:confused:A bottle of vintage port wine has a volume of 750 ml and contains 18% ethanol (C2H6O) by volume. Ethanol has a density of 0.789 g/ml. At 20 OC port wine has a density of 0.990 g/ml. The alcohol in port wine is made when grape sugar (C6H12O6) undergoes fermentation (no oxygen) to ethanol and carbon dioxide. The weight of 150 grapes is 1.5 lb. and contains 26 g of grape sugar (1 ton = 2,000 lb.)

a) Calculate the percent concentration of ethanol by mass.

b) Write the balanced chemical equation for the fermentation of grape sugar.

c) How many grams of grape sugar are required to produce one bottle of port wine?

d) How many grapes are needed to make one bottle of port wine?

e) How many bottles of port wine can be produced from 1.0 ton of grapes?

:confused: A bottle of vintage port wine has a volume of 750 ml and contains 18% ethanol (C2H6O) by volume. Ethanol has a density of 0.789 g/ml. At 20 degrees C port wine has a density of 0.990 g/ml. The alcohol in port wine is made when grape sugar (C6H12O6) undergoes fermentation (no oxygen) to ethanol and carbon dioxide. The weight of 150 grapes is 1.5 lb. and contains 26 g of grape sugar (1 ton = 2,000 lb.)

a) Calculate the percent concentration of ethanol by mass.

b) Write the balanced chemical equation for the fermentation of grape sugar.

c) How many grams of grape sugar are required to produce one bottle of port wine?

d) How many grapes are needed to make one bottle of port wine?

e) How many bottles of port wine can be produced from 1.0 ton of grapes?..

..

I think I remember something about percent composition. Dividing by the smallest number and making sure all percents add up to 100, but I'm clueless as to how to start this, to balance the chemical equation, am I balancing the one from the paragraph or one of my own? I don't know how to convert very well. So the number of grams of grape sugar is really throwing me for a loop. Can someone lead me in the right direction?

:confused:A bottle of vintage port wine has a volume of 750 ml and contains 18% ethanol (C2H6O) by volume. Ethanol has a density of 0.789 g/ml. At 20 OC port wine has a density of 0.990 g/ml. The alcohol in port wine is made when grape sugar (C6H12O6) undergoes fermentation (no oxygen) to ethanol and carbon dioxide. The weight of 150 grapes is 1.5 lb. and contains 26 g of grape sugar (1 ton = 2,000 lb.)

a) Calculate the percent concentration of ethanol by mass.

b) Write the balanced chemical equation for the fermentation of grape sugar.

c) How many grams of grape sugar are required to produce one bottle of port wine?

d) How many grapes are needed to make one bottle of port wine?

e) How many bottles of port wine can be produced from 1.0 ton of grapes?

18% Ethanol by volume and a volume of 750 mL yields

0.18 \cdot 750\, mL\,=\,135\,mL of Ethanol.

a. Since the density of Ethanol is known (0.789), you can calculate the mass of Ethanol. The total mass of port wine is found by multiplying the volume of port wine, 750 mL, by its density, 0.990 g/mL. With that information, you can easily calculate the percent ethanol by mass

b. You know the formulas of the compounds. Simply write the equation and balance it.

c. 1.5 lbs of grapes contains 26 grams of grape sugar. The formula for Grape Sugar is also known so you can calculate the molecular weight of the sugar. Divide the weight of the sugar (26 grams) by the molecular weight (grams/mole). If you're ever unsure whether to multiply or divide, use "dimensional analysis":

\large \frac {grams}{(\frac {grams}{mole})} = mole

d. You know, or have calculated, the volume and weight of 1 bottle of wine. You also know the percentage of alcohol in the wine. In "c", you calculated the number of moles of sugar and in "b", you balanced the chemical equation. Look at the chemical equation and figure out how many moles of alcohol are produced from one mole of sugar. Calculate the molecular weight of ethanol (or look it up). Multiply the number of moles of alcohol by the molecular weight of alcohol to figure out how many grams of alcohol are being produced. You also have calculated the number of moles of sugar 150 grapes, so you can divide by 150 to get the number of moles of sugar in 1 grape. From this, figure out how many grapes are required.

e. Once you know d, e is easy.


\large moles (\frac {grams}{mole}) = grams

Thank you for your help, I think I might be able to understand some of this now.

Thank you for your help, I think I might be able to understand some of this now.


There's only one more thing I don't understand. When I balance the equation Is it set up as

C2H6O + C6H12O6 --->
OR
C2H6O ----> C6H12O6 AND THEN BALANCE IT FROM THERE

The equation is "sugar" to "ethanol", so it's

C6H12O6\, --> CH3CH2OH

CH3CH2OH is the same as C6H6O but there are potentially many compounds with the formula C6H6O. I prefer to write it as CH3CH2OH because it reflects its chemical structure and differentiates it from other oxgenated hydrocarbons of the same formula. That is meaningful to chemists. The carbons are bonded using single bonds, thus:

CH_3 - CH_2 - OH

The equation is "sugar" to "ethanol", so it's

C6H12O6\, --> CH3CH2OH

CH3CH2OH is the same as C6H6O but there are potentially many compounds with the formula C6H6O. I prefer to write it as CH3CH2OH because it reflects its chemical structure and differentiates it from other oxgenated hydrocarbons of the same formula. That is meaningful to chemists. The carbons are bonded using single bonds, thus:

CH_3 - CH_2 - OH



I'm sorry, I don't understand how to balance that. How am I supposed to have the same number on both sides?

I thought you understood more about the process and only needed the position of the sugar and the alcohol. CO2 is also formed. That's what makes the fizz.

C_6H_{12}O_6\, -->\, 2\,CH_3CH_2OH\, +\, 2 CO_2