(0.79 color(red)(cancel(color(black)("g ethanol")))) = "170 mL ethanol}$

Step 5. Calculate the volume % of ethanol

$\text{Volume %" = "volume of ethanol"/"total volume" × 100 % = (170 color(red)(cancel(color(black)("mL"))))/(1000 color(red)(cancel(color(black)("mL")))) × 100 % = "17 % v/v}$