Question

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

Answer 1

Here's what I got.

Explanation:

Start by writing the chemical equation that describes this reaction

`"C"_ 6"H"_ 12"O"_ (6(aq)) -> 2"C"_ 2"H"_ 5"OH"_ ((aq)) + 2"CO"_ (2(g)) uarr`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)`
SIDE NOTE As you can see, I am assuming that by `"250 g"` of sugar you mean `"250 g"` of glucose, not sucrose, `"C"_12"H"_22"O"_11`.

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

`"C"_ 12"H"_ 22"O"_ (11(aq)) + "H"_ 2"O"_ ((l)) -> 2"C"_ 6"H"_ 12 "O"_ (6(aq))`

and redo the calculations with the mass of glucose that you'll get from this reaction.
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(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

`color(darkgreen)(ul(color(black)("theoretical yield = 130 g")))`

To convert this to milliliters, use the density of ethanol

`127.9 color(red)(cancel(color(black)("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, `"% v/v"`, you should assume that you're working with the volume of ethanol produced by the reaction in a total volume of `"1 L" = 10^3` `"mL"` of wine.

Now, the solution's volume by volume percent concentration tells you the volume of solute present for every `"100 mL"` of solution.

In your case, `"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

`color(darkgreen)(ul(color(black)("% 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.

Answer 2

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, `"C"_12"H"_22"O"_11`.

`M_text(r):color(white)(mmm)342.30color(white)(mmmmmmmmll)46.07`
`color(white)(mmmm)"C"_12"H"_22"O"_11 + "H"_2"O" → 4"CH"_3"CH"_2"OH" + 4"CO"_2`
`"Mass/g:"color(white)(mll)250`

Step 1. Calculate the moles of sugar

`"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

`"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

`"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

`"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

`"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"`