# 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

#### Answer:

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

#### Answer:

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

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